a b  R C I I t q RC 2 RC 0 CC C a b + --  R + I I RC Circuits q RC2RC 0 t CC

Kirchoff’s Laws with a capacitor  R C I I Up to now have only considered circuits with batteries and resistors Let’s try adding a Capacitor to our simple circuit Remember Voltage “drop” on C KVL gives a Differential Equation ! Write KVL: Problem: We have two variables I and Q Consider that and substitute. Now eqn has only “ Q ”.

What is different with Capacitors? Previously: –Analysed multi-loop circuits with batteries and resistors. –Currents are attained essentially instantaneously and do not vary with time!! Now: –What changes when we add a capacitor to the circuit? KVL yields a differential equation with a term proportional to Q and a term proportional to I = dQ/dt. Things are varying with time

What does the differential eq n mean? Physically, what’s happening is that the final charge cannot be placed on a capacitor instantly. Initially, the voltage drop across an uncharged capacitor = 0 because the charge on it is zero ! (V=Q/C) As current starts to flow, charge builds up on the capacitor, the voltage drop is proportional to this charge and increases; it then becomes more difficult to add more charge so the current slows When the voltage drop across the capacitor equals the emf of the battery the current stops  R C I I

Question 1 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ = e /2R (c) I 0+ = 2e /R  a b R C I I R

Question 1 1.a 2.b 3.c

Question 1 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ = e /2R (c) I 0+ = 2e /R  a b R C I I R Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0! Applying KVL to the loop at t=0+,

Question 2 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged.  a b R C I I R (a) I  = 0 (b) I  =  /2 R (c) I  > 2  / R – What is the value of the current I  after a very long time?

Question 2 1.a 2.b 3.c

Question 2 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged.  a b R C I I R (a) I  = 0 (b) I  =  /2 R (c) I  > 2  / R – What is the value of the current I  after a very long time? As the current flows, the charge on the capacitor grows. As the charge on the capacitor grows, the voltage across the capacitor will increase. The voltage across the capacitor cannot be bigger than  ; when it reaches that the current goes to 0.

Behavior of Capacitors 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.

a) V 1 = 0 and V 2 = E b) V 1 = 0 and V 2 = 0 c) V 1 = 1/2 E and V 2 = E d) V 1 = E and V 2 = 0 e) V 1 = E and V 2 = E E In the circuit shown the capacitor is initially uncharged, and the two switches are open. The switch S 1 is closed and the voltage across the capacitor immediately after is V 1. After a long time the voltage is V 2. Is Question 3

1.a 2.b 3.c 4.d 5.e

a) V 1 = 0 and V 2 = E b) V 1 = 0 and V 2 = 0 c) V 1 = 1/2 E and V 2 = E d) V 1 = E and V 2 = 0 e) V 1 = E and V 2 = E E In the circuit shown the capacitor is initially uncharged, and the two switches are open. The switch S 1 is closed and the voltage across the capacitor immediately after is V 1. After a long time the voltage is V 2. Is Question 3 Initially: Q = 0 V 1 = 0 I = E/(2R) After a long time: V 2 = E Q = E C I = 0

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? E a) I R = 0 b) I R =E/(3R) c) I R =E/(2R) d) I R =E/R Question 4

1.a 2.b 3.c 4.d

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? E a) I R = 0 b) I R =E/(3R) c) I R =E/(2R) d) I R =E/R Question 4 Now, the battery and the resistor 2R are disconnected from the circuit and the new circuit consists of the capacitor C and the resistor R. Since C is fully charged, V 2 = E. Initially, C acts like a battery, and I = E/R. Eventually after a long time all of the charge on C will have flowed from the positive to the negative plate and the current will be 0

RC Circuits (Time-varying currents) Charge capacitor: C initially uncharged; connect switch to a at t=0 Loop theorem  Convert to differential equation for Q : a b  R C I I Calculate current and charge as function of time.  Would it matter where R is placed in the loop??

RC Circuits (Time-varying currents) Guess solution: Check that it is a solution: Differentiate it Note that this “guess” incorporates the boundary conditions: a b  R C I I Add Q/C to each side

RC Circuits (Time-varying currents) Current is found by differentiation: Charge on capacitor: a b  R C I I  Conclusion: Capacitor reaches its final charge( Q = C  ) exponentially with time constant  = RC. Current decays from max (=  / R ) with same time constant.

Charging Capacitor Q 0 CC Charge on C Max = C  63% Max at t= RC t RC 2 RC I 0 t  R Current Max =  / R 37% Max at t= RC

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

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

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

Discharging Capacitor Charge on C Max = C  37% Max at t= RC Q = C  e -t/RC Current Max = -  /R 37% Max at t=RC t Q 0 CC RC 2 RC 0 -  / R I t zero

Charging Discharging RC 2RC t t Q 0 CC I 0  / R RC t 2RC 0 -  / R I t Q 0 CC Q = C  e -t/RC

Summary Kirchoff’s Laws apply to time dependent circuits they give differential equations! Exponential solutions –from form of differential equation time constant  = RC series RC charging solution series RC discharging solution Try Fishbane Chapter 27 #57, 59, 78