Lecture 11 – 1/18 Capacitor in RC Circuits

I  I charging discharging

• Switch closed at t=0 • Switch closed at t=0 • Initial • Initial ▪ Capacitor (C) initially ▪ Capacitor (C) initially charged uncharged ▪ V = Q /C across C ▪ 0 0 V0 = 0 across C ▪ I = V /R ▪ I = Emf/R 0 0 0 • Final (after a long time) • Final (after a long time) ▪ Capacitor (C) fully discharged ▪ Capacitor (C) fully charged ▪ V∞ = 0 across C ▪ V∞ = Emf across C ▪ I∞ = 0 ▪ I∞ = 0 Lecture 11 – 2/18 Discharging a Capacitor in RC Circuits 1. Switch closed at t=0.

Initially C is fully charged with Q0 Q 2. Loop Rule:  I R  0 C 3. Convert to a differential equation dQ Q dQ I I     R  0 dt C dt 4. Solve it! t dQ dt    RC  Q  RC  Q  Q0 e t   Q  t dQ Q0 RC ln     I    e RC is the time   dt RC constant (τ)  Q0  RC Lecture 11 – 3/18 Charge Q(t) during Discharging • Charge on the capacitor as a function of time: t / RC t / Q(t)  Q0 e  Q0 e

units of the time ▪ Q0 is the initial charge constant (τ) are seconds C C [ ] [RC]     s time constant V A

Charge after time τ 0.37 Q0 is given by:

1 Q( )  Q0 e  0.37Q0 Lecture 11 – 4/18 Current I(t) during Discharging

t / Starting with Q(t)  Q0 e • Current from the capacitor dQ as a function of time: I   dt t / I(t)  I0 e Q0 t / t / I  e  I0 e ▪  I0 is the initial current Q Q V I  0  0  0 0  RC R

Charge after time τ is given by:

1 I( )  I0 e  0.37 I0 Lecture 11 – 5/18 Charging a Capacitor in RC Circuits 1. Switch closed at t=0. Initially C is uncharged with V=0 Q 2. Loop Rule:    I R  0 C 3. Convert to a differential equation dQ Q dQ I       R  0 dt C dt 4. Solve it! t t        Q  C 1e RC   Q 1e RC    f       t t RC is the time dQ    I   e RC  I e RC constant (τ) dt R 0 Lecture 11 – 6/18 Charge Q(t) during Charging • Charge on the capacitor as a function of time: t / RC t / Q(t)  C  (1e )  Q f (1 e )

▪ Qf is the final charge Qf  CV

Charge after time τ is given by: time constant 1 Q( )  Qf (1e )

Q( )  0.63Qf Lecture 11 – 7/18 Current I(t) during Charging Q(t)  Q (1et / ) Starting with f • Current to the capacitor dQ as a function of time: I  dt t / I(t)  I0 e Qf t / t / I  e  I0 e ▪  I0 is the initial current Q Q  I  f  f  0  RC R

Charge after time τ is given by: 0.37 I0 1 I( )  I0 e  0.37 I0 Lecture 11 – 8/18 Capacitor Equations • Which equation is for charging and which is discharging?

t / t / Q(t)  Q0 e Q(t)  Q f (1 e ) Discharging Charging • Check the charge on the capacitor initially (at t=0) 0/ Q(0)  Q (1 e0/ ) Q(0)  Q0 e f 0 Q(0)  Q (1 e0 ) Q(0)  Q0 e f

Q(0)  Q0 (1) Q(0)  Q f (11)

Q(0)  Q0 Q(0)  0

Initially has charge Q0 Initially has zero charge Lecture 11 – 9/18 Quiz Question 1 Which curve best represents the charge on the capacitor in a charging RC circuit as a function of time? Lecture 11 – 10/18 Quiz Question 2 What’s the time constant in the following circuit. All resistors have the same resistance R, and all capacitors have the same capacitance C. R R a) (3/5)RC b) 2RC R c) RC C C d) (4/9)RC V e) (6/4)RC S C Lecture 11 – 11/18 Behavior of Capacitors in Circuits • Charging ▪ Initially, the capacitor behaves like a wire

▪ After a long time, the capacitor behaves like an open switch

• Discharging ▪ Initially, the capacitor behaves like a battery

▪ After a long time, the capacitor behaves like an open switch Lecture 11 – 12/18 Example of a More Complex Circuit

R2 • Switch S is open for a long I2 time and then closed. I1 I R1 3 Junction Rule: I  I  I R3 1 2 3 C •What is the initial current? V (when S is just closed) R S eq12 Q Loop Rule: V  I R   0 0 eq12 C V R3 C V  I0 (R1  R2 ) 0  0 V  I0  I3  0  I1  I2  I0 R1  R2 Lecture 11 – 13/18 Example of a More Complex Circuit

R2 • Switch S is open for a long I2 time and then closed. I1 I R1 3 Junction Rule: I  I  I R3 1 2 3 C •What is the final current? V (when S closed for long time) S Req12

Loop Rule: V  I f Req12  I f R3  0

V R3 C V  I f (R1  R2  R3 )  0 V  I f  I2  0  I1  I3  I f R1  R2  R3 Lecture 11 – 14/18 Example of a More Complex Circuit R2 • Switch S is closed for a long I2 time and then opened. I1 I R1 3 Junction Rule: I1  I2  I3 R3 •What is the initial current? C V (when S is just opened) R eq12 S

Loop Rule: VC  I0, S opened R3  0 V R3 C I f R3  I0, S opened R3  0 V  I0, S opened  I f  I1  0  I2  I3  I0, S opened R1  R2  R3 • What is the final current?  I  0 (when S is open f , S opened for long time) Lecture 11 – 15/18 Quiz Question 3 A RC circuit consists of a battery with potential V = 100 V, a resistor with R = 10 Ω, and a capacitor with C = 14 μF R

V C

S

A) I0 = 10 A If = 0 A D) I0 = 25 A If = 3 A

B) I0 = 15 A If = 2 A E) I0 = 3 A If = 25 A

C) I0 = 0 A If = 10 A Lecture 11 – 16/18 Quiz Question 4 The switch S is initially at position a for a long time. It is then switched to position b. Describe what happens to the light bulb as a function of time after the switch is a bulb flipped from a to b. S  b V  C

a) The light bulb goes on but goes off immediately. b) The light bulb goes off and stays off. c) The light bulb goes on but its brightness decreases with time and eventually goes off. d) The light bulb goes on and stays on at a constant brightness. e) The light bulb goes on but its brightness increases with time. Lecture 11 – 17/18 Energy Conservation in Discharging a Capacitor • Discharging: Energy lost by capacitor (C) 1 1 U  Q V  CV 2 2 0 0 2 0 Power dissipated by resistor (R) dW P   I(t)2 R  I 2e2t / R R dt 0 Total Energy dissipated by resistor (R)   W  P (t) dt  I 2 R e2t / dt R  R 0  Energy Conservation 0 0  U  W 2 2 R 2  V0  R C 1 2 WR  I0 R     CV0 2  R  2 2 Lecture 11 – 18/18 Energy Conservation in Charging a Capacitor

• Charging: Energy stored in capacitor (C) Work done by 1 1 U  Q V  CV 2 battery (V) 2 f 2 2 W  Q f V  CV Power dissipated by resistor (R) dW P   I(t)2 R  I 2e2t / R R dt 0 Total Energy dissipated by resistor (R)   W  P (t) dt  I 2 R e2t / dt R  R 0  0 0 2 2 2  V  R C 1 2 WR  I0 R     CV Independent of R 2  R  2 2