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RC Circuits SPH4UW

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Capacitors Charge on Capacitors cannot change instantly. Short term behavior of Capacitor: If the capacitor starts with no charge, it has no potential difference across it and acts as a wire. If the capacitor starts with charge, it has a potential difference across it and acts as a battery. Current through a Capacitor eventually goes to zero. Long term behavior of Capacitor: If the capacitor is charging, when fully charged no current flows and capacitor acts as an open circuit. If capacitor is discharging, potential difference goes to zero and no current flows.

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Combine R+C Circuits Gives time dependence Current is not constant I(t) Charge is not constant q(t) Used for timing Pacemaker Intermittent windshield wipers Models nervous system include R, C Sports Trivia: How soon after starting gun can you run w/o getting a False Start?

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What is the time constant? The time constant = RC. Given a capacitor starting with no charge, the time constant is the amount of time an RC circuit takes to charge a capacitor to about 63.2% of its final value. The time constant is the amount of time an RC circuit takes to discharge a capacitor by about 63.2% of its original value.

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RC Circuits: Charging KVL: q =q 0 Just after S 1 is closed: q =q 0 Capacitor is uncharged (no time has passed so charge hasn’t changed yet) - I 0 R = 0 I 0 = / R - I 0 R = 0 I 0 = / R I c = 0 Long time after: I c = 0 Capacitor is fully charged + - q /C =0 q = C + - q /C =0 q = C Intermediate (more complex) q(t) = q (1-e -t/RC ) I(t) = I 0 e -t/RC C R S1S1 S2S2 + + + I - - - The switches are originally open and the capacitor is uncharged. Then switch S 1 is closed. t q RC 2RC 0 qq + - I(t)R - q(t) / C = 0

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Capacitor “Rules of Thumb” Initially uncharged capacitor: acts like a wire (short circuit) at t = 0 acts like an open circuit (broken wire) as t Initially charged capacitor: acts like a battery at t = 0 Energy stored by a capacitor is

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Time Constant Demo Which system will be brightest? Which lights will stay on longest? Which lights consumes more energy? 2 Each circuit has a 1 F capacitor charged to 100 Volts. When the switch is closed: 1 = 2RC = RC/2 Same:

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Practice! Calculate current immediately after switch is closed: Calculate current after switch has been closed for 0.5 seconds: Calculate current after switch has been closed for a long time: Calculate charge on capacitor after switch has been closed for a long time: R C E S1S1 R=10 C=30 mF E =20 Volts - I + + + - - - + - I 0 R - q 0 /C = 0 + - I 0 R - 0 = 0 I 0 = /R After a long time current through capacitor is zero! + - IR - q ∞ /C = 0 + - 0 - q ∞ /C = 0 q ∞ = C

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Understanding After switch 1 has been closed for a long time, it is opened and switch 2 is closed. What is the current through the right resistor just after switch 2 is closed? 1) I R = 02) I R = V/(3R) 3) I R = V/(2R)4) I R = V/R KVL: +q 0 /C-IR = 0 Recall q is charge on capacitor after charging: q 0 =VC (since charged w/ switch 2 open!) +V - IR = 0 I = V/R 2R C R S2S2 S1S1 IRIR + - + + - - + -

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Understanding Both switches are closed. What is the final charge on the capacitor after the switches have been closed a long time? 1) Q = 02) Q = C ε /3 3) Q = C ε /24) Q = C ε KVL (right loop): +Q/C-IR = 0 KVL (outside loop): + ε - I(2R) - IR = 0 I = ε /(3R) KVL: +Q/C - ε /(3R) R = 0 Q = C ε /3 R 2R C S2S2 S1S1 IRIR + - + + - - + -

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Charging: Intermediate Times C R1R1 S2S2 S1S1 IbIb + - + + - - Calculate the charge on the capacitor 3 10 -3 seconds after switch 1 is closed. q(t) = q (1-e -t/R 2 C ) = q (1-e - 3 10 -3 /(40 100 10 -6) ) ) = q (0.53) Recall q = ε C = (50)(100x10 -6 ) (0.53) = 2.7 x10 -3 Coulombs R 1 = 20 R 2 = 40 ε = 50 Volts C = 100 F R2R2

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RC Circuits: Discharging - q(t) / C - I(t) R = 0 KVL: - q(t) / C - I(t) R = 0 q=q 0 Just after…: q=q 0 Capacitor is still fully charged Capacitor is still fully charged +q 0 / C + I 0 R = 0 +q 0 / C + I 0 R = 0 I 0 = q 0 /(RC) I 0 = q 0 /(RC) : I c =0 Long time after: I c =0 Capacitor is discharged Capacitor is discharged q / C = 0 q = 0 q / C = 0 q = 0 Intermediate (more complex) q(t) = q 0 e -t/RC q(t) = q 0 e -t/RC I c (t) = I 0 e -t/RC I c (t) = I 0 e -t/RC C R S1S1 - + + I - + - S2S2 q RC2RC t

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RC Challenge After being closed for a long time, the switch is opened. What is the charge Q on the capacitor 0.06 seconds after the switch is opened? 1) 0.368 q 0 2) 0.632 q 0 3) 0.135 q 0 4) 0.865 q 0 2R C E R S1S1 E = 24 Volts R = 4 C = 15 mF q(t) = q 0 e -t/RC = q 0 (e - 0.06 /(4 (15 10 -3 )) ) = q 0 (0.368)

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RC Summary ChargingDischarging q(t) = q (1-e -t/RC )q(t) = q 0 e -t/RC V(t) = V (1-e -t/RC )V(t) = V 0 e -t/RC I(t) = I 0 e -t/RC Short term: Charge doesn’t change (often zero or max) Long term: Current through capacitor is zero. Time Constant = RC Large means long time to charge/discharge

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