# RC Circuits SPH4UW. Capacitors Charge on Capacitors cannot change instantly. Short term behavior of Capacitor: If the capacitor starts with no charge,

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RC Circuits SPH4UW

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.

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?

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.

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 qq +  - I(t)R - q(t) / C = 0

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

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:

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

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 + - + + - - + -

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 + - + + - - + -

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

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

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)

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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