RC Circuits C a b + - e R I a b e R C I RC 2RC Ce RC 2RC Ce q q t t

Today… Calculate Charging of Capacitor through a Resistor Calculate Discharging of Capacitor through a Resistor Text Reference: Chapter 26.6 Examples: 26.17,18 and 19

Last time--Behavior of Capacitors (from Lect. 10) 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 a wire.

The capacitor is initially uncharged, and the two switches are open. Preflight 11: E The capacitor is initially uncharged, and the two switches are open. 3) What is the voltage across the capacitor immediately after switch S1 is closed? a) Vc = 0 b) Vc = E c) Vc = 1/2 E 4) Find the voltage across the capacitor after the switch has been closed for a very long time. a) Vc = 0 b) Vc = E c) Vc = 1/2 E

Initially: Q = 0 VC = 0 I = E/(2R) After a long time: VC = E Q = E C I = 0

Preflight 11: E 6) After being closed a long time, switch 1 is opened and switch 2 is closed. What is the current through the right resistor immediately after the switch 2 is closed? a) IR= 0 b) IR=E/(3R) c) IR=E/(2R) d) IR=E/R

After C is fully charged, S1 is opened and S2 is closed. Now, the battery and the resistor 2R are disconnected from the circuit. So we now have a different circuit. Since C is fully charged, VC = E. Initially, C acts like a battery, and I = VC/R.

RC Circuits (Time-varying currents) b e R C I Charge capacitor: C initially uncharged; connect switch to a at t=0 Calculate current and charge as function of time. Would it matter where R is placed in the loop?? • Loop theorem Þ Convert to differential equation for Q: No! 

RC Circuits (Time-varying currents) b e R C I Charge capacitor: • Guess solution: Check that it is a solution: Note that this “guess” incorporates the boundary conditions:  !

RC Circuits (Time-varying currents) Charge capacitor: a b e R C I Current is found from differentiation: Þ Conclusion: Capacitor reaches its final charge(Q=Ce ) exponentially with time constant t = RC. Current decays from max (=e /R) with same time constant.

Charging Capacitor Current Charge on C Max = Ce 63% Max at t=RC Q t Current Max = e /R 37% Max at t=RC e /R I t

Lecture 11, ACT 1 (a) Q2 < 2Q1 (b) Q2 = 2Q1 (c) Q2 > 2Q1 I I At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. At time t=t1=t, the charge Q1 on the capacitor is (1-1/e) of its asymptotic charge Qf=Ce. What is the relation between Q1 and Q2 , the charge on the capacitor at time t=t2=2t ? R b e C R (a) Q2 < 2Q1 (b) Q2 = 2Q1 (c) Q2 > 2Q1 From the graph at the right, it is clear that the charge increase is not as fast as linear. In fact the rate of increase is just proportional to the current (dQ/dt) which decreases with time. Therefore, Q2 < 2Q1. t Q Q2 2Q1 Q1 2t The point of this ACT is to test your understanding of the exact time dependence of the charging of the capacitor. So the question is: how does this charge increase differ from a linear increase? Charge increases according to:

RC Circuits (Time-varying currents) Discharge capacitor: C initially charged with Q=Ce Connect switch to b at t=0. Calculate current and charge as function of time. C a b + - e R I Loop theorem  • Convert to differential equation for Q: 

RC Circuits (Time-varying currents) b + - e R I Discharge capacitor: • Guess solution: Q = C e -t/RC • Check that it is a solution: Note that this “guess” incorporates the boundary conditions: Þ ! R dQ dt Q C e t RC + = - /

RC Circuits (Time-varying currents) b + - e R I Discharge capacitor: a Q = C e -t/RC Þ Current is found from differentiation: Conclusion: Capacitor discharges exponentially with time constant t = RC Current decays from initial max value (= -e/R) with same time constant Minus sign: original definition of current “I” direction

Discharging Capacitor RC 2RC Ce Charge on C Max = Ce 37% Max at t=RC Q = C e -t/RC Q zero t -e /R I Current Max = -e/R 37% Max at t=RC t

8) Compare the charge on the two capacitors a short time after t = 0 Preflight 11: The two circuits shown below contain identical fully charged capacitors at t=0. Circuit 2 has twice as much resistance as circuit 1. 8) Compare the charge on the two capacitors a short time after t = 0 a) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q2

Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: t1 = RC and t2 = 2RC. Since t2 > t1 it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor is bigger than that on capacitor 1.

Lecture 11, ACT 2 (a) (b) (c) R 2R C e At t=0 the switch is connected to position a in the circuit shown: The capacitor is initially uncharged. At t = t0, the switch is thrown from position a to position b. Which of the following graphs best represents the time dependence of the charge on C? a b R 2R C e (a) (b) (c) For 0 < t < t0, the capacitor is charging with time constant t = RC For t > t0, the capacitor is discharging with time constant t = 2RC (a) has equal charging and discharging time constants (b) has a larger discharging t than a charging t (c) has a smaller discharging t than a charging t

Charging Discharging e Q = C Q Q t t I I t t RC 2RC RC 2RC Ce Ce -t/RC t t -e /R I e /R I t t

A very interesting RC circuit First consider the short and long term behavior of this circuit. Short term behavior: Initially the capacitor acts like an ideal wire. Hence, and Long term behavior: Exercise for the student!!

a) I1 = 0 b) I1 = E/R1 c) I1 = E/(R1+ R2) Preflight 11: The circuit below contains a battery, a switch, a capacitor and two resistors 10) Find the current through R1 after the switch has been closed for a long time. a) I1 = 0 b) I1 = E/R1 c) I1 = E/(R1+ R2)

After the switch is closed for a long time ….. The capacitor will be fully charged, and I3 = 0. (The capacitor acts like an open switch). So, I1 = I2, and we have a one-loop circuit with two resistors in series, hence I1 = E/(R1+R2) What is voltage across C after a long time? C is in parallel with R2 !! VC = I1R2 = E R2/(R1+R2) < E

Very interesting RC circuit continued Loop 1 Loop 2 Node: Loop 1: Loop 2: Eliminate I1 in L1 and L2 using Node equation: Loop 1: Loop 2: eliminate I2 from this Final differential eqn:

Very interesting RC circuit continued Loop 1 Loop 2 Final differential eqn: time constant: t parallel combination of R1 and R2 Try solution of the form: and plug into ODE to get parameters A and τ Obtain results that agree with initial and final conditions:

Very interesting RC circuit continued Loop 1 Loop 2 What about discharging? Open the switch... Loop 1 and Loop 2 do not exist! I2 is only current only one loop start at x marks the spot... I2 e R2 C R1 Different time constant for discharging

Next time: Start Magnetism Summary Kirchoff’s Laws apply to time dependent circuits they give differential equations! Exponential solutions from form of differential equation time constant t = RC what R, what C?? You must analyze the problem! series RC charging solution series RC discharging solution Q = C e e -t/RC Next time: Start Magnetism Reading assignment: Ch. 28.1-2, 28.4 Examples: 28.1,4,5 and 6