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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,

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

1 RC Circuits SPH4UW

2 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.

3 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?

4 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.

5 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 S2S 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

6 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

7 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:

8 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 +  q ∞ /C = 0 q ∞ =  C

9 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

10 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

11 Charging: Intermediate Times  C R1R1 S2S2 S1S1 IbIb Calculate the charge on the capacitor 3  seconds after switch 1 is closed. q(t) = q  (1-e -t/R 2 C ) = q  (1-e - 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

12 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 S1S I S2S2 q RC2RC t

13 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) q 0 2) q 0 3) q 0 4) 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 /(4  (15  )) ) = q 0 (0.368)

14 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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