I. I. I. a. I. a. R. R. b. b. +. +. C. C. e. . . e. RC Circuits. RC. 2 RC. C e. RC. 2 RC. C e. q. q. 0. 0. t. t. Text Reference: Chapter 26.6 . Examples: 26.17,18 and 19. Today…. Calculate Charging of Capacitor through a Resistor
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I
I
I
a
I
a
R
R
b
b
+
+
C
C
e


e
RC Circuits
RC
2RC
Ce
RC
2RC
Ce
q
q
0
0
t
t
Text Reference: Chapter 26.6
Examples: 26.17,18 and 19
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 = EQ = E CI = 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.
I
I
a
R
b
C
e
Would it matter where R is placed in the loop??
Calculate current and
charge as function of time.
No!
I
I
a
R
b
C
e
Note that this “guess” incorporates the boundary conditions:
!
I
I
a
R
b
C
e
Þ
Charge on C
Max = Ce
63% Max at t=RC
Current
Max =e /R
37% Max at t=RC
RC
2RC
Ce
Q
0
t
e /R
I
0
t
(c) Q2 > 2Q1
(b) Q2 = 2Q1
(a) Q2 < 2Q1
2Q1
Q2
Q1
Q
t
2t
a
I
I
R
b
e
C
R
I
I
a
R
b
+
+
C
e


• Convert to differential equation forQ:
I
I
a
R
b
+
+
C
• Guess solution:
e


e
t/RC
Q = C
e
Note that this “guess” incorporates the boundary conditions:
Þ
!
dQ
Q


t
/
RC
t
/
RC
e
e
+
=

e
+
e
=
R
0
dt
C
• Check that it is a solution:
I
I
a
t/RC
R
e
e
Q = C
b
+
+
C
e


Þ
Minus sign:
original definition
of current “I” direction
e
t/RC
Q = C
e
zero
0
Current
Max= e/R
37%Max att=RC
I
e /R
RC
2RC
Ce
Charge on C
Max = Ce
37% Max at t=RC
Q
0
t
t
Preflight 11:
The two circuits shown below contain identical fully charged capacitors att=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 > t1it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor is bigger than that on capacitor 1.
(c)
(b)
(a)
a
b
R
2R
C
e
e
t/RC
Q = C
e
0
I
e /R
RC
RC
2RC
2RC
Ce
Ce
Q
Q
0
0
t
t
e /R
I
0
t
t
I1
I2
I3
e
C
R2
R1
First consider the short and long term behavior of this circuit.
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 oneloop circuit with two resistors in series,
hence I1 = E/(R1+R2)
What is voltage across C after a long time?
Cis in parallel withR2 !!
VC = I1R2 = E R2/(R1+R2) < E
Loop 2
I1
I2
I3
Loop 1
e
C
R2
R1
eliminate I2 from this
Loop 2
I1
I2
I3
Loop 1
e
C
R2
R1
time constant: t
parallel combination
of R1 and R2
Loop 2
I1
I2
I3
Loop 1
e
C
R2
R1
I2
e
C
R2
R1
Different time constant for discharging
Reading assignment: Ch. 28.12, 28.4
Examples: 28.1,4,5 and 6
e
e
t/RC
Q = C
Next time: Start Magnetism