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Transcript
DC CIRCUITS 19 Responses to Questions 1. Even though the bird’s feet are at high potential with respect to the ground, there is very little potential difference between them, because they are close together on the wire. The resistance of the bird is much greater than the resistance of the wire between the bird’s feet. These two resistances are in parallel, so very little current will pass through the bird as it perches on the wire. However, when you put a metal ladder up against a power line, you provide a direct connection between the high potential line and ground. The ladder will have a large potential difference between its top and bottom. A person standing on the ladder will also have a large potential difference between his or her hands and feet. Even if the person’s resistance is large, the potential difference will be great enough to produce a current through the person’s body large enough to cause substantial damage or death. 2. Series: The main disadvantage of Christmas tree lights connected in series is that when one bulb burns out, a gap is created in the circuit and none of the bulbs remains lit. Finding the burned-out bulb requires replacing each individual bulb one at a time until the string of bulbs comes back on. As an advantage, the bulbs are slightly easier to wire in series. Only one wire needs to be attached from bulb to bulb. Also, a “blinker bulb” can make the entire string flash on and off by cutting off the current. Parallel: The main advantage of connecting the bulbs in parallel is that one burned-out bulb does not affect the rest of the strand and is easy to identify and replace. As a disadvantage, wiring the bulbs in parallel is slightly more difficult. Two wires must be attached from bulb to bulb. 3. If 20 of the 6-V lamps were connected in series and then connected to the 120-V line, there would be a voltage drop of 6 V for each of the lamps, and they would not burn out due to too much voltage. Being in series, if one of the bulbs went out for any reason, then they would all turn off. 4. If the lightbulbs are in series, then each will have the same current. The power dissipated by the bulb as heat and light is given by P = I 2 R. Thus the bulb with the higher resistance ( R2 ) will be brighter. If the bulbs are in parallel, then each will have the same voltage. The power dissipated by the bulb as heat and light is given by P = V 2 /R. Thus the bulb with the lower resistance ( R1 ) will be brighter. 5. The outlets are connected in parallel to each other, because you can use one outlet without using the other. If they were in series, then both outlets would have to be used at the same time to have a completed circuit. Also, both outlets supply the same voltage to whatever device is plugged in to the outlet, which indicates that they are wired in parallel to the voltage source. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-1 19-2 Chapter 19 6. The power output from a resistor is given by P = V 2 /R. To maximize this value, the voltage needs to be as large as possible and the resistance as small as possible. That can be accomplished by putting the two batteries in series and then connecting the two resistors in parallel to each other, across the full two-battery voltage. 7. The battery has to supply less power when the two resistors are connected in series than it has to supply when only one resistor is connected. P = V 2 /R, so if V is constant and R increases, then the power decreases. 8. We assume that the bulbs are in parallel with each other. The overall resistance decreases and more current is drawn from the source. A bulb rated at 60 W and 120 V has a resistance of 240 Ω. A bulb rated at 100 W and 120 V has a resistance of 144 Ω. When only the 60-W bulb is on, the total resistance is 240 Ω . When both bulbs are lit, the total resistance is the combination of the two resistances in parallel, which is only 90 Ω. 9. The energy stored in a capacitor network can be calculated by PE = 12 CV 2 . Since the voltage for the capacitor network is the same in this problem for both configurations, the configuration with the highest equivalent capacitance will store the most energy. The parallel combination has the highest equivalent capacitance, so it stores the most energy. Another way to consider this is that the total stored energy is the sum of the quantity PE = 12 CV 2 for each capacitor. Each capacitor has the same capacitance, but in the parallel circuit, each capacitor has a larger voltage than in the series circuit. Thus the parallel circuit stores more energy. 10. No, the sign of the battery’s emf does not depend on the direction of the current through the battery. The sign of the battery’s emf depends on the direction you go through the battery in applying the loop rule. If you go from negative pole to positive pole, then the emf is added. If you go from positive pole to negative pole, then the emf is subtracted. But the terminal voltage does depend on the direction of the current through the battery. If current is flowing through the battery in the normal orientation (leaving the positive terminal, flowing through the circuit, and arriving at the negative terminal), then there is a voltage drop across the internal resistance, and the terminal voltage is less than the emf. If the current flows in the opposite sense (as in charging the battery), then there is a voltage rise across the terminal resistance, and the terminal voltage is higher than the emf. 11. (a) (b) With the batteries in series, a greater voltage is delivered to the lamp, and the lamp will burn brighter. With the batteries in parallel, the voltage across the lamp is the same as for either battery alone. Each battery supplies only half of the current going through the lamp, so the batteries will last longer (and the bulb stay lit longer) as compared to just having one battery. 12. When batteries are connected in series, their emfs add together, producing a larger potential. For instance, if there are two 1.5-V batteries in series in a flashlight, then the potential across the bulb will be 3.0 V, and there will be more current in the bulb. The batteries do not need to be identical in this case. When batteries are connected in parallel, the currents they can generate add together, producing a larger current over a longer time period. Batteries in this case need to be nearly identical, or the battery with the larger emf will end up charging the battery with the smaller emf. 13. Yes. When a battery is being charged, current is forced through it “backward” and Vterminal = Ᏹ + Ir , so Vterminal > Ᏹ. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 14. 19-3 Refer to Fig. 19–2. Connect the battery to a known resistance R and measure the terminal voltage Vab . Vab . It is also true that Vab = Ᏹ − Ir , so the R Ᏹ − Vab =R . Vab The current in the circuit is given by Ohm’s law to be I = internal resistance can be calculated by r = Ᏹ − Vab I 15. No. As current passes through the resistor in the RC circuit, energy is dissipated in the resistor. Therefore, the total energy supplied by the battery during the charging is the combination of the energy dissipated in the resistor and the energy stored in the capacitor. 16. (a) (f) 17. The following is one way to accomplish the desired task. stays the same; (b) decreases; (g) increases; (c) decreases; decreases; (h) increases; (d) increases; (e) increases; (i) stays the same. Light In the current configuration, the light would be on. If either switch is moved, then the light will go out. But if either switch were moved again, then the light would come back on. 18. The soles of your shoes are made of a material which has a relatively high resistance, and there is relatively high resistance flooring material between your shoes and the literal ground (the Earth). With that high resistance, a malfunctioning appliance would not be able to cause a large current flow through your body. The resistance of bare feet is much less than that of shoes, and the feet are in direct contact with the ground, so the total resistance is much lower and a larger current would flow through your body. 19. In an analog ammeter, the internal resistor, or shunt resistor, has a small value and is in parallel with the galvanometer, so that the overall resistance of the ammeter is very small. In an analog voltmeter, the internal resistor has a large value and is in series with the galvanometer, and the overall resistance of the voltmeter is very large. 20. If you mistakenly use an ammeter where you intend to use a voltmeter, then you are inserting a short in parallel with some resistance. That means that the resistance of the entire circuit has been lowered, and almost all of the current will flow through the low-resistance ammeter. Ammeters usually have a fairly small current limit, so the ammeter will very likely be damaged in such a scenario. Also, if the ammeter is inserted across a voltage source, then the source will provide a large current, and again the meter will almost certainly be damaged or at least disabled by burning out a fuse. 21. An ammeter is placed in series with a given circuit element in order to measure the current through that element. If the ammeter did not have very low (ideally, zero) resistance, then its presence in the circuit would change the current it is attempting to measure by adding more resistance in series. An ideal ammeter has zero resistance and thus does not change the current it is measuring. A voltmeter is placed in parallel with a circuit element in order to measure the voltage difference across that element. If the voltmeter does not have a very high resistance, then its presence in parallel will lower the overall resistance and affect the circuit. An ideal voltmeter has infinite resistance so that when placed in parallel with circuit elements it will not change the value of the voltage it is reading. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-4 Chapter 19 22. When the voltmeter is connected to the circuit, it reduces the resistance of that part of the circuit. That will make the resistor + voltmeter combination a smaller fraction of the total resistance of the circuit than the resistor was alone, which means that it will have a smaller fraction of the total voltage drop across it. 23. A voltmeter has a very high resistance. When it is connected to the battery, very little current will flow. A small current results in a small voltage drop due to the internal resistance of the battery, and the emf and terminal voltage (measured by the voltmeter) will be very close to the same value. However, when the battery is connected to the lower-resistance flashlight bulb, the current will be higher and the voltage drop due to the internal resistance of the battery will also be higher. As a battery is used, its internal resistance increases. Therefore, the terminal voltage will be significantly lower than the emf: Vterminal = Ᏹ − Ir. A lower terminal voltage will result in a dimmer bulb and usually indicates a “usedup” battery. Responses to MisConceptual Questions 1. (a, c) Resistors are connected in series when there are no junctions between the resistors. With no junctions between the resistors, any current flowing through one of the resistors must also flow through the other resistor. 2. (d) It might be easy to think that all three resistors are in parallel because they are on three branches of the circuit. However, following a path from the positive terminal of the battery, all of the current from the battery passes through R2 before reaching the junction at the end of that branch. The current then splits and part of the current passes through R1 while the remainder of the current passes through R3 before meeting at the right junction. 3. (c) A common misconception is that the smaller resistor would have the larger current. However, the two resistors are in series, so they must have the same current flowing through them. It does not matter which resistor is first. 4. (a) It might seem that current must flow through each resistor and lightbulb. However, current will only flow through the lightbulb if there is a potential difference across the bulb. If we consider the bottom branch of the circuit to be at ground, then the left end of the lightbulb will be at a potential of 10 V. The top branch will be at 20 V. Since the two resistors are identical, the voltage drop across each will be half the voltage from the top to bottom, or 10 V each. This makes the right side of the lightbulb also at 10 V. Since both sides of the lightbulb are at the same potential, no current will flow through the lightbulb. 5. (d) Resistors R1 and R2 are in parallel, so each has half of the current from the battery. R3 and R4 are in series and add to produce twice as much resistance as R5 . Since they are in parallel with R5 , one-third of the current from the battery goes through them, while two-thirds goes through R5 . The greatest current therefore goes through R5 . 6. (d) Current takes the path of least resistance. Since bulb A is in parallel with the short circuit, all of the current will pass through the short circuit, causing bulb A to go out. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-5 7. (b) A common misconception is that adding another parallel branch would have no effect on the voltage across R4 . However, when the switch is closed, the additional parallel resistor makes the effective resistance of the parallel resistors smaller, and therefore the resistance of the entire circuit gets smaller. With a smaller effective resistance, a greater current flows through the battery and through R1 , resulting in a greater voltage drop across R1. Since the voltage from the battery is equal to the sum of the voltages across R1 and R4 , increasing the voltage across R1 results in a decrease in voltage across R4 . 8. (a) As explained in the answer to Question 7, when the switch is closed, it adds an additional resistor in parallel to R3 and R4 , making the effective resistance of the entire circuit smaller. With less effective resistance, a greater current flows through the circuit, increasing the potential difference across R1. 9. (b) As the capacitor charges, the voltage drop across the capacitor increases, thereby diminishing the voltage drop across the resistor. As the voltage drop across the resistor decreases, the current decreases. 10. (c) Even though steady current cannot flow through a capacitor, charge can build up on the capacitor, allowing current to initially flow in the circuit. As the charge builds on the capacitor, the voltage drop across the capacitor increases, and the current decreases. The rate of charging is determined by the time constant, which is the product of R and C. 11. (b) It might seem that a capacitor would discharge linearly in time, losing one-fourth of the charge every second. This is incorrect, as the capacitor discharges exponentially. That is, every 2.0 seconds, half of the remaining charge on the capacitor will discharge. After 2.0 seconds, half of the charge remains. After 4.0 seconds, half of the half, or one-fourth, of the charge remains. After 6.0 seconds, one-eighth of the charge remains. 12. (a) To double the heart rate, the time of discharging must be shorter, so the discharge rate must be faster. The resistance limits the rate at which the current can flow through the circuit. Decreasing the resistance will increase the current flow, causing the capacitor to discharge faster. 13. (c) In a two-prong cord, one prong is at high voltage and the other is grounded. Electricity flows through the appliance between these two wires. However, if there is a short between the highvoltage wire and the casing, then the casing can become charged and electrocute a person touching the case. The third prong connects the external case to ground, so that the case cannot become charged. 14. (c) Connecting capacitors in series effectively increases the plate separation distance, decreasing the net capacitance. Connecting capacitors in parallel effectively increases the plate area, increasing the net capacitance. Therefore, when capacitors are connected in series, the effective capacitance will be less than the capacitance of the smallest capacitor, and when connected in parallel, the effective capacitance will be greater than the capacitance of the largest capacitance. 15. (b) The ammeter is placed in series with the circuit and therefore should have a small resistance, so there is minimal voltage drop across the ammeter. The voltmeter is placed in parallel with the circuit. It should have a large resistance so that minimal current from the circuit passes through the voltmeter instead of passing through the circuit. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-6 Chapter 19 Solutions to Problems 1. See Fig. 19–2 for a circuit diagram for this problem. Using the same analysis as in Example 19–1, the Ᏹ . Use Eq. 19–1 to calculate the terminal voltage. current in the circuit is I = R+r R 71.0 Ω Ᏹ(R + r ) − Ᏹr ⎛ Ᏹ ⎞ =Ᏹ = (6.00 V) = 5.92 V (a) Vab = Ᏹ − Ir = Ᏹ − ⎜ ⎟r = R r R r R r (71 0 + + + . + 0.900)Ω ⎝ ⎠ (b) 2. Vab = Ᏹ R 710 Ω = (6.00 V) = 5.99 V R+r (710 + 0.900)Ω See the circuit diagram. The current in the circuit is I. The voltage Vab is given by Ohm’s law to be Vab = IR. That same voltage is the terminal voltage of the series emf. b r Ᏹ r r Ᏹ r Ᏹ a Ᏹ I R Vab = ( Ᏹ − Ir ) + ( Ᏹ − Ir ) + ( Ᏹ − Ir ) + ( Ᏹ − Ir ) = 4( Ᏹ − Ir ) and Vab = IR 4( Ᏹ − Ir ) = IR → r = Ᏹ − 14 IR I = (1.50 V) − 14 (0.45 A)(12.0 Ω) 0.45 A 3. See Fig. 19–2 for a circuit diagram for this problem. Use Eq. 19–1. Ᏹ − Vab 12.0 V − 8.8 V Vab = Ᏹ − Ir → r = = = 0.034 Ω I 95 A V 8.8 V Vab = IR → R = ab = = 0.093 Ω I 95 A 4. The equivalent resistance is the sum of the two resistances: Req = R1 + R2 . The current in the circuit is then the voltage divided by the equivalent resistance: I = Ᏹ Ᏹ = . The Req R1 + R2 = 0.333 Ω ≈ 0.33 Ω R2 voltage across the 1800-Ω resistor is given by Ohm’s law. R2 Ᏹ 1800 Ω V2200 = IR2 = R2 = Ᏹ = (12 V) = 8.8 V R1 + R2 R1 + R2 650 Ω + 1800 Ω 5. I Ᏹ R1 (a) For the resistors in series, use Eq. 19–3, which says the resistances add linearly. Req = 3(45 Ω) + 3(65 Ω) = 330 Ω (b) For the resistors in parallel, use Eq. 19–4, which says the resistances add reciprocally. 1 1 1 1 1 1 1 3 3 3(65 Ω) + 3(45 Ω) = + + + + + = + = Req 45 Ω 45 Ω 45 Ω 65 Ω 65 Ω 65 Ω 45 Ω 65 Ω (65 Ω)(45 Ω) Req = → (65 Ω)(45 Ω) = 8.9 Ω 3(65 Ω) + 3(45 Ω) © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 6. (a) 19-7 The maximum resistance is made by combining the resistors in series. Req = R1 + R2 + R3 = 580 Ω + 790 Ω + 1200 Ω = 2.57 kΩ (b) The minimum resistance is made by combining the resistors in parallel. 1 1 1 1 = + + Req R1 R2 R3 → ⎛ 1 1 1 ⎞ Req = ⎜ + + ⎟ R R R 2 3⎠ ⎝ 1 7. ⎛ 1 1 1 1 1 ⎞ =⎜ + + + + ⎟ ⎝ R1 R2 R3 R4 R5 ⎠ = 260 Ω −1 ⎛ 5 ⎞ =⎜ ⎟ ⎝ 100 Ω ⎠ −1 = 20 Ω = n(10 Ω) → n = 20 Ω = 2 10 Ω Ᏹ R1 = 40 Ω I R2 = 10 Ω a b Ᏹ Ᏹ (40) = 0.8%, and the voltage across R2 would be V2 = IR2 = (10) = 0.2%. 50 50 Connecting 18 of the resistors in series will enable you to make a voltage divider with a 3.5-V output. To get the desired output, measure the voltage across 7 consecutive series resistors. In the diagram that would be the voltage between points a and b. Ᏹ Ᏹ Req = 18(1.0 Ω) I = = Req 18.0 Ω Vab = (7.0 Ω) I = (7.0 Ω) 10. −1 The ratio R1 /R2 must be 4/1. Since both resistors have the same current, the voltage drop across R1 (4/5 of the battery voltage) will be 4 times that across R2 (1/5 of the battery voltage). So if R1 = 40 Ω and R2 = 10 Ω, then the current would be Ᏹ/50. The voltage across R1 would be found from Ohm’s law to be V1 = IR1 = 9. ⎛ 1 1 1 ⎞ =⎜ + + ⎟ 580 790 1200 Ω Ω Ω⎠ ⎝ The equivalent resistance of five 100-Ω resistors in parallel is found, and then that resistance is divided by 10 Ω to find the number of 10-Ω resistors needed. Req 8. −1 Ᏹ 11.0 Ω I 7.0 Ω b a Ᏹ 9.0 V = (7.0 Ω) = 3.5 V 18.0 Ω 18.0 Ω The resistors can all be connected in series. Req = R + R + R = 3(1.70 kΩ) = 5.10 kΩ The resistors can all be connected in parallel. 1 1 1 1 = + + Req R R R ⎛3⎞ → Req = ⎜ ⎟ ⎝R⎠ −1 = R 1.70 kΩ = = 567 Ω 3 3 Two resistors in series can be placed in parallel with the third. 1 1 1 1 1 3 = + = + = Req R R + R R 2 R 2 R → Req = 2 R 2(1.70 kΩ) = = 1.13 kΩ 3 3 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-8 Chapter 19 Two resistors in parallel can be placed in series with the third. ⎛1 1⎞ Req = R + ⎜ + ⎟ ⎝R R⎠ 11. −1 = R+ R 3 = (1.70 kΩ) = 2.55 kΩ 2 2 The resistance of each bulb can be found from its power rating. V2 R P= → R= V 2 (12.0 V) 2 = = 36 Ω 4.0 W P r Ᏹ I R Find the equivalent resistance of the two bulbs in parallel. 1 1 1 2 = + = Req R R R → Req = R 36 Ω = = 18 Ω 2 2 R The terminal voltage is the voltage across this equivalent resistance. Use that to find the current drawn from the battery. Vab = IReq → I= Vab Vab 2Vab = = Req R /2 R Finally, use the terminal voltage and the current to find the internal resistance, as in Eq. 19–1. Vab = Ᏹ − Ir 12. (a) → r= Ᏹ − Vab Ᏹ − Vab Ᏹ − Vab 12.0 V − 11.8 V = =R = (36 Ω) = 0.305 Ω ≈ 0.3 Ω 2 V 2Vab 2(11.8 V) I ⎛ ab ⎞ ⎜ R ⎟ ⎝ ⎠ Each bulb should get one-eighth of the total voltage, but let us prove that instead of assuming it. Since the bulbs are identical, the net resistance is Req = 8 R. The current flowing through the bulbs is then Vtot = IReq → I= Vtot Vtot = . The voltage across one bulb is found from Ohm’s Req 8 R law. V = IR = (b) I= Vtot 8R Vtot V 120 V R = tot = = 15 V 8R 8 8 → R= Vtot 120 V = = 33.33 Ω ≈ 33 Ω 8I 8(0.45 A) P = I 2 R = (0.45 A) 2 (33.33 Ω) = 6.75 W ≈ 6.8 W 13. We model the resistance of the long leads as a single resistor r. Since the bulbs are in parallel, the total current is the sum of the current in each bulb, so I = 8I R . The voltage drop across the long leads is Vleads = Ir = 8I R r = 8(0.21 A)(1.4 Ω) = 2.352 V. Thus the voltage across each of the parallel resistors is VR = Vtot − Vleads = 120 V − 2.352 V = 117.6 V. Since we have the current through each resistor, and the voltage across each resistor, we calculate the resistance using Ohm’s law. VR = I R R → R = VR 117.6 V = = 560 Ω IR 0.21 A The total power delivered is P = Vtot I , and the “wasted” power is I 2 r. The fraction wasted is the ratio of those powers. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits fraction wasted = 19-9 I 2r Ir 8(0.21 A)(1.4 Ω) = = = 0.0196 ≈ 0.020 120 V IVtot Vtot So about 2% of the power is wasted. 14. To fix this circuit, connect another resistor in parallel with the 480-Ω resistor so that the equivalent resistance is the desired 350 Ω. ⎛ 1 1 ⎞ → R2 = ⎜ − ⎟ ⎜ Req R1 ⎟ ⎝ ⎠ 1 1 1 = + Req R1 R2 −1 ⎛ 1 1 ⎞ =⎜ − ⎟ ⎝ 350 Ω 480 Ω ⎠ −1 = 1292 Ω ≈ 1300 Ω So solder a 1300-Ω resistor in parallel with the 480-Ω resistor. 15. Vtot . That could be proven using a 8 similar argument to that used in Problem 12. Use that voltage and the power dissipated by each bulb to calculate the resistance of a bulb. Each bulb will get one-eighth of the total voltage, so Vbulb = Pbulb = 16. (a) 2 Vbulb R → R= 2 Vbulb V2 (120 V) 2 = tot = = 32 Ω P 64 P 64(7.0 W) The equivalent resistance is found by combining the 750-Ω and 680-Ω resistors in parallel and then adding the 990-Ω resistor in series with that parallel combination. ⎛ 1 1 ⎞ Req = ⎜ + ⎟ ⎝ 750 Ω 680 Ω ⎠ (b) −1 + 990 Ω = 357 Ω + 990 Ω = 1347 Ω ≈ 1350 Ω The current delivered by the battery is I = V 12.0 V = = 8.909 × 10−3 A. This is the current in Req 1347 Ω the 990-Ω resistor. The voltage across that resistor can be found by Ohm’s law. V990 = IR = (8.909 × 10−3 A)(990 Ω) = 8.820 V ≈ 8.8 V Thus the voltage across the parallel combination must be 12.0 V − 8.8 V = 3.2 V . This is the voltage across both the 750-Ω and 680-Ω resistors, since parallel resistors have the same voltage across them. Note that this voltage value could also be found as follows. Vparallel = IRparallel = (8.909 × 10−3 A)(357 Ω) = 3.181 V ≈ 3.2 V (c) The current in the 990-Ω resistor was already seen to be 8.91 mA. The other currents can be found from Ohm’s law. I 750 = V750 3.18 V = ≈ 4.2 mA ; R750 470 Ω I 680 = V750 3.18 V = ≈ 4.7 mA R680 680 Ω Note that these last two currents add to be the current in the 990-Ω resistor. 17. The resistance of each bulb can be found by using Eq. 18–6, P = V 2 /R. The two individual resistances are combined in parallel. We label the bulbs by their wattage. P = V 2 /R → 1 P = 2 R V © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-10 Chapter 19 Req 18. (a) ⎛ 1 1 ⎞ =⎜ + ⎟ ⎝ R75 R25 ⎠ −1 ⎛ 75 W 25 W =⎜ + ⎜ (120 V) 2 (120 V) 2 ⎝ ⎞ ⎟⎟ ⎠ −1 = 144 Ω ≈ 140 Ω The three resistors on the far right are in series, so their equivalent resistance is 3R. That combination is in parallel with the next resistor to the left, as shown in the dashed box in figure (b). The equivalent resistance of the dashed box is found as follows. ⎛1 1 ⎞ Req1 = ⎜ + ⎟ ⎝ R 3R ⎠ −1 = 34 R This equivalent resistance of 3 4 R is in series with the next two resistors, as shown in the dashed box in figure (c). The equivalent R. This resistance of that dashed box is Req2 = 2 R + 34 R = 11 4 11 R 4 is in parallel with the next resistor to the left, as shown in figure (d). The equivalent resistance of that dashed box is found as follows. 4 ⎞ ⎛1 Req2 = ⎜ + ⎟ ⎝ R 11R ⎠ −1 11 R = 15 This is in series with the last two resistors, the ones connected directly to A and B. The final equivalent resistance is then 11 R = 41 R = 41 (175 Ω) = 478.33 Ω ≈ 478 Ω Req = 2 R + 15 15 15 (b) The current flowing from the battery is found from Ohm’s law. V 50.0 V = = 0.10453 A ≈ 0.105 A Req 478.33 Ω I total = This is the current in the top and bottom resistors. There will be less current in the next resistor because the current splits, with some current passing through the resistor in question, and the rest of the current passing through the equivalent resistance of 11 R, as shown in figure (d). The 4 voltage across R and across 11 R 4 must be the same, since they are in parallel. Use this to find the desired current. VR = V11 R 4 → I R R = I 11 R 4 ( 114 R ) = (I total − I R ) ( 114 R ) → 11 I 11 I R = 15 total = 15 (0.10453 A)I total = 0.0767 A 19. The resistors have been numbered in the accompanying diagram to help in the analysis. R1 and R2 are in series with an equivalent resistance of R12 = R + R = 2 R. This combination is in parallel with R3 , with an equivalent −1 ⎛1 1 ⎞ 2 resistance of R123 = ⎜ + ⎟ = 3 R. This combination is in series with R4 , ⎝ R 2R ⎠ with an equivalent resistance of R1234 = 32 R + R = 53 R. This combination is in ⎛1 3 ⎞ parallel with R5 , with an equivalent resistance of R12345 = ⎜ + ⎟ ⎝ R 5R ⎠ −1 = 85 R. B R2 R1 A R3 R5 R4 C R6 V © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-11 Finally, this combination is in series with R6 , and we calculate the final equivalent resistance. Req = 85 R + R = 13 R 8 A more detailed solution with more circuit diagrams is given in Problem 20. 20. We reduce the circuit to a single loop by combining series and parallel combinations. We label a combined resistance with the subscripts of the resistors used in the combination. See the successive diagrams. The initial diagram has been rotated by 90 degrees. R1 and R2 are in series. B R2 R1 A R3 R4 C R5 B R12 R3 A R5 R6 R4 C R6 R12 = R1 + R2 = R + R = 2 R R123 ⎛ 1 1 ⎞ =⎜ + ⎟ ⎝ R12 R3 ⎠ −1 Ᏹ Ᏹ R12 and R3 are in parallel. ⎛ 1 1⎞ =⎜ + ⎟ ⎝ 2R R ⎠ −1 B = 23 R R123 A R123 and R4 are in series. R1234 = R123 + R4 = 23 R + R = 53 R R1234 and R5 are in parallel. ⎛ 1 1 ⎞ + R12345 = ⎜ ⎟ R R 5⎠ ⎝ 1234 −1 ⎛ 1 1⎞ =⎜5 + ⎟ ⎜ R R⎟ ⎝3 ⎠ Req = R12345 + R6 = R + R = 13 8 A R5 R6 = 85 R A R12345 C R6 Ᏹ C R6 Ᏹ Ᏹ −1 R12345 and R6 are in series, producing the equivalent resistance. 5 8 C R5 R1234 R4 Req Ᏹ R Now work “backward” from the simplified circuit. Resistors in series have the same current as their equivalent resistance, and resistors in parallel have the same voltage as their equivalent resistance. To avoid rounding errors, we do not use numeric values until the end of the problem. I eq = Ᏹ Ᏹ 8Ᏹ = = = I 6 = I12345 Req 13 13 R R 8 ⎛ 8Ᏹ ⎞ 5 5 V5 = V1234 = V12345 = I12345 R12345 = ⎜ ⎟ ( R) = 13 Ᏹ; ⎝ 13R ⎠ 8 I1234 = I3 = V1234 = R1234 5 Ᏹ 13 5R 3 = V3 2Ᏹ = = I3 ; R3 13R 3Ᏹ = I 4 = I123 ; 13R I12 = I5 = 5 V5 13 Ᏹ 5Ᏹ = = = I5 R5 R 13R ⎛ 3Ᏹ ⎞ 2 2 V123 = I123 R123 = ⎜ ⎟ ( R) = 13 Ᏹ = V12 = V3 ⎝ 13R ⎠ 3 2 V12 13 Ᏹ Ᏹ = = = I1 = I 2 R12 2R 13R © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-12 Chapter 19 Now substitute in numerical values. I1 = I 2 = I5 = 21. Ᏹ 12.0 V 2Ᏹ 3Ᏹ = = 0.284 mA ; I3 = = 0.568 mA ; I 4 = = 0.852 mA ; 13R 13(3250 Ω) 13R 13R 5Ᏹ 8Ᏹ 2 Ᏹ = 1.85 V = 1.42 mA ; I 6 = = 2.27 mA ; VAB = V3 = 13 13R 13R It is given that the power used when the resistors are in series is one-fourth the power used when the resistors are in parallel. The voltage is the same in both cases. Use Eq. 18–6b, along with the definitions of series and parallel equivalent resistance. Pseries = 14 Pparallel → V2 = Rseries ( R1 + R2 ) 2 = 4 R1 R2 1 4 V2 Rparallel → Rseries = 4 Rparallel → ( R1 + R2 ) = 4 → R12 + 2 R1 R2 + R22 − 4 R1 R2 = 0 = ( R1 − R2 ) 2 R1 R2 ( R1 + R2 ) → → R1 = R2 Thus the two resistors must be the same, so the “other” resistor is 4.8 kΩ . 22. We label identical resistors from left to right as Rleft , Rmiddle , and Rright . When the switch is opened, the equivalent resistance of the circuit increases from the battery decreases, from (a) 3 2 3 2 R + r to 2 R + r. Thus the current delivered by Ᏹ Ᏹ to . Note that this is LESS than a 50% decrease. 2R + r R+r Because the current from the battery has decreased, the voltage drop across Rleft will decrease, since it will have less current than before. The voltage drop across Rright decreases to 0, since no current is flowing in it. The voltage drop across Rmiddle will increase, because even though the total current has decreased, the current flowing through Rmiddle has increased since before the switch was opened, only half the total current was flowing through Rmiddle . Vleft decreases; Vmiddle increases; Vright goes to 0 (b) By Ohm’s law, the current is proportional to the voltage for a fixed resistance. I left decreases; I middle increases; I right goes to 0 (c) Since the current from the battery has decreased, the voltage drop across r will decrease, and thus the terminal voltage increases. (d) With the switch closed, the equivalent resistance is I closed = 3 2 3 2 R + r. Thus the current in the circuit is Ᏹ , and the terminal voltage is given by Eq. 19–1. R+r Vterminal = Ᏹ − I closed r = Ᏹ − closed 3 2 ⎛ ⎛ ⎞ Ᏹ r ⎞ 0.50 Ω r = Ᏹ ⎜1 − 3 ⎟ = (9.0 V) ⎜ 1 − 3 ⎟ ⎜ ⎜ R+r R + r ⎟⎠ (5.50 Ω) + 0.50 Ω ⎟⎠ ⎝ ⎝ 2 2 = 8.486 V ≈ 8.5 V © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits (e) 19-13 With the switch open, the equivalent resistance is 2 R + r. Thus the current in the circuit is Ᏹ I closed = , and again the terminal voltage is given by Eq. 19–1. 2R + r Vterminal = Ᏹ − I closed r = Ᏹ − closed ⎛ ⎞ Ᏹ r ⎞ 0.50 Ω ⎛ r = Ᏹ ⎜1 − ⎟ ⎟ = (9.0 V) ⎜ 1 − + . Ω + . Ω 2R + r 2 R r 2(5 50 ) 0 50 ⎝ ⎠ ⎝ ⎠ = 8.609 V ≈ 8.6 V 23. Find the maximum current and resulting voltage for each resistor under the power restriction. P = I 2R = I1400 = I 2500 = I 3700 = V2 R → I= 0.5 W 1.4 × 103 Ω 0.5 W 2.5 × 103 Ω 0.5 W 3.7 × 103 Ω P , V = RP R = 0.0189 A V1800 = (0.5 W)(1.4 × 103 Ω) = 26.5 V = 0.0141 A V2500 = (0.5 W)(2.8 × 103 Ω) = 35.4 V = 0.0116 A V3700 = (0.5 W)(3.7 × 103 Ω) = 43.0 V The parallel resistors have to have the same voltage, so the voltage across that combination is limited to 35.4 V. That would require a current given by Ohm’s law and the parallel combination of the two resistors. I parallel = Vparallel Rparallel ⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ = Vparallel ⎜ + + ⎟ = (35.4 V) ⎜ ⎟ = 0.0237 A ⎝ 2500 Ω 3700 Ω ⎠ ⎝ R2500 R3700 ⎠ This is more than the maximum current that can be in R1400 . Thus the maximum current that R1400 can carry, 0.0189 A, is the maximum current for the circuit. The maximum voltage that can be applied across the combination is the maximum current times the equivalent resistance. The equivalent resistance is the parallel combination of R2500 and R3700 added to R1400 . Vmax −1 −1 ⎡ ⎡ ⎛ 1 ⎛ 1 1 ⎞ ⎤ 1 ⎞ ⎤ ⎥ ⎢ = I max Req = I max ⎢ R1800 ⎜ + = . Ω + (0 0189 A) 1400 ⎜ ⎟ ⎟ ⎥ R2800 R3700 ⎠ ⎥ 2500 Ω 3700 Ω ⎠ ⎥ ⎢ ⎢ ⎝ ⎝ ⎣ ⎦ ⎣ ⎦ = 54.66 V ≈ 55 V 24. (a) Note that adding resistors in series always results in a larger resistance, and adding resistors in parallel always results in a smaller resistance. Closing the switch adds another resistor in parallel with R3 and R4 , which lowers the net resistance of the parallel portion of the circuit and thus lowers the equivalent resistance of the circuit. That means that more current will be delivered by the battery. Since R1 is in series with the battery, its voltage will increase. Because of that increase, the voltage across R3 and R4 must decrease so that the total voltage drops around the loop are equal to the battery voltage. Since there was no voltage across R2 until the switch was closed, its voltage will increase. To summarize: V1 and V2 increase; V3 and V4 decrease © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-14 Chapter 19 (b) By Ohm’s law, the current is proportional to the voltage for a fixed resistance. Thus I1 and I 2 increase; I 3 and I 4 decrease (c) Since the battery voltage does not change and the current delivered by the battery increases, the power delivered by the battery, found by multiplying the voltage of the battery by the current delivered, increases . (d) Before the switch is closed, the equivalent resistance is R3 and R4 in parallel, combined with R1 in series. ⎛ 1 1 ⎞ + Req = R1 + ⎜ ⎟ ⎝ R3 R4 ⎠ −1 ⎛ 2 ⎞ = 155 Ω + ⎜ ⎟ ⎝ 155 Ω ⎠ −1 = 232.5 Ω The current delivered by the battery is the same as the current through R1. I total = Vbattery Req = 22.0 V = 0.09462 A = I1 232.5 Ω The voltage across R1 is found by Ohm’s law. V1 = IR1 = (0.09462 A)(155 Ω) = 14.666 V The voltage across the parallel resistors is the battery voltage less the voltage across R1. Vp = Vbattery − V1 = 22.0 V − 14.666 V = 7.334 V The current through each of the parallel resistors is found from Ohm’s law. I3 = Vp R2 = 7.334 V = 0.04732 A = I 4 155 Ω Notice that the current through each of the parallel resistors is half of the total current, within the limits of significant figures. The currents before closing the switch are as follows. I1 = 0.0946 A I 3 = I 4 = 0.0473 A After the switch is closed, the equivalent resistance is R2 , R3 , and R4 in parallel, combined with R1 in series. Do a similar analysis. Req ⎛ 1 1 1 ⎞ = R1 + ⎜ + + ⎟ R R R 3 4⎠ ⎝ 2 I total = Vbattery Req = −1 ⎛ 3 ⎞ = 155 Ω + ⎜ ⎟ ⎝ 155 Ω ⎠ −1 = 206.7 Ω 22.0 V = 0.1064 A = I1 V1 = IR1 = (0.1064 A)(155 Ω) = 16.49 V 206.7 Ω Vp = Vbattery − V1 = 22.0 V − 16.49 V = 5.51 V I 2 = Vp R2 = 5.51 V = 0.0355 A = I 3 = I 4 155 Ω Notice that the current through each of the parallel resistors is one-third of the total current, within the limits of significant figures. The currents after closing the switch are as follows. I1 = 0.106 A I 2 = I3 = I 4 = 0.0355 A Yes, the predictions made in part (b) are all confirmed. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 25. 19-15 All of the resistors are in series, so the equivalent resistance is just the sum of the resistors. Use Ohm’s law to find the current, and show all voltage changes starting at the negative pole of the battery and going counterclockwise. I= Ᏹ 9.0 V = = 0.3529 A ≈ 0.35 A Req (9.5 + 14.0 + 2.0)Ω ∑ voltages = 9.0 V − (9.5 Ω)(0.3529 A) − (14.0 Ω)(0.3529 A) − (2.0 Ω)(0.3529 A) = 9.0 V − 3.35 V − 4.94 V − 0.71 V = 0 26. Apply Kirchhoff’s loop rule to the circuit, starting at the upper left corner of the circuit diagram, in order to calculate the current. Assume that the current is flowing clockwise. − I (2.0 Ω) + 18 V − I (4.8 Ω) − 12 V − I (1.0 Ω) = 0 → I = 6V = 0.769 A 7.8 Ω The terminal voltage for each battery is found by summing the potential differences across the internal resistance and emf from left to right. Note that for the 12-V battery, there is a voltage gain going across the internal resistance from left to right. 18-V battery: Vterminal = − I (2.0 Ω) + 18 V = −(0.769 A)(2.0 Ω) + 18 V = 16.46 V ≈ 16 V 12-V battery: Vterminal = I (1.0 Ω) + 12 V = (0.769 A)(1.0 Ω) + 12 V = 12.769 V ≈ 13 V 27. To find the potential difference between points a and b, the current must be found from Kirchhoff’s loop law. Start at point a and go counterclockwise around the entire circuit, taking the current to be counterclockwise. − IR + Ᏹ − IR − IR + Ᏹ − IR = 0 → I = Ᏹ 2R Vab = Va − Vb = − IR + Ᏹ − IR = Ᏹ − 2 IR = Ᏹ − 2 Ᏹ R= 0 2R Notice that the actual values for the battery voltages and the resistances were not used. 28. (a) There are three currents involved, so there must be three independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction of the three branches on the right of the circuit. I 2 = I1 + I 3 I1 Ᏹ1 → I1 = I 2 − I 3 Another equation comes from Kirchhoff’s loop rule applied to the top loop, starting at the negative terminal of the battery and progressing clockwise. Ᏹ1 − I1 R1 − I 2 R2 = 0 → 9 = 25I1 + 68I 2 The final equation comes from Kirchhoff’s loop rule applied to the bottom loop, starting at the negative terminal of the battery and progressing counterclockwise. R1 R2 I2 R3 Ᏹ2 I3 Ᏹ 2 − I 3 R3 − I 2 R2 = 0 → 12 = 35I 3 + 68I 2 Substitute I1 = I 2 − I 3 into the top loop equation, so that there are two equations with two unknowns. 9 = 25 I1 + 68 I 2 = 25( I 2 − I 3 ) + 68 I 2 = 93I 2 − 25 I 3 ; 12 = 35 I 3 + 68I 2 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-16 Chapter 19 Solve the bottom loop equation for I 2 and substitute into the top loop equation, resulting in an equation with only one unknown, which can be solved. 12 = 35 I 3 + 68 I 2 → I2 = 12 − 35 I 3 68 ⎛ 12 − 35I 3 ⎞ 9 = 93I 2 − 25 I 3 = 93 ⎜ − 25I 3 68 ⎟⎠ ⎝ I3 = → 612 = 1116 − 3255I 3 − 1700 I 3 → 12 − 35I 3 504 = 0.1017 A ≈ 0.10 A, up ; I 2 = = 0.1241 A ≈ 0.12 A, left 4955 68 I1 = I 2 − I 3 = 0.0224 A ≈ 0.02 A, right (b) We can include the internal resistances simply by adding 1.0 Ω to R1 and R3 . So let R1 = 26 Ω and let R3 = 36 Ω. Now re-work the problem exactly as in part (a). I 2 = I1 + I 3 → I1 = I 2 − I 3 Ᏹ1 − I1 R1 − I 2 R2 = 0 → 9 = 26 I1 + 68 I 2 Ᏹ 2 − I 3 R3 − I 2 R2 = 0 → 12 = 36 I 3 + 68 I 2 9 = 26 I1 + 68 I 2 = 26( I 2 − I 3 ) + 68I 2 = 94 I 2 − 26 I 3 ; 12 = 36 I 3 + 68I 2 12 = 36 I 3 + 68I 2 → I2 = 12 − 36 I 3 3 − 9 I 3 = 68 17 ⎛ 3 − 9 I3 ⎞ 9 = 94 I 2 − 26 I 3 = 94 ⎜ ⎟ − 26 I 3 → 153 = 282 − 846 I 3 − 442 I 3 → ⎝ 17 ⎠ 3 − 12 I 3 129 I3 = = 0.1002 A ≈ 0.10 A, up ; I 2 = = 0.1234 A ≈ 0.12 A, left 1288 17 I1 = I 2 − I 3 = 0.0232 A ≈ 0.02 A, right The currents are unchanged to the nearest 0.01 A by the inclusion of the internal resistances. 29. This circuit is identical to Example 19–8 and Fig. 19–13 except for the numeric values. So we may copy the equations developed in that Example but use the current values. Eq. (i): I 3 = I1 + I 2 Eq. (ii): −34 I1 + 45 − 48I 3 = 0 Eq. (iii): −34 I1 + 19 I 2 − 85 = 0 Eq. (iv): I 2 = 85 + 34 I1 = 4.474 + 1.789 I1 19 45 − 34 I1 = 0.938 − 0.708 I1 48 I 3 = I1 + I 2 → 0.938 − 0.708 I1 = I1 + 4.474 + 1.789 I1 → I1 = −1.011 A Eq. (v): I 3 = I 2 = 4.474 + 1.789 I1 = 2.665 A; I 3 = 0.938 − 0.708 I1 = 1.654 A (a) To find the potential difference between points a and d, start at point a and add each individual potential difference until reaching point d. The simplest way to do this is along the top branch. Vad = Vd − Va = − I1 (34 Ω) = −(−1.011 A)(34 Ω) = 34.37 V ≈ 34 V Slight differences might be obtained in the final answer depending on the branch used, due to rounding. For example, using the bottom branch, we get the following. Vad = Vd − Va = Ᏹ1 − I 2 (19 Ω) = 85 V − (2.665 A)(19 Ω) = 34.365 V ≈ 34 V © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits (b) 19-17 For the 85-V battery, the terminal voltage is the potential difference from point g to point e. For the 45-V battery, the terminal voltage is the potential difference from point d to point b. 85 = V battery: Vterminal = Ᏹ1 − I 2 r = 85 V − (2.665 A)(1.0 Ω) = 82 V 45 = V battery: Vterminal = Ᏹ 2 − I 3 r = 45 V − (1.654 A)(1.0 Ω) = 43 V 30. I1 There are three currents involved, so there must be three independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction of the three branches at the top center of the circuit. 58V 120Ω I1 = I 2 + I 3 I3 25Ω 74Ω 3.0V I2 56Ω 110Ω Another equation comes from Kirchhoff’s loop rule applied to the left loop, starting at the negative terminal of the battery and progressing counterclockwise. 58 V − I1 (120 Ω) − I1 (74 Ω) − I 2 (56 Ω) = 0 → 58 = 194 I1 + 56 I 2 The final equation comes from Kirchhoff’s loop rule applied to the right loop, starting at the negative terminal of the battery and progressing counterclockwise. 3.0 V − I 3 (25 Ω) + I 2 (56 Ω) − I 3 (110 Ω) = 0 → 3 = −56 I 2 + 135I 3 Substitute I1 = I 2 + I 3 into the left loop equation, so that there are two equations with two unknowns. 58 = 194( I 2 + I 3 ) + 56 I 2 = 250 I 2 + 194 I 3 Solve the right loop equation for I 2 and substitute into the left loop equation, resulting in an equation with only one unknown, which can be solved. 3 = −56 I 2 + 135 I 3 → I2 = 135 I 3 − 3 ⎛ 135 I 3 − 3 ⎞ ; 58 = 250 I 2 + 194 I 3 = 250 ⎜ ⎟ + 194 I 3 56 ⎝ 56 ⎠ → 3998 = 44614 I 3 I 3 = 0.08961 A; I2 = 135 I3 − 3 = 0.1625 A; I1 = I 2 + I 3 = 0.2521 A 56 The current in each resistor is as follows: 120 Ω: 0.25 A 31. 74 Ω: 0.25 A 56 Ω: 0.16 A 25 Ω: 0.090 A 110 Ω: 0.090 A Because there are no resistors in the bottom branch, it is possible to write Kirchhoff loop equations that only have one current term, making them easier to solve. To find the current through R1 , go around the outer loop counterclockwise, starting at the lower left corner. V3 − I1 R1 + V1 = 0 → I1 = V3 + V1 6.0 V + 9.0 V = = 0.68 A, left R1 22 Ω © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-18 Chapter 19 To find the current through R2 , go around the lower loop counterclockwise, starting at the lower left corner. V3 − I 2 R2 = 0 → I 2 = 32. V3 6.0 V = = 0.33 A, left R2 18 Ω There are three currents involved, so there must be three independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction of the three branches on the left of the circuit. I1 = I 2 + I 3 1.4Ω 22Ω I2 I3 Another equation comes from Kirchhoff’s loop rule applied to the outer loop, starting at the lower left corner and progressing counterclockwise. I1 9.0V 18Ω 6.0V 1.4Ω − I 3 (1.4 Ω) + 6.0 V − I1 (22 Ω) − I1 (1.4 Ω) + 9.0 V = 0 → 15 = 23.4 I1 + 1.4 I 3 The final equation comes from Kirchhoff’s loop rule applied to the bottom loop, starting at the lower left corner and progressing counterclockwise. − I 3 (1.4 Ω) + 6.0 V + I 2 (18 Ω) = 0 → 6 = −18 I 2 + 1.4 I 3 Substitute I1 = I 2 + I 3 into the top loop equation, so that there are two equations with two unknowns. 15 = 23.4 I1 + 1.4 I 3 = 23.4( I 2 + I 3 ) + 1.4 I 3 = 23.4 I 2 + 24.8 I 3 ; 6 = −18I 2 + 1.4 I 3 Solve the bottom loop equation for I 2 and substitute into the top loop equation, resulting in an equation with only one unknown, which can be solved. −6 + 1.4 I 3 18 ⎛ −6 + 1.4 I 3 ⎞ 15 = 23.4 I 2 + 24.8I 3 = 23.4 ⎜ ⎟ + 24.8 I 3 → 270 = −140.4 + 32.76 I 3 + 446.4 I 3 → 18 ⎝ ⎠ 6 = −18I 2 + 1.4 I 3 → I 2 = −6 + 1.4 I 3 −6 + 1.4(0.8565) 410.4 = 0.8565 A ; I 2 = = = −0.2667 A ≈ 0.27 A, left 479.16 18 18 I1 = I 2 + I 3 = 0.5898 A ≈ 0.59 A, left I3 = 33. (a) We label each of the currents as shown in the accompanying figure. Using Kirchhoff’s junction rule and the first three junctions (a–c), we write equations relating the entering and exiting currents. I = I1 + I 2 [1] I 2 = I3 + I 4 [2] I1 + I 4 = I5 [3] © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-19 We use Kirchhoff’s loop rule to write equations for loops abca, abcda, and bdcb. 0 = − I 2 R − I 4 R + I1 R [4] 0 = − I 2 R − I3 R + Ᏹ [5] 0 = − I3 R + I5 R + I 4 R [6] We have six unknown currents and six equations. We solve these equations by substitution. First, insert Eq. [3] into [6] to eliminate current I5 . Next insert Eq. [2] into Eqs. [1], [4], and [5] to eliminate I 2 . 0 = − I 3 R + ( I1 + I 4 ) R + I 4 R → 0 = − I 3 R + I1 R + 2 I 4 R [6′] I = I1 + I 3 + I 4 [1′] 0 = −( I 3 + I 4 ) R − I 4 R + I1 R → 0 = − I 3 R − 2 I 4 R + I1 R [4′] 0 = −( I 3 + I 4 ) R − I 3 R + Ᏹ → 0 = − I 4 R − 2 I 3 R + Ᏹ [5′] Next we solve Eq. [4′] for I 4 and insert the result into Eqs. [1′], [5′], and [6′]. 0 = − I 3 R − 2 I 4 R + I1 R → I 4 = 12 I1 − 12 I 3 I = I1 + I 3 + 12 I1 − 12 I 3 → I = 32 I1 + 12 I 3 [1″] 0 = − I 3 R + I1 R + 2( 12 I1 − 12 I 3 ) R = −2 I 3 R + 2 I1R → I1 − I 3 [6″] 0 = −( 12 I1 − 12 I 3 ) R − 2 I 3 R + Ᏹ → 0 = − 12 I1 R − 32 I 3 R + Ᏹ [5″] Finally we substitute Eq. [6″] into Eq. [5″] and solve for I1. We insert this result into Eq. [1″] to write an equation for the current through the battery in terms of the battery emf and resistance. 0 = − 12 I1 R − 32 I1 R + Ᏹ → I1 = (b) (a) → I= Ᏹ R We divide the battery emf by the current to determine the effective resistance. Req = 34. Ᏹ ; I = 32 I1 + 12 I1 = 2 I1 2R Ᏹ Ᏹ = = R I Ᏹ/R Since there are three currents to determine, there must be three independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction near the negative terminal of the middle battery. I1 = I 2 + I 3 Another equation comes from Kirchhoff’s loop rule applied to the top loop, starting at the negative terminal of the middle battery and progressing counterclockwise. We add series resistances. 12.0 V − I 2 (12 Ω) + 12.0 V − I1 (35 Ω) = 0 → 24 = 35 I1 + 12 I 2 The final equation comes from Kirchhoff’s loop rule applied to the bottom loop, starting at the negative terminal of the middle battery and progressing clockwise. 12.0 V − I 2 (12 Ω) − 6.0 V + I 3 (28 Ω) = 0 → 6 = 12 I 2 − 28I 3 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-20 Chapter 19 Substitute I1 = I 2 + I 3 into the top loop equation, so that there are two equations with two unknowns. 24 = 35 I1 + 12 I 2 = 35( I 2 + I 3 ) + 12 I 2 = 47 I 2 + 35 I 3 Solve the bottom loop equation for I 2 and substitute into the top loop equation, resulting in an equation with only one unknown, which can be solved for I 3 . 6 + 28 I 3 3 + 14 I 3 ⎛ 3 + 14 I 3 ⎞ ; 24 = 47 I 2 + 35I 3 = 47 ⎜ = + 35 I 3 12 6 6 ⎟⎠ ⎝ 6 + 28I 3 I 3 = 3.46 mA ; I 2 = = 0.508 A ; I1 = I 2 + I3 = 0.511 A 12 6 = 12 I 2 − 28I 3 (b) → I2 = → The terminal voltage of the 6.0-V battery is 6.0 V − I 3 r = 6.0 V − (3.46 × 10−3 A)(1.0 Ω) = 5.997 V ≈ 6.0 V . 35. This problem is the same as Problem 34, except the total resistance in the top branch is now 23 Ω instead of 35 Ω. We simply reproduce the adjusted equations here without the prose. I1 = I 2 + I 3 12.0 V − I 2 (12 Ω) + 12.0 V − I1 (23 Ω) = 0 → 24 = 23I1 + 12 I 2 12.0 V − I 2 (12 Ω) − 6.0 V + I 3 (28 Ω) = 0 → 6 = 12 I 2 − 28 I 3 24 = 23I1 + 12 I 2 = 23( I 2 + I 3 ) + 12 I 2 = 35 I 2 + 23I3 6 = 12 I 2 − 28 I 3 → I2 = 6 + 28I 3 3 + 14 I 3 ⎛ 3 + 14 I 3 ⎞ ; 24 = 35I 2 + 23I 3 = 35 ⎜ = + 23I 3 12 6 6 ⎟⎠ ⎝ → 39 = 628 I 3 I 3 = 0.0621 A; I 2 = 36. 6 + 28 I 3 = 0.6449 A; I1 = I 2 + I 3 = 0.707 A ≈ 0.71 A 12 Define I1 to be the current to the right through the 2.00-V battery ( Ᏹ1 ), and I 2 to be the current to the right through the 3.00 V battery ( Ᏹ 2 ). At the junction, they combine to give current I = I1 + I 2 to the left through the top branch. Apply Kirchhoff’s loop rule, first to the upper loop and then to the outer loop, and solve for the currents. Ᏹ1 − I1r − ( I1 + I 2 ) R = 0 → Ᏹ1 − ( R + r ) I1 − RI 2 = 0 R Ᏹ I1 + I 2 r I1 r Ᏹ 2 I2 Ᏹ 2 − I 2 r − ( I1 + I 2 ) R = 0 → Ᏹ 2 − RI1 − ( R + r ) I 2 = 0 Solve the first equation for I 2 and substitute into the second equation to solve for I1. Ᏹ1 − ( R + r ) I1 2.00 − 4.350 I1 = = 0.500 − 1.0875I1 R 4.00 Ᏹ 2 − RI1 − ( R + r ) I 2 = 3.00 − 4.00 I1 − (4.35)(0.500 − 1.0875I1 ) = 0 → Ᏹ1 − ( R + r ) I1 − RI 2 = 0 → I 2 = 0.825 = −0.7306 I1 I1 = −1.129 A; I 2 = 0.500 − 1.0875 I1 = 1.728 A © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-21 The voltage across R is its resistance times I = I1 + I 2 . VR = R ( I1 + I 2 ) = (4.00 Ω)(−1.129 A + 1.728 A) = 2.396 V ≈ 2.40 V Note that the top battery is being charged—the current is flowing through it from positive to negative. 37. Take 100 of the batteries and connect them in series, which would give a total voltage of 300 volts. Do that again with the next 100 batteries, and again with the last 100 batteries. That gives three sets, each with a total voltage of 300 volts. Then those three sets can be connected in parallel with each other. The total combination of batteries would have a potential difference of 300 volts. Another possibility is to connect three batteries in parallel, which would provide a potential difference of 3 volts. Make 100 sets of those three-battery combinations and connect those 100 sets in series. That total combination would also have a potential difference of 300 volts. 38. (a) Capacitors in parallel add according to Eq. 19–5. Ceq = C1 + C2 + C3 + C4 + C5 + C6 = 6(4.8 × 10−6 F) = 2.88 × 10−5 F = 28.8 μ F (b) Capacitors in series add according to Eq. 19–6. Ceq 39. ⎛ 1 1 1 1 1 1 ⎞ =⎜ + + + + + ⎟ ⎝ C1 C2 C3 C4 C5 C6 ⎠ ⎛ ⎞ 6 =⎜ ⎜ 4.8 × 10−6 F ⎟⎟ ⎝ ⎠ −1 = 4.8 × 10−6 F = 8.0 × 10−7 F 6 The series capacitors add reciprocally, and then the parallel combination is found by adding linearly. ⎛ 1 1 ⎞ Ceq = ⎜ + ⎟ ⎝ C1 C2 ⎠ 40. −1 (a) −1 ⎛ ⎞ 1 1 + C3 = ⎜ + − − 6 6 ⎜ 3.00 × 10 F 4.00 × 10 F ⎟⎟ ⎝ ⎠ −1 + 2.00 × 10−6 F = 3.71× 10−6 F = 3.71 μ F The full voltage is across the 2.00-μ F capacitor, so V3 = 21.0 V . To find the voltage cross the two capacitors in series, find their equivalent capacitance and the charge stored. That charge will be the same for both of the series capacitors. Finally, use that charge to determine the voltage on each capacitor. The sum of the voltages across the series capacitors is 26.0 V. ⎛ 1 1 ⎞ Ceq = ⎜ + ⎟ C C 2 ⎠ ⎝ 1 −1 ⎛ ⎞ 1 1 =⎜ + ⎟ −6 −6 ⎟ ⎜ ⎝ 3.00 × 10 F 4.00 × 10 F ⎠ −1 = 1.714 × 10−6 F Qeq = CeqV = (1.714 × 10−6 F)(21.0 V) = 3.599 × 10−5 C V1 = (b) Qeq C = 3.599 × 10−5 C 3.00 × 10−6 F = 12.0 V V2 = Qeq C = 3.599 × 10−5 C 4.00 × 10−6 F = 9.0 V We have found two of the three charges already. The charge Qeq = 3.599 × 10−5 C is the charge on the two capacitors in series. The other charge is found by using the full voltage. Q3 = C3V = (2.00 × 10−6 F)(21.0 V) = 4.20 × 10−5 C Q1 = Q2 = 3.60 × 10−5 C; Q3 = 4.20 × 10−5 C © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-22 41. Chapter 19 To reduce the net capacitance, another capacitor must be added in series. 1 1 1 = + Ceq C1 C2 C2 = C1Ceq C1 − Ceq → = 1 1 1 C1 − Ceq = − = C2 Ceq C1 C1Ceq (2.9 × 10−9 F)(1.2 × 10−9 F) (2.9 × 10 −9 F) − (1.2 × 10 −9 F) → = 2.047 × 10−9 F ≈ 2.0 nF Yes, an existing connection needs to be broken in the process. One of the connections of the original capacitor to the circuit must be disconnected in order to connect the additional capacitor in series. 42. Capacitors in parallel add linearly, so adding a capacitor in parallel will increase the net capacitance without removing the 5.0-μ F capacitor. 7.0 μ F + C = 16 μ F → C = 9 μ F connected in parallel 43. The maximum capacitance is found by connecting the capacitors in parallel. Cmax = C1 + C2 + C3 = 3.2 × 10−9 F + 5.8 × 10−9 F + 1.00 × 10−8 F = 1.90 × 10−8 F in parallel The minimum capacitance is found by connecting the capacitors in series. Cmin 44. ⎛ 1 1 1 ⎞ =⎜ + + ⎟ ⎝ C1 C2 C3 ⎠ −1 ⎛ ⎞ 1 1 1 =⎜ + + ⎜ 3.2 × 10−9 F 5.8 × 10−9 F 1.00 × 10−8 F ⎟⎟ ⎝ ⎠ −1 = 1.7 × 10−9 F in series From the diagram, we see that C1 and C2 are in parallel and C3 and C4 are in parallel. Those two combinations are then in series with each other. Use those combinations to find the equivalent capacitance. We use subscripts to indicate which capacitors have been combined. C12 = C1 + C2 ; C34 = C3 + C4 ; 1 C1234 = C1234 = 1 1 1 1 + = + C12 C34 C1 + C2 C3 + C4 → (C1 + C2 )(C3 + C4 ) (C1 + C2 + C3 + C4 ) 45. The voltage across C1 is the full 10 V from the battery. Since the other two capacitors are identical, they will each have 5 V across them so that the total voltage across the two of them is 10 V. Thus V1 /V2 = 10 V/5 V = 2/1. 46. When the capacitors are connected in series, they each have the same charge as the net capacitance. −1 −1 (a) Q1 = Q2 = Qeq ⎛ ⎞ ⎛ 1 1 ⎞ 1 1 = CeqV = ⎜ + + ⎟⎟ (9.0 V) ⎟ V = ⎜⎜ − 6 − 6 ⎝ C1 C2 ⎠ ⎝ 0.50 × 10 F 1.40 × 10 F ⎠ = 3.316 × 10−6 C V1 = Q1 3.316 × 10−6 C = = 6.632 V ≈ 6.6 V C1 0.50 × 10−6 F V2 = Q2 3.316 × 10−6 C = = 2.369 V ≈ 2.4 V C2 1.4 × 10−6 F © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits (b) 19-23 Q1 = Q2 = Qeq = 3.316 × 10−6 C ≈ 3.3 × 10−6 C When the capacitors are connected in parallel, they each have the full potential difference. (c) V1 = 9.0 V V2 = 9.0 V Q1 = C1V1 = (0.50 × 10−6 F)(9.0 V) = 4.5 × 10−6 C Q2 = C2V2 = (1.4 × 10−6 F)(9.0 V) = 1.3 × 10−5 C 47. The energy stored by a capacitor is given by Eq. 17–10, PE = 12 CV 2 . = 4PEinitial PE final → 1C V2 2 final final 2 = 4 12 CinitialVinitial One simple way to accomplish this is to have Cfinal = 4Cinitial and Vfinal = Vinitial . In order to keep the voltage the same for both configurations, any additional capacitors must be connected in parallel to the original capacitor. In order to multiply the capacitance by a factor of 4, we recognize that capacitors added in parallel add linearly. Thus if a capacitor of value 3C = 750 pF were connected in parallel to the original capacitor, then the final capacitance would be 4 times the original capacitance with the same voltage, so the potential energy would increase by a factor of 4. 48. The capacitors are in parallel, so the potential is the same for each capacitor, and the total charge on the capacitors is the sum of the individual charges. We use Eqs. 17–7 and 17–8. Q1 = C1V = ε 0 A A1 A V ; Q2 = C2V = ε 0 2 V ; Q3 = C3V = ε 0 3 V d1 d2 d3 Qtotal = Q1 + Q2 + Q3 = ε 0 Cnet = 49. Qtotal V ⎛ A A A ⎞ A1 A A V + ε 0 2 V + ε 0 3 V = ⎜ ε 0 1 + ε 0 2 + ε 0 3 ⎟V d1 d2 d3 d2 d3 ⎠ ⎝ d1 ⎛ A1 A ⎞ A + ε 0 2 + ε 0 3 ⎟V ⎜ ε0 d1 d2 d3 ⎠ ⎛ A A ⎞ A =⎝ = ⎜ ε 0 1 + ε 0 2 + ε 0 3 ⎟ = C1 + C2 + C3 V d2 d3 ⎠ ⎝ d1 We have CP = C1 + C2 = 35.0 μ F and 1 1 1 1 . Solve for C1 and C2 in terms of CP = + = CS C1 C2 4.8 μ F and CS . (C − C1 ) + C1 CP 1 1 1 1 1 = + = + = P = CS C1 C2 C1 CP − C1 C1 (CP − C1 ) C1 (CP − C1 ) CP 1 = CS C1 (CP − C1 ) C1 = → → C12 − CP C1 + CP CS = 0 → CP ± CP2 − 4CP CS 2 = 29.3 μ F, 5.7 μ F = 35.0 μ F ± (35.0 μ F) 2 − 4(35.0 μ F)(4.8 μ F) 2 C2 = CP − C1 = 35.0 μ F − 29.3 μ F = 5.7 μ F or 35.0 μ F − 5.7 μ F = 29.3 μ F So the two values are 29.3 μ F and 5.7 μ F . © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-24 50. Chapter 19 We want a small voltage drop across C1. Since V = Q /C , if we put the smallest capacitor in series with the battery, there will be a large C3 voltage drop across it. Then put the two larger capacitors in parallel, C1 so that their equivalent capacitance is large and therefore will have a small voltage drop across them. So put C1 and C3 in parallel with V0 C2 each other, and then put that combination in series with C2 . See the diagram. To calculate the voltage across C1 , find the equivalent capacitance and the net charge. That charge is used to find the voltage drop across C2 , and then that voltage is subtracted from the battery voltage to find the voltage across the parallel combination. C + C2 + C3 1 1 1 = + = 1 Ceq C2 C1 + C3 C2 (C1 + C3 ) V1 = V0 − V2 = V0 − Qeq C2 = V0 − → Ceq = CeqV0 C2 Qeq C2 (C1 + C3 ) Q ; Qeq = CeqV0 ; V2 = 2 = ; C1 + C2 + C3 C2 C2 C2 (C1 + C3 ) V0 C1 + C2 + C3 C2 1.5 μ F V0 = = V0 − = (12 V) C2 C1 + C2 + C3 6.5 μ F = 2.8 V 51. (a) From the diagram, we see that C1 and C2 are in series. That combination is in parallel with C3 , and then that combination is in series with C4 . Use those combinations to find the equivalent capacitance. We use subscripts to indicate which capacitors have been combined. 1 1 1 = + C12 C C 1 C1234 (b) = → C12 = 12 C ; C123 = C12 + C3 = 12 C + C = 32 C 1 1 2 1 5 + = + = C123 C4 3C C 3C → C1234 = 3C 5 The charge on the equivalent capacitor C1234 is given by Q1234 = C1234V = 53 CV . This is the charge on both of the series components of C1234 . Q123 = 53 CV = C123V123 = 32 CV123 Q4 = 53 CV = C4V4 → V123 = 52 V → V4 = 53 V The voltage across the equivalent capacitor C123 is the voltage across both of its parallel components. Note that the sum of the charges across the two parallel components of C123 is the same as the total charge on the two components, 3 CV . 5 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-25 ( 12 C ) ( 52 V ) = 15 CV V123 = 52 V = V3 ; Q3 = C3V3 = C ( 52 V ) = 52 CV V123 = 52 V = V12 ; Q12 = C12V12 = Finally, the charge on the equivalent capacitor C12 is the charge on both of the series components of C12 . Q12 = 15 CV = Q1 = C1V1 → V1 = 15 V ; Q12 = 15 CV = Q2 = C1V2 → V2 = 15 V Here are all the results, gathered together. Q1 = Q2 = 15 CV ; Q3 = 52 CV ; Q4 = 53 CV V1 = V2 = 15 V ; V3 = 52 V ; V4 = 53 V 52. We take each of the given times as the time constant of the RC combination. τ = RC → R = τ R1s = C R4s = R15s = = τ C τ C = τ C 1s 1× 10 −6 F = 1× 106 Ω ; R2s = 4s 1× 10 = −6 F 15 s 1× 10−6 F τ C = 4 × 106 Ω ; R2s = τ C = 2s 1× 10−6 F = = 2 × 106 Ω ; 8s 1× 10 −6 F = 8 × 106 Ω ; = 15 × 106 Ω So we estimate the range of resistance to be 1 MΩ − 15 MΩ . 53. From Eq. 19–7, the product RC is equal to the time constant. τ = RC → R = 54. (a) τ C = 3.0 × 10−6 F = 1.0 × 106 Ω From Eq. 19–7, the product RC is equal to the time constant. τ = RC → C = (b) 3.0 s τ R = 18.0 × 10−6 s 3 15.0 × 10 Ω = 1.20 × 10−9 F Since the battery has an EMF of 24.0 V, if the voltage across the resistor is 16.0 V, then the voltage across the capacitor will be 8.0 V as it charges. Use the expression for the voltage across a charging capacitor. t ⎛ V ⎞ ⎛ V ⎞ VC = Ᏹ(1 − e−t /τ ) → e−t /τ = ⎜ 1 − C ⎟ → − = ln ⎜ 1 − C ⎟ → τ Ᏹ⎠ Ᏹ⎠ ⎝ ⎝ ⎛ V t = −τ ln ⎜ 1 − C Ᏹ ⎝ ⎛ 16.0 V ⎞ ⎞ −6 −5 ⎟ = −(18.0 × 10 s)ln ⎜ 1 − 24.0 V ⎟ = 1.98 × 10 s ⎠ ⎝ ⎠ © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-26 55. Chapter 19 The current for a capacitor-charging circuit is given by Eq. 19–7d, with R the equivalent series resistance and C the equivalent series capacitance. I= Ᏹ −(t /Req Ceq ) e Req R1 R2 C2 – + S → C1 – + – + I % ⎛ IReq ⎞ ⎛ C1C2 ⎞ ⎡ I (R1 + R2 ) ⎤ t = − Req Ceq ln ⎜ ⎟ = −(R1 + R2 ) ⎜ ⎟ ln ⎢ ⎥ Ᏹ ⎦ ⎝ C1 + C2 ⎠ ⎣ ⎝ Ᏹ ⎠ ⎡ (3.8 × 10−6 F) 2 ⎤ ⎡ (1.50 × 10−3 A)(4400 Ω) ⎤ −3 = −(4400 Ω) ⎢ ⎥ ln ⎢ ⎥ = 7.4 × 10 s −6 (16.0V) ⎣⎢ 7.6 × 10 F ⎦⎥ ⎣⎢ ⎦⎥ 56. The voltage of the discharging capacitor is given by VC = V0 e−t /RC . The capacitor voltage is to be 0.0025 V0 . VC = V0 e−t /RC → 0.0010 V0 = V0 e−t /RC → 0.0010 = e−t /RC → ln (0.010) = − t RC → t = − RC ln (0.010) = −(8.7 × 103 Ω)(3.0 × 10−6 F) ln (0.0025) = 0.16 s 57. (a) At t = 0, the capacitor is uncharged, so there is no voltage difference across it. The capacitor is a “short,” so a simpler circuit can be drawn just by eliminating the capacitor. In this simpler circuit, the two resistors on the right are in parallel with each other, and then in series with the resistor by the switch. The current through the resistor by the switch splits equally when it reaches the junction of the equal, parallel resistors. ⎛1 1⎞ Req = R + ⎜ + ⎟ ⎝R R⎠ I1 = (b) −1 = 32 R → Ᏹ Ᏹ Ᏹ 2Ᏹ = = ; I 2 = I 3 = 12 I1 = Req 32 R 3R 3R At t = ∞, the capacitor will be fully charged and there will be no current in the branch containing the capacitor, so a simpler circuit can be drawn by eliminating that branch. In this simpler circuit, the two resistors are in series, and they both have the same current. Req = R + R = 2 R → I1 = I 2 = (c) Ᏹ Ᏹ = ; I3 = 0 Req 2R At t = ∞, since there is no current through the branch containing the capacitor, there is no potential drop across that resistor. Therefore, the voltage difference across the capacitor equals the voltage difference across the resistor through which I 2 flows. ⎛ Ᏹ ⎞ VC = VR2 = I 2 R = ⎜ ⎟R = ⎝ 2R ⎠ 1 2 Ᏹ © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 58. (a) 19-27 With the switch open, the resistors are in series with each other, so have the same current. Apply the loop rule clockwise around the left loop, starting at the negative terminal of the source, to find the current. V − IR1 − IR2 = 0 → I = V 24 V = = 1.818 A R1 + R2 8.8 Ω + 4.4 Ω The voltage at point a is the voltage across the 4.4 Ω -resistor. Va = IR2 = (1.818 A)(4.4 Ω) = 8.0 V (b) With the switch open, the capacitors are in series with each other. Find the equivalent capacitance. The charge stored on the equivalent capacitance is the same value as the charge stored on each capacitor in series. 1 1 1 = + Ceq C1 C2 → Ceq = C1C2 (0.48 μ F)(0.36 μ F) = = 0.2057 μ F C1 + C2 (0.48 μ F + 0.36 μ F) Qeq = VCeq = (24.0 V)(0.2057 μ F) = 4.937 μ C = Q1 = Q2 The voltage at point b is the voltage across the 0.24-μ F capacitor. Vb = (c) (d) Q2 4.937 μ C = = 13.7 V ≈ 14 V C2 0.36 μ F The switch is now closed. After equilibrium has been reached for a long time, there is no current flowing in the capacitors, so the resistors are again in series, and the voltage of point a must be 8.0 V. Point b is connected by a conductor to point a, so point b must be at the same potential as point a, 8.0 V . This also means that the voltage across C2 is 8.0 V, and the voltage across C1 is 16 V. Find the charge on each of the capacitors, which are no longer in series. Q1 = V1C1 = (16 V)(0.48 μ F) = 7.68 μ C Q2 = V2C2 = (8.0 V)(0.36 μ F) = 2.88 μ C When the switch was open, point b had a net charge of 0, because the charge on the negative plate of C1 had the same magnitude as the charge on the positive plate of C2 . With the switch closed, these charges are not equal. The net charge at point b is the sum of the charge on the negative plate of C1 and the charge on the positive plate of C2 . Qb = −Q1 + Q2 = −7.68 μ C + 2.88 μ C = −4.80 μ C ≈ −4.8 μ C Thus 4.8 μ C of charge has passed through the switch, from right to left. 59. (a) The full-scale current is the reciprocal of the sensitivity. I full = scale (b) 1 = 2.9 × 10−5 A or 29 μ A 35,000 Ω /V The resistance is the full-scale voltage multiplied by the sensitivity. R = Vfull (sensitivity) = (250 V)(35,000 Ω /V) = 8.75 × 106 Ω ≈ 8.8 × 106 Ω scale © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-28 60. Chapter 19 The total resistance with the ammeter present is Req = 720 Ω + 480 Ω + 53 Ω = 1253 Ω. The voltage supplied by the battery is found from Ohm’s law to be Vbattery = IReq = (5.25 × 10−3 A)(1253 Ω) = 6.578 V. When the ammeter is removed, we assume that the battery voltage does not change. The ′ = 1200 Ω, and the new current is again found from Ohm’s law. equivalent resistance changes to Req I= 61. Vbattery ′ Req = To make a voltmeter, a resistor Rser must be placed in series with the existing meter so that the desired full-scale voltage corresponds to the full-scale current of the galvanometer. We know that 35 mA produces full-scale deflection of the galvanometer, so the voltage drop across the total meter must be 25 V when the current through the meter is 35 mA. Vfull = I full Req scale scale Vfull Rser = scale I full scale (a) Rser scale Rshunt Vfull scale −1 = 25 V 35 × 10−3 ⎛ 1 1 ⎞ −⎜ + ⎟ A ⎝ 33 Ω 0.20 Ω ⎠ −1 = 714.1 Ω ≈ 710 Ω To make an ammeter, a shunt resistor must be placed in parallel with the galvanometer. The voltage across the shunt resistor must be the voltage across the galvanometer. See Fig. 19–31 for a circuit diagram. Rshunt = → (I − I G ) Rshunt = I G r → −6 IG r (55 × 10 A)(32 Ω) = = 7.0 × 10−5 Ω (I − I G ) (25 A − 55 × 10−6 A) To make a voltmeter, a resistor must be placed in series with the galvanometer, so that the desired full-scale voltage corresponds to the full-scale current of the galvanometer. See Fig. 19–32 for a circuit diagram. Vfull scale = I G (Rser + r ) → Rser = 63. I full 714 Ω = 29 Ω /V . 25 V Vshunt = VG (b) G r −1 ⎡ ⎛1 1 ⎞ ⎤ ⎢ = I full Rser + ⎜ + ⎟ ⎥ → r Rshunt ⎠ ⎥ scale ⎢ ⎝ ⎣ ⎦ ⎛1 1 ⎞ −⎜ + ⎟ ⎝ r Rshunt ⎠ The sensitivity is 62. 6.578 V = 5.48 × 10−3 A = 5.48 mA 1200 Ω Vfull scale 250 V −r = − 32 Ω = 4.5 × 106 Ω IG 55 × 10−6 A Find the equivalent resistance for the entire circuit, and then find the current drawn from the source. That current will be the ammeter reading. The ammeter and voltmeter symbols in the diagram below are each assumed to have resistance. r % A 0.50Ω 7.5kΩ 7.5kΩ V 15kΩ © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits Req = 1.0 Ω + 0.50 Ω + 7500 Ω + 19-29 (7500 Ω)(15, 000 Ω) (7500 Ω + 15, 000 Ω) = 12501.5 Ω ≈ 12500 Ω; I source = Ᏹ 12.0 V = = 9.60 × 10−4 A Req 12500 Ω The voltmeter reading will be the source current times the equivalent resistance of the resistor– voltmeter combination. Vmeter = I source Req = (9.60 × 10−4 A) (7500 Ω)(15, 000 Ω) = 4.8 V (7500 Ω + 15, 000 Ω) If there were no meters at all, then the equivalent resistance, delivered current, and voltage across the resistor would be as follows. Req = 1.0 Ω + 7500 Ω + 7500 Ω = 15,001 Ω I source = 12.0 V Ᏹ = = 8.00 × 10−4 A Req 15,001 Ω V7500Ω = I source R = (8.00 × 10−4 A)(7500 Ω) = 6.00 V We also calculated the % difference from the readings with realistic meters. ⎛ 9.60 × 10−4 A − 8.00 × 10−4 A ⎞ (% diff) I = ⎜ ⎟⎟ × 100 = 20 % ⎜ 8.00 × 10−4 A ⎝ ⎠ ⎛ 4.8 V − 6.0 V ⎞ (% diff)V = ⎜ ⎟ × 100 = −20 % 6.0 V ⎝ ⎠ The current reads 20% higher with the meters present, and the voltage reads 20% lower. 64. We know from Example 19–17 that the voltage across the resistor without the voltmeter connected is 4.0 V. Thus the minimum acceptable voltmeter reading is 95% of that: (0.95)(4.0 V) = 3.8 V. The voltage across the other resistor would then be 4.2 V, which is used to find the current in the circuit. I= V2 4.2 V = = 2.8 × 1024 A R2 15, 000 Ω This current is used with the voltmeter reading to find the equivalent resistance of the meter–resistor combination, from which the voltmeter resistance can be found. V 3.8 V = 13,570 Ω Rcomb = comb = I 2.8 × 10−4 A 1 Rcomb = Rmeter = 65. 1 1 + R1 Rmeter → 1 Rmeter = 1 Rcomb − 1 R1 → R1 Rcomb (15,000 Ω)(13,570 Ω) = = 142,300 Ω ≈ 140 kΩ R1 − Rcomb 15,000 Ω − 13,570 Ω The sensitivity of the voltmeter is 1000 ohms per volt on the 3.0volt scale, so it has a resistance of 3000 ohms. The circuit is shown in the diagram. Find the equivalent resistance of the meter–resistor parallel combination and the entire circuit. % R R V RV © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-30 Chapter 19 −1 ⎛1 RV R 1 ⎞ (3000 Ω)(9400 Ω) Rp = ⎜ + = = 2274 Ω ⎟ = R R R R 3000 Ω + 9400 Ω + V ⎠ V ⎝ Req = R + Rp = 2274 Ω + 9400 Ω = 11, 674 Ω Using the meter reading of 1.9 volts, calculate the current into the parallel combination, which is the current delivered by the battery. Use that current to find the emf of the battery. I= 1.9 V V = = 8.355 × 10−4 A Rp 2274 Ω Ᏹ = IReq = (8.355 × 10−4 A)(11, 674 Ω) = 9.754 V ≈ 9.8 V 66. Because the voltmeter is called “high resistance,” we can assume it has no current passing through it. Write Kirchhoff’s loop rule for the circuit for both cases, starting with the negative pole of the battery and proceeding counterclockwise. Case 1: Vmeter = V1 = I1 R1 Ᏹ − I1r − I1 R1 = 0 → Ᏹ = I1 (r + R1 ) = Case 2: Vmeter = V2 = I 2 R2 V1 (r + R1 ) R1 Ᏹ − I 2 r − I 2 R2 = 0 → Ᏹ = I 2 (r + R2 ) = V2 (r + R2 ) R2 Solve these two equations for the two unknowns of Ᏹ and r. Ᏹ= V1 V (r + R1 ) = 2 (r + R2 ) → R1 R2 ⎛ V2 − V1 ⎞ ⎛ ⎞ 8.1 V − 9.7 V r = R1 R2 ⎜ ⎟ = (35 Ω)(14.0 Ω) ⎜ ⎟ = 5.308 Ω ≈ 5.3 Ω ⎝ (9.7 V)(14.0 Ω) − (8.1 V)(35 Ω) ⎠ ⎝ V1 R2 − V2 R1 ⎠ V 9.7 V Ᏹ = 1 (r + R1 ) = (5.308 Ω + 35 Ω) = 11.17 V ≈ 11.2 V R1 35 Ω 67. We can use a voltage divider circuit, as discussed in Example 19–3b and illustrated in Fig. 19–6b. The current to be passing through the body is given by Ohm’s law. I= Vbody Rbody = 0.25 V = 1.389 × 10−4 A 1800 Ω This is the current in the entire circuit. Use this to find the resistor to put in series with the body. Vbattery = I (Rbody + Rseries ) → Rseries = 68. (a) Vbattery I − Rbody = 1.5 V 1.389 × 10−4 A − 1800 Ω = 8999 Ω = 9.0 kΩ Since P = V 2 /R and the voltage is the same for each combination, the power and resistance are inversely related to each other. So for the 50-W output, use the higher-resistance filament. For the 100-W output, use the lower-resistance filament. For the 150-W output, use the filaments in parallel. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits (b) P = V 2 /R → R = V2 P R1 = (120 V) 2 = 288 Ω ≈ 290 Ω 50 W R2 = 19-31 (120 V) 2 = 144 Ω ≈ 140 Ω 100 W As a check, the parallel combination of the resistors gives the following. Rp = 69. R1 R2 (288 Ω)(144 Ω) V 2 (120 V) 2 = = 96 Ω P = = = 150 W R1 + R2 288 Ω + 144 Ω R 96 Ω There are nine values of effective capacitance that can be obtained from the four capacitors. All four in parallel: All four in series: Ceq = C + C + C + C = 4C ⎛1 1 1 1⎞ Ceq = ⎜ + + + ⎟ ⎝C C C C ⎠ −1 = 1C 4 (Three in parallel) in series with one: 1 1 1 1 1 4 = + = + = Ceq C + C + C C 3C C 3C → Ceq = 3C 4 (Two in parallel) in series with (two in parallel): 1 1 1 1 1 2 = + = + = Ceq C + C C + C 2C 2C 2C → Ceq = C (Two in parallel) in series with (two in series): 1 1 1 1 1 2 5 = + + = + = Ceq C + C C C 2C C 2C → Ceq = 2C 5 (Two in series) in parallel with (two in series): ⎛1 1⎞ Ceq = ⎜ + ⎟ ⎝C C ⎠ −1 ⎛1 1⎞ +⎜ + ⎟ ⎝C C ⎠ −1 ⎛2⎞ =⎜ ⎟ ⎝C⎠ −1 ⎛2⎞ +⎜ ⎟ ⎝C⎠ −1 = C C + =C 2 2 (not a new value) (Three in series) in parallel with one: ⎛1 1 1⎞ Ceq = ⎜ + + ⎟ ⎝C C C ⎠ −1 ⎛3⎞ +C = ⎜ ⎟ ⎝C⎠ −1 +C = 4C 3 (Two in series) in parallel with (two in parallel): ⎛1 1⎞ Ceq = ⎜ + ⎟ ⎝C C ⎠ −1 ⎛2⎞ +C +C = ⎜ ⎟ ⎝C ⎠ −1 + 2C = 5C 2 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-32 Chapter 19 ((Two in series) in parallel with one) in series with one: −1 −1 ⎡⎛ 1 1 ⎞ −1 ⎤ 1 1 ⎡C 1 ⎤ = ⎢⎜ + ⎟ + C ⎥ + = ⎢ + C ⎥ + Ceq ⎢⎣⎝ C C ⎠ C ⎣2 C ⎦ ⎥⎦ 2 1 5 = + = → Ceq = 53 C 3C C 3C ((Two in parallel) in series with one) in parallel with one: 1 ⎞ ⎛1 Ceq = ⎜ + ⎟ ⎝ C 2C ⎠ 70. −1 +C = C + 2C = 3 5C 3 The equivalent resistance is the sum of all the resistances. The current is found from Ohm’s law. I= Ᏹ 240 V = = 8 × 10−3 A Req 3 × 104 Ω This is 8 milliamps, and 10 milliamps is considered to be a dangerous level in that it can cause sustained muscular contraction. The 8 milliamps could certainly be felt by the patient and could be painful. 71. The capacitor will charge up to 75% of its maximum value and then discharge. The charging time is the time for one heartbeat. tbeat = 1 min 60 s × = 0.8333 s 72 beats 1 min ( ) t ⎛ − beat → 0.75 V0 = V0 ⎜ 1 − e RC ⎝ tbeat 0.8333 s =− = R=− C ln(0.25) (6.5 × 10−6 F)( − 1.3863) V = V0 1 − e 72. (a) t − RC t t ⎞ − beat RC = 0.25 → ⎛ − beat → e ⎟ ⎜ RC ⎝ ⎠ ⎞ ⎟ = ln(0.25) → ⎠ 9.2 × 104 Ω Apply Ohm’s law to find the current. I= Vbody Rbody = 120 V = 0.126 A ≈ 0.13 A 950 Ω (b) The description of “alternative path to ground” is a statement that the 25-Ω path is in parallel with the body. Thus the full 110 V is still applied across the body, so the current is the same: 0.13 A . (c) If the current is limited to a total of 1.5 A, then that current will be divided between the person and the parallel path. The voltage across the body and the parallel path will be the same, since they are in parallel. Vbody = Valternate I body = I total → I body Rbody = I alternate Ralternate = (I total − I body )Ralternate → Ralternate 25 Ω = (1.5 A) = 0.03846 A ≈ 38 mA (Rbody + Ralternate ) 950 Ω + 25 Ω This is still a very dangerous current. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 73. The original resistance of the motor is found from Ohm’s law. V = IR → Roriginal = (a) V 12 V = = 2.4 Ω I 5.0 A Now another resistor is put in series to reduce the current to 2.0 A. We assume the battery voltage has not changed. V = IR → Roriginal + Rseries = 74. V 12 V = = 6.0 Ω → Rseries = 6.0 Ω − 2.4 Ω = 3.6 Ω I 2.0 A (b) P = I 2 R = (2.0 A) 2 (3.6 Ω) = 14.4 W ≈ 14 W (a) If the ammeter shows no current with the closing of the switch, then points B and D must be at the same potential, because the ammeter has some small resistance. Any potential difference between points B and D would cause current to flow through the ammeter. Thus the potential drop from A to B must be the same as the drop from A to D. Since points B and D are at the same potential, the potential drop from B to C must be the same as the drop from D to C. Use these two potential relationships to find the unknown resistance. (b) 75. 19-33 VBA = VDA → I 3 R3 = I1 R1 → VCB = VCD → I 3 Rx = I1 R2 Rx = R2 R3 I1 = R1 I 3 → Rx = R2 I1 = R2 R3 /R1 I3 ⎛ 78.6 Ω ⎞ R3 = (972 Ω) ⎜ ⎟ = 129 Ω ≈ 130 Ω R1 ⎝ 590 Ω ⎠ Divide the power by the required voltage to determine the current drawn by the hearing aid. I= P 2.5 W = = 0.625 A V 4.0 V Use Eq. 19–1 to calculate the terminal voltage across the series combinations of three batteries for both mercury (Hg) and dry (D) cells. VHg = 3(Ᏹ − Ir ) = 3[1.35 V − (0.625 A)(0.030 Ω)] = 3.99 V VD = 3(Ᏹ − Ir ) = 3[1.50 V − (0.625 A)(0.35 Ω)] = 3.84 V The terminal voltage of the mercury cell batteries is closer to the required 4.0 V than the voltage from the dry cell. 76. One way is to connect N resistors in series. If each resistor can dissipate 0.5 W, then it will take 7 resistors in series to dissipate 3.5 W. Since the resistors are in series, each resistor will be 1/7 of the total resistance. R= Req 7 = 3200 Ω = 457 Ω ≈ 460 Ω 7 So connect 7 resistors of 460 Ω each, rated at ½ W, in series. Each resistor would have to have a voltage of about 15 V in order to not exceed the power rating. Thus the total voltage would be limited to about 105 V. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-34 Chapter 19 Alternately, the resistors could be connected in parallel. Again, if each resistor watt can dissipate 0.5 W, then it will take 7 resistors in parallel to dissipate 3.5 W. Since the resistors are in parallel, the equivalent resistance will be 1/7 of each individual resistance. 1 ⎛1⎞ = 7 ⎜ ⎟ → R = 7 Req = 7(3200 Ω) = 22.4 kΩ Req ⎝R⎠ So connect 7 resistors of 22.4 kΩ each, rated at ½ W, in parallel. Each resistor could have about 105 V across it in this configuration. 77. To build up a high voltage, the cells will have to be put in series. 120 V is needed from a series of 120 V 0.80 V cells. Thus = 150 cells are needed to provide the desired voltage. Since these cells 0.80 V/cell are all in series, their current will all be the same at 350 mA. To achieve the higher current desired, banks made of 150 cells each can be connected in parallel. Then their voltage will still be at 120 V, but 1.3 A the currents would add, making a total of = 3.71 banks ≈ 4 banks. So the total 350 × 10−3 A/bank number of cells is 600 cells . The panel area is 600 cells (9.0 × 10−4 m 2 /cell) = 0.54 m 2 . The cells should be wired in 4 banks of 150 cells in series per bank, with the banks in parallel. This will produce 1.4 A at 120 V. 78. There are two answers because it is not known which direction the given current is flowing through the 4.0-kΩ resistor. Assume the current is to the right. The voltage across the 4.0-kΩ resistor is given by Ohm’s law as V = IR = (3.10 × 10−3 A)(4000 Ω) = 12.4 V. The voltage drop across the 8.0 kΩ must be the same, so the current through it is I = V 12.4 V = = 1.55 × 10−3 A. The total current in the circuit is R 8000 Ω the sum of the two currents: I tot = 4.65 × 10−3 A. That current can be used to find the terminal voltage of the battery. Write a loop equation, starting at the negative terminal of the unknown battery and going clockwise. Vba − (3200 Ω)I tot − 12.4 V − 12.0 V − (1.0 Ω)I tot Vba = 24.4V + (3201 Ω)(4.65 × 10 −3 → A) = 39.28 V ≈ 39 V If the current is to the left, then the voltage drop across the parallel combination of resistors is still 12.4 V, but with the opposite orientation. Again write a loop equation, starting at the negative terminal of the unknown battery and going clockwise. The current is now to the left. Vab + (3200 Ω)I tot + 12.4 V − 12.0 V + (1.0 Ω)I tot → Vab = −0.4V − (3201 Ω)(4.65 × 10−3 A) = −15.28 V ≈ −15 V 79. (a) If the terminal voltage is to be 3.5 V, then the voltage across R1 will be 8.5 V. This can be used to find the current, which then can be used to find the value of R2 . V V1 = IR1 → I = 1 V2 = IR2 → R1 R2 = V2 V 3.5 V = R1 2 = (14.5 Ω) = 5.971 Ω ≈ 6.0 Ω I V1 8.5 V © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits (b) 19-35 If the load has a resistance of 7.0 Ω, then the parallel combination of R2 and the load must be used to analyze the circuit. The equivalent resistance of the circuit can be found and used to calculate the current in the circuit. Then the terminal voltage can be found from Ohm’s law, using the parallel combination resistance. R2+ load = I= R2 Rload (5.971 Ω)(7.0 Ω) = = 3.222 Ω Req = 3.222 Ω + 14.5 Ω = 17.722 Ω R2 + Rload (5.971 Ω + 7.0 Ω) V 12.0 V = = 0.6771 A VT = IR2+ load = (0.6771 A)(3.222 Ω) = 2.182 V ≈ 2.2 V Req 17.722 Ω The presence of the load has affected the terminal voltage significantly. 80. The terminal voltage and current are given for two situations. Apply Eq. 19–1 to both of these situations and solve the resulting two equations for the two unknowns. V1 = Ᏹ − I1r ; V2 = Ᏹ − I 2 r → Ᏹ = V1 + I1r = V2 + I 2 r → r= V2 − V1 47.3 V − 40.8 V = = 1.161 Ω ≈ 1.2 Ω I1 − I 2 8.40 A − 2.80 A Ᏹ = V1 + I1r = 40.8 V + (8.40 A)(1.161 Ω) = 50.6 V 81. The current in the circuit can be found from the resistance and the power dissipated. Then the product of that current and the equivalent resistance is equal to the battery voltage. P = I 2R → I = P33 0.80 W = = 0.1557 A 33 Ω R33 ⎛ 1 1 ⎞ + Req = 33 Ω + ⎜ ⎟ Ω Ω⎠ 68 85 ⎝ 82. −1 = 70.78 Ω V = IReq = (0.1557 A)(70.78 Ω) = 11.02 V ≈ 11 V There are three loops, so there are three loop equations necessary to solve the circuit. We identify five currents in the diagram, so we need two junction equations to complete the analysis. We also assume that the 10-kΩ resistor is known to 2 significant digits, like the other resistors. Finally, note that if we use the resistances in “kiloohms,” the currents will be in “milli-amps.” Lower left junction: I1 I4 I5 I2 I3 I 4 = I1 + I 3 Lower right junction: I 2 = I3 + I 5 Left loop, clockwise: 16 V − 10I1 − 12I 4 = 0 Right loop, counterclockwise: 21 V − 13I 2 − 18I 5 = 0 Bottom loop, counterclockwise: 12 V + 18I 5 − 12I 4 = 0 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-36 Chapter 19 16 − 12I 3 22 21 − 13I 3 21 − 13(I 3 + I 5 ) − 18I5 = 0 → 21 − 13I 3 − 31I5 = 0 → I 5 = 31 12 + 18I 5 − 12(I1 + I 3 ) = 0 → 12 − 12 I1 − 12 I 3 + 18 I5 = 0 16 − 10I1 − 12(I1 + I 3 ) = 0 → 16 − 22I1 − 12I 3 = 0 → I1 = Substitute the results of the first two equations into the third equation. ⎡16 − 12 I 3 ⎤ ⎡ 21 − 13I 3 ⎤ 12 − 12 ⎢ ⎥ − 12 I 3 + 18 ⎢ 31 ⎥ = 0 22 ⎣ ⎦ ⎣ ⎦ 12 − 12 (16 − 12 I 3 ) − 12 I 3 + 18 (21 − 13I 3 ) = 0 22 31 6 16 + 18 21 = (12 + 18 13 − 6 12)I 12 − 11 3 31 31 11 15.466 = 13.003I 3 → I3 = 15.466 = 1.189 mA 13.003 16 − 12I 3 16 − 12(1.189) = = 0.0787 mA ≈ 7.9 × 10−5 A, upward 22 22 (a) I1 = (b) Reference all voltages to the lower left corner of the circuit diagram. Then Va = 16 V, Vb = 12 V + 21 V = 33 V, and Va − Vb = 16 V − 33 V = −17 V . 83. If the switches are both open, then the circuit is a simple series circuit. Use Kirchhoff’s loop rule to find the current in that case. 6.0 V − I (50 Ω + 20 Ω + 10 Ω) = 0 → I = 6.0 V/80 Ω = 0.075 A If the switches are both closed, then the 20-Ω resistor is in parallel with R. Apply Kirchhoff’s loop rule to the outer loop of the circuit, with the 20-Ω resistor having the current found previously. 6.0 V − I (50 Ω) − (0.075 A)(20 Ω) = 0 → I = 6.0 V − (0.075 A)(20 Ω) = 0.090 A 50 Ω This is the current in the parallel combination. Since 0.075 A is in the 20-Ω resistor, 0.015 A must be in R. The voltage drops across R and the 20-Ω resistor are the same since they are in parallel. V20 = VR 84. (a) → I 20 R20 = I R R → R = R20 I 20 0.075 A = (20 Ω) = 100 Ω IR 0.015 A The 12-Ω and the 22-Ω resistors are in parallel, with a net resistance R1− 2 as follows. ⎛ 1 1 ⎞ R1− 2 = ⎜ + ⎟ ⎝ 12 Ω 22 Ω ⎠ −1 = 7.765 Ω R1− 2 is in series with the 4.5-Ω resistor, for a net resistance R1− 2−3 as follows. R1− 2−3 = 4.5 Ω + 7.765 Ω = 12.265 Ω © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-37 That net resistance is in parallel with the 14-Ω resistor, for a final equivalent resistance as follows. ⎛ 1 1 ⎞ Req = ⎜ + ⎟ Ω Ω⎠ 12.265 14 ⎝ (b) −1 Find the current in the 14-Ω resistor by using Kirchhoff’s loop rule for the loop containing the battery and the 14-Ω resistor. Ᏹ − I14 R14 = 0 → I18 = (c) = 6.538 Ω ≈ 6.5 Ω Ᏹ 6.0 V = = 0.4286 A ≈ 0.43 A R18 14 Ω Find the current in R1− 2 and the 4.5-Ω resistor by using Kirchhoff’s loop rule for the outer loop containing the battery and the resistors R1− 2 and the 4.5-Ω resistor. Those resistances are in series. Ᏹ − I1− 2 R1− 2 − I1− 2 R4.5 = 0 → I1− 2 = Ᏹ 6.0 V = = 0.4892 A R1− 2 + R4.5 7.765 Ω + 4.5 Ω This current divides to go through the 12-Ω and 22-Ω resistors in such a way that the voltage drop across each of them is the same. Use that to find the current in the 12-Ω resistor. I1− 2 = I12 + I 25 VR12 = VR25 I12 = I1− 2 (d) → I 25 = I1− 2 − I12 → I12 R12 = I 25 R25 = (I1− 2 − I12 )R25 → R25 22 Ω = (0.4892 A) = 0.3165 A ≈ 0.32 A (R12 + R25 ) 34 Ω The current in the 4.5-Ω resistor was found above to be I1− 2 = 0.4892 A. Find the power accordingly. P4.5 = I12− 2 R4.5 = (0.4892 A) 2 (4.5 Ω) = 1.077 W ≈ 1.1 W 85. (a) We assume that the ammeter is ideal so has 0 resistance but that the voltmeter has resistance RV . Then apply Ohm’s law, using the equivalent resistance. We also assume the voltmeter is accurate, so it is reading the voltage across the battery. V = IReq = I (b) 1 1 1 + R RV ⎛1 1 ⎞ → V⎜ + ⎟=I ⎝ R RV ⎠ → 1 1 I + = R RV V → 1 I 1 = − R V RV We now assume the voltmeter is ideal, so has an infinite resistance but that the ammeter has resistance RA . We also assume that the voltmeter is accurate and is reading the voltage across the battery. V = IReq = I (R + RA ) → R + RA = V I → R= V − RA I © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-38 86. Chapter 19 (a) The light will first flash when the voltage across the capacitor reaches 90.0 V. t ⎞ ⎛ − RC ⎜ ⎟ → V = Ᏹ0 1 − e ⎜ ⎟ ⎝ ⎠ ⎛ 90 ⎞ V ⎞ ⎛ 6 −6 t = − RC ln ⎜1 − ⎟ = −(2.35 × 10 Ω)(0.150 × 10 F) ln ⎜1 − ⎟ = 0.686 s ⎝ 105 ⎠ ⎝ Ᏹ0 ⎠ (b) We see from the equation that t ∝ R, so if R increases, then the time will increase. (c) The capacitor discharges through a very low resistance (the lamp filled with ionized gas), so the discharge time constant is very short. Thus the flash is very brief. Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp “goes out,” the lamp resistance increases, and the capacitor starts to recharge. It charges again for 0.686 second, and the process will repeat. (d) 87. We find the current for the bulb from the bulb’s ratings. Also, since the bulb will have 3.0 V across it, the resistor R will have 6.0 V across it. Pbulb = I bulbVbulb 88. (a) → I bulb = Pbulb 2.0 W V 6.0 V = = 0.667 A; R = R = = 9.0 Ω Vbulb 3.0 V I R 0.667 A As shown, the equivalent capacitance is found from adding the capacitance of C1 to the series combination of C1 and C1. That net capacitance is used to calculate the stored energy. ⎛ 1 1 ⎞ Cnet = C1 + ⎜ + ⎟ C C 3⎠ ⎝ 2 −1 ⎛ ⎞ 2 = 25.4 μ F + ⎜ ⎟ 25.4 F μ ⎝ ⎠ −1 = 38.1 μ F PE = 12 CnetV 2 = 12 (38.1× 10−6 F)(10.0 V) 2 = 1.905 × 10−3 J ≈ 1.91× 10−3 J (b) If all the capacitors were in series, then the equivalent capacitance is 1/3 the value of one of the capacitors. ⎛ 1 1 1 ⎞ Cnet = ⎜ + + ⎟ C C C 2 3⎠ ⎝ 1 −1 ⎛ ⎞ 3 =⎜ ⎟ μ 25.4 F ⎝ ⎠ −1 = 8.467 μ F PE = 12 CnetV 2 = 12 (8.467 × 10−6 F)(10.0 V) 2 ≈ 4.24 × 10−4 J (c) If all the capacitors were in parallel, then the equivalent capacitance is 3 times the value of one of the capacitors. Cnet = C1 + C2 + C3 = 3(25.4 μ F) = 76.2 μ F PE = 12 CnetV 2 = 12 (76.2 × 10−6 F)(10.0 V) 2 ≈ 3.81× 10−4 J 89. After a long time, the only current is in the resistors. The voltage across the 3.3-kΩ resistor is found using the left loop. 12.0V − I (4600 Ω) = 0 → I = 12.0V = 2.609 × 10−3 Ω → 4600 Ω V3.3 kΩ = IR = (2.609 × 10−3 Ω)(3300 Ω) = 8.609 V © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-39 The voltage across the 3.3 kΩ is the voltage across each capacitor. Q12 μ F = CV = (12 × 10−6 F)(8.609 V) = 1.033 × 10−4 C ≈ 1.0 × 10−4 C Q48 μ F = CV = (48 × 10−6 F)(8.609 V) = 4.132 × 10−4 C ≈ 4.1× 10−4 C 90. a We have labeled currents through the resistors with the value of the specific resistance and the emfs with the appropriate voltage value. We apply the junction rule to points a and b and then apply the loop rule to the left loop, the middle loop, and the right loop. This enables us to solve for all of the currents. I 5 = I 6 + I top ; I top + I 6.8 = I12 → I 5 − I 6 = I12 − I 6.8 17 V − I 5 (5.00 Ω) − I 6 (6.00 Ω) = 0 I6 I5 I top b I12 I 6.8 → I 5 + I 6.8 = I12 + I 6 [1] [2] (left loop) 10 V − I12 (12.00 Ω) − I 6.8 (6.80 Ω) = 0 [3] (right loop) I12 (12.00 Ω) − I 6 (6.00 Ω) = 0 [4] (center loop) Solve Eq. [4] to get I 6 = 2 I12 . Substitute that into the other equations. I 5 + I 6.8 = 3I12 [1] 17 V − I5 (5.00 Ω) − 2 I12 (6.00 Ω) = 0 10 V − I12 (12.00 Ω) − I 6.8 (6.80 Ω) = 0 [ 2] [3] Use Eq. [1] to eliminate I 6.8 by I 6.8 = 3I12 − I 5 . 17 V − I5 (5.00 Ω) − I12 (12.00 Ω) = 0 [ 2] 10 V − I12 (12.00 Ω) − (3I12 − I 5 )(6.80 Ω) = 0 → 10 V − I12 (32.40 Ω) + I 5 (6.80 Ω) = 0 Use Eq. [2] to eliminate I 5 by I 5 = [3] 17 V − I12 (12.00 Ω) , and then substitute into Eq. [3] and solve for (5.00 Ω) solve for I12 . Then back-substitute to find the other currents. ⎡17 V − I12 (12.00 Ω) ⎤ ⎥ (6.80 Ω) = 0 → (5.00 Ω) ⎣ ⎦ 10 V − I12 (32.40 Ω) + ⎢ 10 V(5.00 Ω) − I12 (32.40 Ω)(5.00 Ω) + [17 V − I12 (12.00 Ω) ] (6.80 Ω) = 0 I12 = I5 = 10 V(5.00 Ω) + 17 V(6.80 Ω) (12.00 Ω)(6.80 Ω) + (32.40 Ω)(5.00 Ω) = 0.6798 A → → I12 = 0.680 A 17 V − I12 (12.00 Ω) 17 V − (0.6798 A)(12.00 Ω) = = 1.768 A → I 5 = 1.77 A (5.00 Ω) (5.00 Ω) I 6.8 = 3I12 − I 5 = 3(0.6798 A) − 1.768 A = 0.2714 A → I 6.8 = 0.271 A I 6 = 2 I12 = 2(0.6798 A) = 1.3596 A → I 6 = 1.36 A © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-40 91. Chapter 19 (a) When the capacitors are connected in parallel, their equivalent capacitance is found from Eq. 19–5. The stored energy is given by Eq. 17–10. Ceq = C1 + C2 = 0.65 μ F; PE = 12 CeqV 2 = 12 (0.65 × 10−6 F)(24 V) 2 = 1.9 × 10−4 J (b) When the capacitors are connected in parallel, their equivalent capacitance is found from Eq. 19–6. The stored energy is again given by Eq. 17–10. 1 1 1 = + Ceq C1 C2 → Ceq = C1C2 (0.45 μ F)(0.20 μ F) = = 0.1385 μ F C1 + C2 0.65 μ F PE = 12 CeqV 2 = 12 (0.1385 × 10−6 F)(24 V) 2 = 4.0 × 10−5 J The charge is found from Eq. 17–7, Q = CV . (c) Qparallel = (Ceq ) parallelV = (0.65 μ F)(24 V) = 16 μ C Qseries = (Ceq )seriesV = (0.1385 μ F)(45 V) = 3.3 μ C 92. Because Q = CV and the capacitors both have the same voltage applied, it will be the case that Q1 > Q2 . Because of conservation of charge, when the capacitors are hooked up, the total charge that is available is Q1 − Q2 . That amount of charge redistributes over the capacitor plates, and once equilibrium has been reached, the total charge on the plates is given by q1 + q2 = Q1 − Q2 . The other constraint on this system is that once equilibrium has been reached, the capacitors must have the same q q voltage, so 1 = 2 . C1 C2 Q1 = C1V = (2.2 μ F)(24 V) = 52.8 μ F; Q2 = C2V = (1.2 μ F)(24 V) = 28.8 μ F Q1 − Q2 = 52.8 μ F − 28.8 μ F = 24.0 μ F = q1 + q2 q1 C1 = q2 C2 = 24.0 μ F − q1 q1 = 15.5 μ C, 93. C2 → q1 = 24.0 μ F ⎛ C2 ⎞ ⎜1 + ⎟ ⎝ C1 ⎠ = → q2 = 24.0 μ F − q1 24.0 μ F ⎛ 1.2 μ F ⎞ ⎜ 1 + 2.2 μ F ⎟ ⎝ ⎠ = 15.5 μ C q2 = 8.5 μ C When the switch is down, the initial charge on C2 is calculated from Eq. 17–7. Q2 = C2V0 When the switch is moved up, charge will flow from C2 to C1 until the voltage across the two capacitors is equal. V= Q2′ Q1′ = C2 C1 → Q2′ = Q1′ C2 C1 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 19-41 The sum of the charges on the two capacitors is equal to the initial charge on C2 . Q2 = Q2′ + Q1′ = Q1′ ⎛ C + C2 ⎞ C2 + Q1′ = Q1′ ⎜ 1 ⎟ C1 ⎝ C1 ⎠ Inserting the initial charge in terms of the initial voltage gives the final charges. ⎛ C + C2 ⎞ Q1′ ⎜ 1 ⎟ = C2V0 ⎝ C1 ⎠ 94. → Q1′ = C1C2 C C22 V0 ; Q2′ = Q1′ 2 = V0 C1 + C2 C1 C1 + C2 The power delivered to the starter is equal to the square of the current in the circuit multiplied by the resistance of the starter. Since the resistors in each circuit are in series, we calculate the currents as the battery emf divided by the sum of the resistances. 2 ⎛ Ᏹ/Req P I 2 RS ⎛ I ⎞ = 2 =⎜ ⎟ =⎜ ⎜ Ᏹ/R0eq P0 I 0 RS ⎝ I 0 ⎠ ⎝ 2 ⎞ ⎛ R0eq ⎟ =⎜ ⎟ ⎜ Req ⎠ ⎝ 2 2 ⎞ ⎛ r + RS ⎞ ⎟ =⎜ ⎟ ⎟ ⎝ r + RS + RC ⎠ ⎠ 2 ⎛ ⎞ 0.02 Ω + 0.15 Ω =⎜ ⎟ = 0.40 0.02 0.15 0.10 Ω + Ω + Ω ⎝ ⎠ 95. (a) From the diagram, we see that one group of four plates is connected together, and the other group of four plates is connected together. This common grouping shows that the capacitors are connected in parallel. Each capacitor would have the same voltage across it. (b) Since they are connected in parallel, the equivalent capacitance is the sum of the individual capacitances. The variable area will change the equivalent capacitance. Ceq = 7C = 7ε 0 A d Cmin = 7ε 0 Amin (2.0 × 10−4 m 2 ) = 7(8.85 × 10−12 C2 /N ⋅ m 2 ) = 7.7 × 10−12 F −3 d (1.6 × 10 m) Cmax = 7ε 0 Amax (9.0 × 10−4 m 2 ) = 7(8.85 × 10−12 C2 /N ⋅ m 2 ) = 3.5 × 10−11 F 3 − d (1.6 × 10 m) So the range is from 7.7 pF to 35 pF . 96. Capacitors in series each store the same amount of charge, so the capacitance on the unknown capacitor is also 125 pC. We calculate the voltage across each capacitor, and then calculate the unknown capacitance. V175 = Q175 125 × 10−12 C = = 0.714 V C175 175 × 10−12 F Vunknown = 25.0 V − V185 = 25.0 V − 0.714 V = 24.286 V Cunknown = Q 125 × 10−12 C = = 5.15 × 10−12 C V 24.286 V © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-42 Chapter 19 Thus the voltage across the unknown capacitor is 25 V − V185 = 25 V − 0.676 V = 24.324 V. The capacitance can be calculated from the voltage across and charge on that capacitor. C= 97. Q 125 × 10−12 C = = 5.14 pF V 24.324 V The first capacitor is charged and has charge Q0 = C1V0 on its plates. Then, when the switch is moved, the capacitors are not connected to a source of charge, so the final charge is equal to the initial charge. Initially treat capacitors C2 and C3 as their equivalent capacitance, C23 = C2C3 (2.0 μ F)(2.4 μ F) = = 1.091 μ F. The final voltage across C1 and C23 must be the same. C2 + C3 4.4 μ F The charge on C2 and C3 must be the same. Use Eq. 17–7. Q0 = C1V0 = Q1 + Q23 = C1V1 + C23V23 = C1V1 + C23V1 → V1 = C1 1.0 μ F (24 V) = 11.48 V = V1 = V23 V0 = 1.0 μ F + 1.091 μ F C1 + C23 Q1 = C1V1 = (1.0 μ F)(11.48 V) = 11.48 μ C Q23 = C23V23 = (1.091 μ F)(11.48 V) = 12.52 μ C = Q2 = Q3 V2 = Q 12.52 μ C Q2 12.52 μ C = = 6.26 V; V3 = 3 = = 5.22 V 2.0 μ F 2.4 μ F C2 C3 To summarize: Q1 = 11 μ C, V1 = 11 V; Q2 = 13 μ C, V2 = 6.3 V; Q3 = 13 μ C, V3 = 5.2 V . Solutions to Search and Learn Problems 1. Req Ceq Series Req = R1 + R2 + " 1 1 1 = + +" Ceq C1 C2 Parallel 1 1 1 = + +" Req R1 R2 Ceq = C1 + C2 + " When resistors are connected in series, the equivalent resistance is the sum of the individual resistances, Req = R1 + R2 + " . The current has to go through each additional resistance if the series resistors are in series and therefore the equivalent resistance is greater than any individual resistance. In contrast, when capacitors are in parallel, the equivalent capacitance is equal to the sum of the = C1 + C2 + ". Charge drawn from the battery can go down any one of individual capacitors, Ceq parallel the different branches and land on any one of the capacitors, so the overall capacitance is greater than that of each individual capacitor. When resistors are connected in parallel, the current from the battery or other source divides into the different branches so the equivalent resistance is less than any individual resistor in the circuit. The © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. DC Circuits 1 Req corresponding expression is = parallel 19-43 1 1 + + ". The formula for the equivalent capacitance of R1 R2 capacitors in series follows this same form, 1 Ceq series = 1 1 + + ". When capacitors are in series, the C1 C2 overall capacitance is less than the capacitance of any individual capacitor. Charge leaving the first capacitor lands on the second rather than going straight to the battery, so on. Compare the expressions defining resistance (R = V /I ) and capacitance (C = Q /V ). Resistance is proportional to voltage, whereas capacitance is inversely proportional to voltage. 2. Property Equivalent resistance Resistors in Series R1 + R2 Resistors in Parallel R1 R2 /(R1 + R2 ) Current through equivalent resistance Voltage across equivalent resistance Voltage across the pair of resistors Voltage across each resistor (VB − VA )/(R1 + R2 ) (VB − VA )(R1 + R2 )/(R1 R2 ) VB − VA VB − VA VB − VA VB − VA V1 = (VB − VA )R1 /(R1 + R2 ) V1 = VB − VA V2 = (VB − VA )R2 /(R1 + R2 ) V2 = VB − VA (VA R2 + VB R1 )/(R1 + R2 ) Not applicable I1 = (VB − VA )/(R1 + R2 ) I1 = (VB − VA )/R1 I 2 = (VB − VA )/(R1 + R2 ) I 2 = (VB − VA )/R2 Voltage at a point between the resistors Current through each resistor 3. (a) Use Eq. 19–7a to solve for the unknown resistance, where the ratio VC /Ᏹ = 0.95. 0.95 = 1 − e−t /RC (b) t 2.0 s =− = 6.7 × 105 Ω − C ln (1 − 0.95) (1.0 × 10 6 F) ln (0.05) The defibrillator paddles are usually moistened so that the resistance of the skin is about 1000 Ω. Use Eq. 19–7c to solve for the capacitance. τ = RC → C = 4. → R =2 τ R = 10 × 10−3 s = 10 × 10−6 F 1000 Ω When the galvanometer gives a null reading, no current is passing through the galvanometer or the emf that is being measured. All of the current is flowing through the slide wire resistance. Application of the loop rule to the lower loop gives Ᏹ − IR = 0, since there is no current through the emf to cause a voltage drop across any internal resistance in the emf or any resistance in the galvanometer. The © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19-44 Chapter 19 current flowing through the slide wire resistor is independent of the emf in the lower loop because no current is flowing through the lower loop. We solve the loop rule equation for the current and set the currents equal. The resulting equation is solved for the unknown emf. Ᏹ x − IRx = 0; Ᏹs − IRs = 0 → I = 5. Ᏹ x Ᏹs = Rx Rs → ⎛R Ᏹx = ⎜ x ⎝ Rs ⎞ ⎟ Ᏹs ⎠ Note that, based on the significant figures of the resistors, that the 1.0-Ω resistor will not change the equivalent resistance of the circuit as determined by the resistors in the switch bank. Case 1: n = 0 switch closed. The effective resistance of the circuit is 16.0 kΩ. The current in the circuit is I = 16 V = 1.0 mA. The voltage across the 1.0-Ω resistor is V = IR 16.0 kΩ = (1.0 mA)(1.0 Ω) = 1.0 mV . Case 2: n = 1 switch closed. The effective resistance of the circuit is 8.0 kΩ. The current in the circuit is I = 16 V = 2.0 mA. The voltage across the 1.0-Ω resistor is V = IR 8.0 kΩ = (2.0 mA)(1.0 Ω) = 2.0 mV . Case 3: n = 2 switch closed. The effective resistance of the circuit is 4.0 kΩ. The current in the circuit is I = 16 V = 4.0 mA. The voltage across the 1.0-Ω resistor is V = IR 4.0 kΩ = (4.0 mA)(1.0 Ω) = 4.0 mV . Case 4: n = 3 and n = 0 switches closed. The effective resistance of the circuit is found by the parallel combination of the 2.0-kΩ and 16.0-kΩ resistors. ⎛ 1 1 ⎞ + Req = ⎜ ⎟ ⎝ 2.0 kΩ 16.0 kΩ ⎠ −1 The current in the circuit is I = = 1.778 kΩ 16 V = 8.999 mA ≈ 9.0 mA. The voltage across the 1.778 kΩ 1.0-Ω resistor is V = IR = (9.0 mA)(1.0 Ω) = 9.0 mV . So in each case, the voltage across the 1.0-Ω resistor, if taken in mV, is the expected analog value corresponding to the digital number set by the switches. © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.