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6/3/2013 DC Circuits: Ch 32 • Voltage – Starts out at highest point at “+” end of battery • Voltage drops across lightbulbs and other sources of resistance. • Voltage increases again at battery. The following circuit uses a 1.5 V battery and has a 15 W lightbulb. a. Calculate the current in the circuit b. Calculate the voltage drop across the lightbulb. c. Sketch a graph of voltage vs. path (battery, top wire, resistor, bottom wire) I + Voltage highest Voltage zero Resistors in Series • Same current (I) passes through all resistors (bulbs) • All bulbs are equally bright (energy loss, not current loss) • Voltage drop across each resistor (V1,V2, V3) 1 6/3/2013 V = V1 + V2 + V3 V = IR1 + IR2 + IR3 V = I(R1 + R2 + R3) Req = R1 + R2 + R3 V = IReq Resistors in Parallel • Current splits at the junction • Same Voltage across all resistors Which combination of auto headlights will produce the brightest bulbs? Assume all bulbs have a resistance of R. I = I1 + I2 + I3 I1 = V R1 I=V Req 1 = 1 Req R1 + 1 R2 + 1 R3 For the Bulbs in Series: Req = R + R = 2R What current flows through each resistor in the following circuit? (R = 100 W) For the Bulbs in Parallel 1 = 1 + 1 Req R R 1 = 2 Req R Req = R/2 The bulbs in parallel have less resistance and will be brighter Req = R1 + R2 Req = 200 W V = IReq I = V/Req I = 24.0 V/ 200 W = 0.120 A 2 6/3/2013 Calculate the current through this circuit, and the voltage drop across each resistor. Req = 400 W + 290 W Req = 690 W V = IR I = V/Req I = 12.0 V/690 W I = 0.0174 A What current flows through each of the resistors in this circuit? (Both are 100 W) Vab = (0.0174A)(400W) Vab = 6.96 V Vbc = (0.0174A)(290W) Vbc = 5.04 V DC Circuits: Ex 4 What current will flow through the circuit shown? 1 = 1 + Rp = 500 W 1 700 W Rp = 290 W (0.48 A, 0.24 A) Req = 400 W + 290 W Req = 690 W V = IR I = V/R I = 12.0 V/690 W I = 0.017 A or 17 mA Example 4 Calculate the equivalent resistance in the following circuit. 3 6/3/2013 DC Circuits: Ex 5 What current is flowing through just the 500 W resistor? The voltage through the resistors in parallel will be: 12.0 V – 6.8 V = 5.2 V To find the current across the 500 W resistor: First we find the voltage drop across the first resistor: V = IR = (0.017 A)(400 W) V = 6.8 V DC Circuits: Ex 6 Which bulb will be the brightest in this arrangement (most current)? V = IR I = V/R I = 5.2 V/500 W = 0.010 A = 10 mA Bulb C (current gets split running through A and B) What happens when the switch is opened? – C and B will have the same brightness (I is constant in a series circuit) DC Circuits: Ex 5 What resistance would be present between points A and B? (ANS: 41/15 R) 4 6/3/2013 EMF and Terminal Voltage EMF: Example 1 • Batteries - source of emf (Electromotive Force), E (battery rating) • All batteries have some internal resistance r A 12-V battery has an internal resistance of 0.1 W. If 10 Amps flow from the battery, what is the terminal voltage? Vab = E – Ir Vab = E – Ir Vab = terminal(useful)voltage E = battery rating Vab = 12 V – (10 A)(0.10 W) Vab = 11 V r = internal resistance EMF: Example 2 Calculate the current in the following circuit. Req = 6 W + 2.7 W Req = 8.7 W 1/Req = 1/8 W + ¼ W Req = 2.7 W 1/Req = 1/10 W + 1/8.7 W Req = 4.8 W Everything is now in series Req = 4.8 W + 5.0 W + 0.50 W Req = 10.3 W V = IR I = V/R I = 9.0 V/10.3 W I = 0.87 A 5 6/3/2013 EMF: Example 2a Now calculate the terminal(useful)voltage. V = E – Ir V = 9.0 V – (0.87 A)(0.50 W) V = 8.6 V Kirchoff’s Rules 1. Junction Rule - The sum of the currents entering a junction must equal the sum of currents leaving 2. Loop Rule - The sum of the changes in potential around any closed path = 0 Kirchoff Conventions Grounded • • • • Wire is run to the ground Houses have a ground wire at main circuit box Does not affect circuit behavior normally Provides path for electricity to flow in emergency Kirchoff Conventions The “loop current” is not a current. Just a direction that you follow around the loop. Kirchoff’s Rule Ex 1 6 6/3/2013 Junction Rule I1 = I2 + I3 Loop Rule Main Loop 6V – (I1)(4W) – (I3)(9W) = 0 Side Loop (-I2)(5W) + (I3)(9W) = 0 I1 = I2 + I3 6V – (I1)(4W) – (I3)(9W) = 0 (-I2)(5W) + (I3)(9W) = 0 Solve Eqn 1 (-I2)(5W) + (I3)(9W) = 0 (I3)(9W) = (I2)(5W) I3 = 5/9 I2 Eqn 3 Eqn 2 Eqn 3 Substitution into Eqn 2 6V – (I2 + I3)(4W) – (I3)(9W) = 0 6 – 4I2 -4I3 - 9I3 = 0 6 – 4I2 - 13I3 = 0 I3 = 5/9 I2 (from last slide) 6 – 4I2 - 13(5/9 I2) = 0 6 = 101/9 I2 I2 = 0.53 A I3 = 5/9 I2 = 0.29 A I1 = I2 + I3 = 0.53 A + 0.29 A = 0.82 A 7 6/3/2013 Kirchoff’s Rule Ex 2 I1 = I2 + I3 Loop Rule Main Loop 9V – (I3)(10W) – (I1)(5W) = 0 Side Loop (-I2)(5W) + (I3)(10W) = 0 (-I2)(5W) + (I3)(10W) = 0 (I2)(5W) = (I3)(10W) I2 = 2I3 I3 = 9/25 = 0.36 A I2 = 2I3 = 2(0.36 A) = 0.72 A I1 = I2 + I3 = 0.36A + 072 A = 1.08 A 9V – (I3)(10W) – (I1)(5W) = 0 9V – (I3)(10W) – (I2 + I3)(5W) = 0 9 –10I3 – 5I2 – 5I3 = 0 9 –15I3 – 5I2 = 0 9 –15I3 – 5(2I3) = 0 9 –25I3 = 0 I3 = 9/25 = 0.36 A 8 6/3/2013 Kirchoff’s Rules: Ex 3 Calculate the currents in the following circuit. Bottom Loop (clockwise) 10V – (6W)I1 – (2W)I3 = 0 Top Loop (clockwise) -14V +(6W)I1 – 10 V -(4W)I2 = 0 Work with Bottom Loop 10V – (6W)I1 – (2W)I3 = 0 I1 + I2 = I3 10 – 6I1 – 2(I1 + I2) = 0 10 – 6I1 – 2I1 - 2I2 = 0 10 - 8I1 - 2I2 = 0 10 = 8I1 + 2I2 5 = 4I1 + I2 I2 = 5 - 4I1 I1 = 22/11 = 2.0 Amps Working with Top Loop -14V +(6W)I1 – 10 V -(4W)I2 = 0 24 = 6I1 - 4I2 12 = 3I1 - 2I2 12 = 3I1 - 2(5 - 4I1) 22 = 11I1 I1 + I2 = I3 I3 = -1.0 A Batteries in Series • If + to -, voltages add (top drawing) • If + to +, voltages subtract (middle drawing = 8V, used to charge the 12V battery as in a car engine) I2 = 5 - 4I1 I2 = 5 – 4(2) = -3.0 Amps Extra Kirchoff Problems I1 = -0.864 A I2 = 2.6 A I3 = 1.73 A Batteries in Parallel • Provide large current when needed (Same voltage) 9 6/3/2013 A Strange Example Calculate I (0.5 Amps) RC Circuits • • • • Capacitors store energy (flash in a camera) Resistors control how fast that energy is released Car lights that dim after you shut them Camera flashes a. Calculate the equivalent resistance. (2.26 W) b. Calculate the current in the upper and lower wires. (3.98 A) c. Calculate I1, I2, and I3 (0.60 A, 2.25 A, 1.13 A) d. Sketch a graph showing the voltage through the circuit starting at the battery. DVc + DVr = 0 Q - IR = 0 C (Divide by R) Q - I =0 RC (I = -dQ/dt) Q + dQ = 0 RC dt = time constant (time to reach 63% of full voltage) = RC 10 6/3/2013 A versatile relationship Charging the Capacitor (for capacitor) V = Vo (1-e-t/RC) Q = Qo (1-e-t/RC) I = Io e-t/RC Discharging the Capacitor V = Vo e-t/RC I = Io(e-t/RC) Q = Qo e-t/RC RC Circuits: Ex 1 What is the time constant for an RC circuit of resistance 200 kW and capacitance of 3.0 mF? = (200,000 W)(3.0 X 10-6 F) = 0.60 s (lower resistance will cause the capacitor to charge more quickly) RC Circuits: Ex 2 What will happen to the bulb (resistor) in the circuit below when the switch is closed (like a car door)? RC Circuits: Ex 3 An uncharged RC circuit has a 12 V battery, a 5.0 mF capacitor and a 800 kW resistor. Calculate the time constant. = RC = (5.0 X 10-6 F)(800,000 W) = 4.0 s Answer: Bulb will glow brightly initially, then dim as capacitor nears full charge. What is the maximum charge on the capacitor? Q = CV Q = (5.0 X 10-6 F)(12 V) Q = 6 X 10-5 C or 60 mC What is the voltage and charge on the capacitor after 1 time constant (when discharging)? 11 6/3/2013 Consider the circuit below, which is being charged. Calculate: a. The time constant (6 ms) b. Maximum charge on the capacitor (3.6 mC) c. Maximum current (600 mA) d. Time to reach 99% of maximum charge (28 ms) e. Current when charge = ½ Qmax (300 mA) f. The charge when the current is 20% of the maximum value. (2.9 mC) RC Circuits: Ex 4 An RC circuit has a charged capacitor C = 35 mF and a resistance of 120W. How much time will elapse until the voltage falls to 10 percent of its original (maximum) value? V = Vo e-t/RC 0.10Vo =Voe-t/RC 0.10 =e-t/RC ln(0.10) = ln(e-t/RC) -2.3 = -t/RC RC Circuits: Ex 5 If a capacitor is discharged in an RC circuit, how many time constants will it take the voltage to drop to ¼ its maximum value? Discharging the RC Circuit V = Vo e-t/RC 2.3 = t/RC t = 2.3RC t = (2.3)(120W)(35 X 10-6 F) t = 0.0097 s or 9.7 ms A fully charged 1.02 mF capacitor is in a circuit with a 20.0 V battery and a resistor. When discharged, the current is observed to decrease to 50% of it’s initial value in 40 ms. a. Calculate the charge on the capacitor at t=0 (20.4 mC) b. Calculate the resistance R (57 W) c. Calculate the charge at t = 60 ms (7.3 mC) (t = 1.39RC) 12 6/3/2013 The capacitor in the drawing has been fully charged. The switch is quickly moved to position b (camera flash). a. Calculate the initial charge on the capacitor. (9 mC) b. Calculate the charge on the capacitor after 5.0 ms. (5.5 mC) c. Calculate the voltage after 5.0 ms (5.5 V d. Calculate the current through the resistor after 5.0 ms (0.55 A) Meters • Galvanometer – Can only handle a small current • Full-scale Current Sensitivity (Im) – Maximum deflection • Ex: – Multimeter – Car speedometer Measuring I and V Measuring Current – Anmeter is placed in series – Current is constant in series Anmeter (Series) Voltmeter (parallel) Measuring Voltage – Voltmeter placed in parallel – Voltage constant in parallel circuits – Measuring voltage drop across a resistor DC Anmeter • Uses “Shunt” (parallel) resistor • Shunt resistor has low resistance • Most of current flows through shunt, only a little through Galvanometer • IRR = IGr Meters: Ex 1 What size shunt resistor should be used if a galvanometer has a full-scale sensitivity of 50 mA and a resistance of r= 30 W? You want the meter to read a 1.0 A current. Voltage same through both (V=IR) IRR = Igr Since most of the current goes through the shunt IR ~ 1 A 13 6/3/2013 IRR = Igr (1 A)(R) = (50 X 10-6 A)(30 W) R = 1.5 X 10-3 W or 1.5 m W Meters: Ex 2 Design an anmeter that can test a 12 A vacuum cleaner if the galvanometer has an internal resistance of 50 W and a full scale deflection of 1 mA. IRR = Igr (12 A)(R) = (1 X 10-3 A)(50 W) R = 4.2 X 10-3 W or 4.2 m W DC Voltmeter • Resistor in series • Large R for resistor (keeps current low in Galvanometer) • V = I(R + r) Meters: Ex 4 Design a voltmeter for a 120 V appliance with and internal galvanometer resistance of 50 W and a current sensitivity of 1 mA. Meters: Ex 3 What resistor should be used in a voltmeter that can read a maximum of 15 V? The galvanometer has an internal resistance of 30W and a full scale deflection of 50 mA. V = I(R + r) 12 V = (50 X 10-6 A)(R + 30W) R + 30W = 12 V 50 X 10-6 A R + 30W = 300,000 W R ~ 300,000 W Electric Power • Watt • 1 Watt = 1 Joule 1 second P = I2R (ANS: R = 120,000 W) 14 6/3/2013 Power: Ex 1 Household Electricity Calculate the resistance of a 40-W auto headlight that operates at 12 V. • Kilowatt-hour • You do not pay for power, you pay for energy P = I2R 1 kWh = (1000 J)(3600 s) = 3.60 X 106 J 1s (3.6 W) Power: Ex 2 An electric heater draws 15.0 A on a 120-V line. How much power does it use? P = I2R V = IR so R = V/I P = I2V I P = IV = (15 A)(120V) = 1800 W or 1.8 kW Power: Ex 4 A lightening bolt can transfer 109 J of energy at a potential difference of 5 X 107V over 0.20 s. What is the charge transferred? V = PE/Q Q = E/V = 109 J/ 5 X 107V = 20 C Power: Ex 3 How much does it cost to run it for 30 days if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh? Hours = 30 days X 3.0 h/day = 90 h Cost = (1.80 kW)(90 h)($0.105/kWh) = $17 What is the current? I = DQ/Dt I = 20 C/0.2 s = 100 A What is the power? P = I2R V = IR so R = V/I P = I2V I P = IV = (100 A)(5 X 107V ) = 5 X 109 W 15 6/3/2013 Household Electricity • Circuit breakers – prevent “overloading” (too much current per wire) • Metal melts or bimetallic strip expands Household Electricity: Ex 1 Determine the total current drawn by all of the appliances shown. P = IV I = P/V Ilight = 100W/120 V = 0.8 A Ilight = 100W/120 V = 0.8 A Iheater = 1800W/120 V = 15 A Istereo = 350W/120 V = 2.9 A Ihair = 1200W/120 V = 10.0 A DC vs AC Itotal = 0.8A + 15.0A + 2.9A + 10.0A = 28.7 A This would blow the 20 A fuse DC vs AC DC AC •Electrons flow constantly •Electrons flow in short burst •Electrons flow in only one direction •Electrons switch directions (60 times a second) •Batteries •House current Jump Starting a Car POSITIVE TO POSITIVE (or your battery could explode) http://www.ibiblio.org/obp/electricCircuits/AC/AC_1.html 16 6/3/2013 2. Clockwise (12 V. 50 W, (10 mF and 100 W parallel)) 4. a) 0.10 A b) Graph page 32-3 6. a) 5 V, 10 V b) Graph page 32-4 8. 1.92 W, 2.9 W 10. 3.6 X 106 J 14. 25 W 18. 3.2 % 20. R/4 22. 12 W 24. 24 W 46. 9.0 V 50.1.0 A, 2.0 A, 15 V 56.a) 8 Amps (8 Vab) 58. a) R =0.505 W 62.Resistor(W) 4 6 8 24(bottom) 24 (right) 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 183 W 13 V, 5 V, 0 V, -2 V 2 ms 6.9 ms 0.87 K W A> D = E > B = C 1.99 m 4.0 W 7W 10 W and 60 V b) 9 Amps (0 Vab) b) Req = 0.500 W (V) Current (A) 8 2 8 1.3 8 1 8 0.33 16 0.66 17

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