Lect07 handout

RC Circuits Physics 102: Lecture 7

Recall …. First we covered circuits with batteries and capacitors series, parallel Then we covered circuits with batteries and resistors series, parallel Kirchhoff’s Loop and Junction Relations Today: circuits with batteries, resistors, and capacitors

RC Circuits RC Circuits Charging Capacitors Discharging Capacitors Intermediate Behavior

RC Circuits Circuits that have both resistors and capacitors : With resistance in the circuits, capacitors do not charge and discharge instantaneously – it takes time (even if only fractions of a second). R K C S + + + R Na R Cl ε K ε Na ε Cl

Capacitors Charge (and therefore voltage) on Capacitors cannot change instantly: remember V C = Q/C 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. Long term behavior of Capacitor: Current through a Capacitor eventually goes to zero. 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

Charging Capacitors Capacitor is initially uncharged and switch is open. Switch is then closed. What is current I 0 in circuit immediately thereafter? What is current I  in circuit a long time later?  R C S

Charging Capacitors: t=0 Capacitor is initially uncharged and switch is open. Switch is then closed. What is current I 0 in circuit immediately thereafter? Capacitor initially uncharged Therefore V C is initially 0 Therefore C behaves as a wire (short circuit) KLR:  – I 0 R = 0 I 0 =  /R  R C S R 

Charging Capacitors: t>0 I 0 =  /R Positive charge flows Onto bottom plate (+Q) Away from top plate (-Q) As charge builds up, V C rises (V C =Q/C) Loop:  – V C – I R = 0 I = (  -V C )/R Therefore I falls as Q rises When t is very large (  ) I  = 0: no current flow into/out of capacitor for t large V C =   R C + - Demo R 

ACT/Preflight 7.1 Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S 1 is closed? 2R C  R S 2 1) I b = 0 2) I b = E /(3R) 3) I b = E /(2R) 4) I b = E /R S 1 I b + - + + - -

ACT/Preflight 7.3 Both switches are initially open, and the capacitor is uncharged. What is the current through the battery after switch 1 has been closed a long time ? 1) I b = 0 2) I b = E /(3R) 3) I b = E /(2R) 4) I b = E /R 2R C  R S 2 S 1 I b + - + + - -

Discharging Capacitors Capacitor is initially charged (Q) and switch is open. Switch is then closed. What is current I 0 in circuit immediately thereafter? What is current I  in circuit a long time later? R C S

Discharging Capacitors Capacitor is initially charged (Q) and switch is open. Switch is then closed. What is current I 0 in circuit immediately thereafter? KLR: Q/C – I 0 R = 0 So, I 0 = Q/RC What is current I  in circuit a long time later? I  = 0 R C + - Demo

ACT/Preflight 7.5 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 = 0 2) I R =  /(3R) 3) I R =  /(2R) 4) I R =  /R 2R C  R S 2 S 1 I R + - + + - - + -

ACT: RC Circuits Both switches are closed. What is the final charge on the capacitor after the switches have been closed a long time? 1) Q = 0 2) Q = C E /3 3) Q = C E /2 4) Q = C E R 2R C  S 2 S 1 I R + - + + - - + -

RC Circuits: Charging Loop:  – I(t)R – q(t) / C = 0 Just after…: q =q 0 Capacitor is uncharged  – I 0 R = 0  I 0 =  / R Long time after: I c = 0 Capacitor is fully charged  – q  /C =0  q  =  C Intermediate (more complex) q(t) = q  (1-e -t/RC ) I(t) = I 0 e -t/RC C  R S 1 S 2 + + + I - - - The switches are originally open and the capacitor is uncharged. Then switch S 1 is closed. t q RC 2RC 0 q 

RC Circuits: Discharging Loop: q(t) / C + I(t) R = 0 Just after…: q=q 0 Capacitor is still fully charged q 0 / C + I 0 R = 0  I 0 = –q 0 / (RC) Long time after: I c =0 Capacitor is discharged (like a wire) q  / C = 0  q  = 0 Intermediate (more complex) q(t) = q 0 e -t/RC I c (t) = I 0 e -t/RC C  R S 1 - + + I - + - S 2 q RC 2RC t

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.

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

Summary of Concepts Charge (and therefore voltage) on Capacitors cannot change instantly: remember V C = Q/C 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. Long term behavior of Capacitor: Current through a Capacitor eventually goes to zero. 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. Intermediate behavior: Charge and current exponentially approach their long-term values  = RC

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: Example E – I 0 R – q 0 /C = 0 + + + - - - E – I 0 R – 0 = 0 I 0 = E /R After a long time current through capacitor is zero! E – IR – q ∞ /C = 0 E – 0 – q ∞ /C = 0 q ∞ = E C R C ε S 1 R=10  C=30 mF ε =20 Volts I

ACT: 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 R C ε 2R S 1 ε = 24 Volts R = 2  C = 15 mF

Charging: Intermediate Times  Calculate the charge on the capacitor 3  10 -3 seconds after switch 1 is closed. q(t) = q  (1-e -t/RC ) = q  (1-e - 3  10 -3 /(20  100  10 -6) ) ) = q  (0.78) Recall q  =  C = (50)(100x10 -6 ) (0.78) = 3.9 x10 -3 Coulombs R = 10  V = 50 Volts C = 100  F Example 2R C R S 2 S 1 I b + - + + - -

RC Summary Charging Discharging 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 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

Lect07 handout

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