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Physics 112 Homework 4 (solutions) (2004 Fall) Solutions to Homework Questions 4 Chapt18, Problem-1: A battery having an emf of 9.00 V delivers 117 mA when connected to a 72.0-! load. Determine the internal resistance of the battery. Solution: From !V = I ( R + r ) , the internal resistance is r= !V 9.00 V "R = " 72.0 # = 4.92 ! I 0.117 A Chapt18, Problem-3: A lightbulb marked “75 W [at] 120 V” is screwed into a socket at one end of a long extension cord in which each of the two conductors has a resistance of 0.800 !. The other end of the extension cord is plugged into a 120-V outlet. Draw a circuit diagram, and find the actual power of the bulb in this circuit. Solution: The ‘rating’ on the bulb is for an ideal case where there is a potential difference of exactly 120V across the two contacts of the bulb. However in this problem, this is not quite the case since there extension cord is sufficiently long to have some resistance (and hence there is a small potential difference between each of the two ends of the cord). So basically we have a circuit with three resistances in series, and a source of emf (the outlet). We are given the resistances of each ‘arm’ of the cord, and we can find the resistance of the bulb from its rating: Rbulb = (!V ) 2 " = (120 V ) 2 75.0 W = 192 # 0.800 ! So we can now calculate the current in the circuit: 120 V !V 120 V I= = = 0.620 A , R eq 0.800 " +192 " + 0.800 " 192 ! 0.800 ! and so the actual power dissipated in the bulb: 2 ! = I 2 Rbulb = ( 0.620 A ) (192 ") = 73.8 W Chapt18, Problem-5: (a) Find the equivalent resistance between points a and b in Figure P18.5. (b) Calculate the current in each resistor if a potential difference of 34.0 V is applied between points a and b. Solution: (1 (a) 1 % " 1 + ' = 4.12 ! # 7.00 ! 10.0 ! & The equivalent resistance of the two parallel resistors is R p = $ Thus, (b) Rab = R4 + R p + R9 = ( 4.00 + 4.12 + 9.00 ) ! = 17.1 ! I ab = (!V )ab Also, ( !V ) p Then, I 7 = 34.0 V = 1.99 A , so I 4 = I 9 = 1.99 A 17.1 " Rab = I ab R p = (1.99 A) ( 4.12 " ) = 8.18 V (!V) p R7 = = 8.18 V = 1.17 A 7.00 " and I 10 = 1 (!V )p R10 = 8.18 V = 0.818 A 10.0 " Physics 112 Homework 4 (solutions) Chapt18, Problem-13: (2004 Fall) Find the current in the 12-! resistor in Figure P18.13. Solution: Clearly the first thing we need to determine is The current in the whole circuit, then we can determine how it branches. So the resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance of Rad = 63 11 ! . ( ) 6.0 ! 3.0 ! a 3.0 ! b I1 c 3.0 ! 6.0 ! 4.0 ! I2 I12 d 3.0 ! a I1 2.0 ! 12 ! b d I2 I e 3.0 ! c e 3.0 ! 2.0 ! I 18 V 18 V Figure 1 Figure 2 6.0 ! 3.0 ! I1 a b d 3! a b 30 — 11 ! d 63 — 11 a ! d I2 I I 5.0 ! I 18 V 18 V Figure 3 18 V Figure 4 So from Figure 5, a simple application of Ohm’s law gives Figure 5 I= (!V)ad = 18 V = 3.14 A . (63 11) " (!V )bd = I R bd = ( 3.14 A )( 30 11 ") = 8.57 V . So now, going back to Figure 4, we know R ad (!V )bd 8.57 V = 1.71 A , 3.0 " + 2 .0 " 5.0 " (!V )be = I2 R be = (1.71 A )(3.0 ") = 5.14 V (!V )be 5.14 V 0.43 A I 12 = = = 12 " R12 Going back again, this time to Figure 2, we see I2 = so So finally, from Figure 1, we can determine that 2 = Physics 112 Homework 4 (solutions) Chapt18, Problem-19: (2004 Fall) Figure P18.19 shows a circuit diagram. Determine (a) the current, (b) the potential of wire A relative to ground, and (c) the voltage drop across the 1 500-! resistor. Solution: (a) Applying Kirchhoff’s loop rule, as you go clockwise around the loop, gives + 20.0 V ! ( 2000) I ! 30.0 V ! ( 2500) I + 25.0 V ! ( 500 ) I = 0 or I = 3.00 !10 "3 A = 3.00 mA (b) Start at the grounded point and move up the left side, recording changes in potential as you go, to obtain ( ) ( or VA = +20.0 V ! (2000 " ) 3.00 # 10!3 A ! 30.0 V ! ( 1000 ") 3.00 #10 !3 A VA = ! 19.0 V (c) (!V )1500 = (1500 " )(3.00 # 10$3 ) A = 4.50 V ) (The upper end is at the higher potential.) Chapt18, Problem-22: Find the current in each of the three resistors of Figure P18.22 (a) by the rules for resistors in series and parallel and (b) by the use of Kirchhoff’s rules. Solution: (a) The resistors can be combined as shown below to yield an equivalent of 6.6 W. I6 6.0 ! 3.0 ! 3.6 ! 3.0 ! a b c a b I9 I3 I3 9.0 ! I3 I3 12 V 12 V Figure 2 Figure 1 (!V)ac c 6.6 ! a c I3 I3 12 V Figure 3 12 V = 1.8 A 6.6 " R ac Then, from Figure 2, (!V )ab = I3 Rab = (1.82 A )( 3.6 " ) = 6.6 V (!V)ab 6.6 V 1.1 A (!V)ab 0.73 A From Figure 1, I6 = = = , and I 9 = = 6.0 " 6.0 " 9.0 " (b) Using Figure 1 above, apply Kirchhoff’s junction rule at point a to obtain I 3 = I 6 + I 9 (1) Next, apply Kirchhoff’s loop rule to the small loop containing the 6.0 - ! and 9.0 - ! resistors to find (2) !( 6.0) I 6 + ( 9.0 ) I 9 = 0 , or I 9 = (2 3 ) I 6 From Figure 3, I3 = = Finally, apply Kirchhoff’s loop rule to the outside perimeter of Figure 1 to obtain !( 6.0) I 6 ! ( 3.0 ) I 3 +12 V = 0 , or 2.0 I 6 + I 3 = 4.0 V (3) ( ) Solving equations (1), (2), and (3) simultaneously yields I 3 = 1.8 A, I 6 = 1.1 A, I 9 = 0.73 A 3 Physics 112 Homework 4 (solutions) Chapt18, Problem-30: (2004 Fall) Show that ! = RC has units of time. Solution: The time constant is ! = R C . Considering the units, we find " Volts % " Coulombs % " Coulombs % RC ! ( Ohms )( Farads ) = $ ' =$ '$ ' # Amperes & # Volts & # Amperes & ! $ Coulombs =# so indeed ! = R C has units of time. & = Second " Coulombs Second % Chapt18, Problem-32: An uncharged capacitor and a resistor are connected in series to a source of emf. If E = 9.00 V, C = 20.0 µF, and R = 100 ", find (a) the time constant of the circuit, (b) the maximum charge on the capacitor, and (c) the charge on the capacitor after one time constant. Solution: ( a ) ! = R C = (100 " ) 20.0 # 10$6 F = 2.00 #10 $3 s = 2 .00 ms ( ) ( ) (b) Q max = C ! = 20.0 " 10#6 F ( 9.00 V ) = 1.80 "10 #4 C = ( ) ( (c) Q = Q max 1 ! e ! t " = Q max 1! e !" " )=Q # $ max % 1! 180 µC 1 & 114 µC (= e' Chapt18, Problem-37: An electric heater is rated at 1 300 W, a toaster is rated at 1 000 W, and an electric grill is rated at 1 500 W. The three appliances are connected in parallel to a common 120-V circuit. (a) How much current does each appliance draw? (b) Is a 30.0-A circuit breaker sufficient in this situation? Explain. Solution: (a) The current drawn by each appliance is ! 1300 W = = 10.8 A "V 120 V ! 1000 W Toaster: I = = = 8.33 A "V 120 V ! 1500 W Grill: I= = = 12.5 A "V 120 V Heater: I= ( b ) If the three appliances are operated simultaneously, they will draw a total current of I total = 10.8 + 8.33 + 12.5 A = 31.7 A . Therefore, a 30!ampere circuit breaker is ( ) insufficient to handle the load . Chapt18, Problem-39: A heating element in a stove is designed to dissipate 3 000 W when connected to 240 V. (a) Assuming that the resistance is constant, calculate the current in this element if it is connected to 120 V. (b) Calculate the power it dissipates at this voltage. Solution: From ! = ( "V ) 2 R , the resistance of the element is R= (!V) " When the element is connected to a 120-V source, we find that (a) (b) !V 120 V = = 6.25 A , and R 19.2 " ! = ( "V) I = ( 120 V ) ( 6.25 A ) = 750 W I= 4 2 = (240 V ) 2 3000 W = 19.2 # Physics 112 Homework 4 (solutions) Chapt18, Problem-41: (2004 Fall) Assume that a length of axon membrane of about 10 cm is excited by an action potential. (Length excited = nerve speed x pulse duration = 50 m/s x 2.0 ms = 10 cm.) In the resting state, the outer surface of the axon wall is charged positively with K+ ions and the inner wall has an equal and opposite charge of negative organic ions as shown in Figure P18.41. Model the axon as a parallel plate capacitor and use C=!"0A/d and Q = C#V to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d = 1.0x10–8 m, axon radius r = 10 µm, and cell wall dielectric constant ! = 3.0. (a) Calculate the positive charge on the outside of a 10-cm piece of axon when it is not conducting an electric pulse. How many K+ ions are on the outside of the axon? Is this a large charge per unit area? [Hint: Calculate the charge per unit area in terms of the number of square angstroms (Å2) per electronic charge. An atom has a cross section of about 1 Å2 (1 Å = 10–10 m).] (b) How much positive charge must flow through the cell membrane to reach the excited state of +30 mV from the resting state of –70 mV? How many sodium ions is this? (c) If it takes 2.0 ms for the Na+ ions to enter the axon, what is the average current in the axon wall in this process? (d) How much energy does it take to raise the potential of the inner axon wall to +30 mV starting from the resting potential of –70 mV? Solution: (a) The area of each surface of this axon membrane is [ ( )] A = l ( 2!r ) = (0.10 m ) 2! 10" 10#6 m = 2! " 10#6 m 2 , and the capacitance is C = ! "0 ' 2& #10 $6 m 2 * A $8 = 3.0 8.85 # 10$12 C2 N % m 2 ) , = 1.67 #10 F -8 d ( 1.0 # 10 m + ( ) In the resting state, the charge on the outer surface of the membrane is ( )( ) "9 Q i = C ( !V) i = 1.67 "10 #8 F 70" 10#3 V = 1.17 " 10#9 C $ 1.2 !10 C The number of potassium ions required to produce this charge is NK + = Q i 1.17 ! 10"9 C 9 + = = 7.3 ! 10 K ions , -19 e 1.6 ! 10 C and the charge per unit area on this surface is != #20 2 1e Q i 1.17 "10 #9 C % 1e 1e ( % 10 m ( = = * ' ' * -6 2 -19 2 4 2 = ) A 2$ "10 m & 1.6 " 10 C & 1 Å ) 8.6 "10 Å (290 Å )2 This corresponds to a low charge density of one electronic charge per square of side 290 Å, compared to a normal atomic spacing of one atom per several Å2 . (b) In the resting state, the net charge on the inner surface of the membrane is ! Q i = !1.17 " 10!9 C , and the net positive charge on this surface in the excited state is ( )( ) Q f = C ( !V) f = 1.67 "10 #8 F +30" 10#3 V = +5.0 " 10#10 C The total positive charge which must pass through the membrane to produce the excited state is therefore !Q = Q f " Q i ( ) "9 = +5.0 ! 10"10 C " "1.17 !10 "9 C = 1.67 ! 10"9 C # 1.7 ! 10 C , corresponding to N Na + !Q 1.67 " 10#9 C 10 + = = = 1.0 !10 Na ions e 1.6 "10 -19 C Na+ ion 5 Physics 112 Homework 4 (solutions) (c) If the sodium ions enter the axon in a time of !t = 2.0 ms , the average current is I= (2004 Fall) !Q 1.67 "10 #9 C = = 8.3 "10 #7 A = 0.83 µ A !t 2.0 " 10#3 s (d) When the membrane becomes permeable to sodium ions, the initial influx of sodium ions neutralizes the capacitor with no required energy input. The energy input required to charge the now neutral capacitor to the potential difference of the excited state is Chapt18, Conceptual-4: How would you connect resistors so that the equivalent resistance is larger than the individual resistances? Give an example involving two or three resistors. Solution: The resistors should be connected in series. For example, connecting three resistors of 5 ! , 7 ! , and 2 ! in series gives a resultant resistance of 14 ! . Chapt18, Conceptual-5: If you have your headlights on while starting your car, why do they dim while the car is starting? Solution: The starter in the car draws a relatively large current from the battery. This large current causes a significant voltage drop across the internal resistance of the battery. As a result the terminal voltage of the battery is reduced, and the headlights dim accordingly. Chapt18, Conceptual-7: Electrical devices are often rated with a voltage and a current – for example, 120V, 5A. Batteries, however are only rated witth a voltage – for example, 1.5 V. Why? Solution: An electrical appliance has a given resistance. Thus, when it is attached to a power source with a known potential difference, a definite current will be drawn. The device can be labeled with both the voltage and the current. Batteries, however, can be applied to a number of devices. Each device will have a different resistance, so the current from the battery will vary with the device. As a result, only the voltage of the battery can be specified. Chapt18, Conceptual-10: If electrical power is transmitted over long distances, the resistance of the wores becomes significant. Why? Which mode of transmission would result in less energy loss- high current and low voltage or low current and high voltage? Solution: A wire or cable in a transmission line is thick and made of material with very low resistivity. Only when its length is very large does its resistance become significant. To transmit power over a long distance it is most efficient to use low current at high voltage, minimizing the I 2 R power loss in the transmission line. Chapt18, Conceptual-13: Why is it possible for a bird to sit on a high-voltage woire without being electrocuted? Solution: The bird is at rest on the wire whose electrical potential is also constant along its length. In order to be electrocuted, a large potential difference is required between the bird’s feet. The potential difference between the bird’s feet is too small to harm the bird. 6