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RC Circuits
I
a
I
a
I
I
R
C
e
RC
Ce1
R
b
b
+ +
C
e
2RC
RC
Ce
1 1
- -
2RC
Q
q  Cee  t / RC

f( x ) q
0.5
q  Ce 1  e
00
0
1
t
 t / RC

f( xq) 0.5
0.0183156 0
2
x
3
4
0
0
1
t
2
x
3
4
4
Kirchoff’s Laws with a capacitor
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
•Write KVL:
I
R
C
e
V
Q
C
Q
 IR   e  0
C
• Problem: We have two variables I and Q
•Consider that I 
dQ
dt
and substitute. Now eqn has only “Q”.
KVL gives a Differential Equation ! e  R dQ  Q  0
dt
C
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.
dQ Q
eR

0
dt
C
• Things are varying with time
What does the differential eqn mean?
I
dQ Q
eR
 0
dt C
I
R
+
+
-
e
C
-
•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
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
I0+ just after the switch is thrown?
(a) I0+ = 0
(b) I0+ = e /2R
a
I
I
R
b
C
e
(c) I0+ = 2e /R
R
Question 1
1. a
2. b
3. c
70%
19%
c
b
a
11%
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
I0+ just after the switch is thrown?
(a) I0+ = 0
(b) I0+ = e /2R
a
I
I
R
b
C
e
R
(c) I0+ = 2e /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+,
e  I 0 R  0  I 0 R  0
e
I 0 
2R
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
I
I
R
b
C
e
What is the value of the current
I after a very long time?
–
(a) I = 0
(b) I = e /2R
(c) I > 2e /R
R
Question 2
98%
1%
c
b
2%
a
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
I
I
R
b
C
e
What is the value of the current
I after a very long time?
–
(a) I = 0
(b) I = e /2R
R
(c) I > 2e /R
•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 e ; 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.
Question 3
In the circuit shown the capacitor is
initially uncharged, and the two
switches are open. The switch S1 is
closed and the voltage across the
capacitor immediately after is V1.
After a long time the voltage is V2. Is
a) V1 = 0 and V2 = E
b) V1 = 0 and V2= 0
c) V1 = 1/2 E and V2 = E
d) V1 = E and V2 = 0
e) V1 = E and V2 = E
E
Question 3
61%
29%
d
e
5%
2%
c
3%
b
a
b
c
d
e
a
1.
2.
3.
4.
5.
Question 3
In the circuit shown the capacitor is
initially uncharged, and the two
switches are open. The switch S1 is
closed and the voltage across the
capacitor immediately after is V1.
After a long time the voltage is V2. Is
a) V1 = 0 and V2 = E
b) V1 = 0 and V2= 0
c) V1 = 1/2 E and V2 = E
d) V1 = E and V2 = 0
e) V1 = E and V2 = E
E
Initially:
Q=0
V1 = 0
I = E/(2R)
After a long time:
V2 = E Q = E C I = 0
Question 4
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)
E
c) IR=E/(2R)
d) IR=E/R
Question 4
a
b
c
d
82%
d
6%
c
5%
b
8%
a
1.
2.
3.
4.
Question 4
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)
E
c) IR=E/(2R)
d) IR=E/R
•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, V2 = 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:
I
a
R
C initially uncharged;
connect switch to a at t=0
Calculate current and
charge as function of
time.
Q
• Loop theorem  e  IR   0
C
b
e

dQ Q
e R 
dt C
C
Would it matter where R
is placed in the loop??
•Convert to differential equation for Q:
dQ
I
dt
I
RC Circuits
(Time-varying currents)
• Guess solution:
Q  Ce (1  e
t
RC
I
a
I
R
b
)
•Check that it is a solution:
Differentiate it
dQ
1 

 Ce e  t / RC  

dt
 RC 
dQ
t
R
 e e RC
dt
•Add Q/C to each side
t
dQ Q
 t / RC
R
  ee
 e (1  e RC )  e
dt C
e
C
dQ Q
e R

dt C
Note that this “guess”
incorporates the
boundary conditions:
t 0Q0
t    Q  Ce
RC Circuits
(Time-varying currents)
• Charge on capacitor:
Q  Ce 1  e
 t / RC
I
a
I
R

b
C
e
• Current is found by
differentiation:
dQ e t / RC
I
 e
dt R

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
RC
Charge on C
Q
Q  Ce 1  et / RC 
Max = Ce
Ce
2RC
1
f( x ) 0.5
Q
63% Max at t=RC
00
0
I
Current
dQ e  t / RC
dt

Max = e /R
R
1
e 1/R1
2
t
3
4
t
3
4
x
t/RC
e
f( x ) 0.5
I
37% Max at t=RC
0.0183156 0
0
0
1
2
x
4
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.
I
a
b
e
Q
0
• Loop theorem  IR 
C
• Convert to differential equation for Q:
dQ 
I
dt
dQ Q
R
 0
dt C
I
R
+ +
C
- -
RC Circuits
(Time-varying currents)
• Discharge capacitor:
R
b
dQ Q
 0
dt C
Q  Ce e
R
+ +
C
- -
RC
• Check that it is a solution:
dQ
1 

 Ce e  t / RC  

dt
 RC 
dQ
t
R
 e e RC
dt
I
e
• Guess solution:
t
I
a
Note that this “guess” incorporates
the boundary conditions:
t  0  Q  Ce
t Q0
dQ Q
t
t
RC
R

 e e
 e e RC  0
dt
C
RC Circuits
(Time-varying currents)
• Discharge capacitor:
Q  Ce e
t
dQ
e t / RC
I
 e
dt
R
b
RC
• Current is found by
differentiation:

Minus sign:
original definition
of current “I” direction
I
a
I
R
+ +
C
e
- -
Conclusion:
• Capacitor discharges
exponentially with time constant
t = RC
• Current decays from initial max
value (= -e/R) with same time
constant
Discharging Capacitor
Ce1
Charge on C
RC
2RC
1
2
1
Q = C e e -t/RC
Max = Ce
f( x ) 0.5
Q
37% Max at t=RC
0.0183156
0
zero
0
01 0
t
3
x
4
4
dQ
e
I
  e t / RC
dt
R
Q
Current
I
f( x ) 0.5
Max = -e/R
37% Max at t=RC
-e /R0
0
1
2
x
t/RC
t
3
4
Charging
RC
Ce1
2RC
Q  Ce 1  et / RC 
2RC
Q = C e e -t/RC
f( x ) Q
0.5
0.0183156 0
00
0
1
1e /R1
2
t
3
0
1
dQ e t / RC
 e
dt R
2
1
01 0
Q
( x ) 0.5
I
0
4
x
t/RC
I
0
RC
Ce 1
1
f( x ) Q
0.5
156
Discharging
t
2
4
4
dQ
e t / RC
I
 e
dt
R
I
f( x ) 0.5
4
3
x
-e /R0
3
t
0
1
2
t
3
4
Summary
• Kirchoff’s Laws apply to time dependent circuits
they give differential equations!
• Exponential solutions
– from form of differential equation
• time constant t = RC
• series RC charging solution
• series RC discharging solution

Q  Ce 1  e
t
Q  Ce e
• Try Fishbane Chapter 27 #57, 59, 78
RC
t

RC