Chapter 32B - RC Circuits
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
RC Circuits: The rise and decay
of currents in capacitive circuits
Optional: Check with Instructor
The calculus is used only for derivation of
equations for predicting the rise and decay
of charge on a capacitor in series with a
single resistance. Applications are not
calculus based.
Check with your instructor to see if this
module is required for your course.
RC Circuit
RC-Circuit: Resistance R and capacitance C
in series with a source of emf V.
a
R
a
R
- -
C
V
b
i
+
+
+
+
V
b
- -
q
C
C
Start charging capacitor. . . loop rule gives:
q
 E   iR; V  C  iR
RC Circuit: Charging Capacitor
R
a
i
+
+
V
b
- -
q
C
C
q
V   iR
C
dq
q
R
V 
dt
C
Rearrange terms to place in differential form:
Multiply by C dt :
dq
dt

(CV  q) RC
RCdq  (CV  q)dt
q
t dt
dq
0 (CV  q)  o RC
RC Circuit: Charging Capacitor
R
a
i
+
+
V
b
- -
q
C
C
t dt
dq

0 (CV  q) o RC
t
q
 ln(CV  q) 0 
RC
q
t
ln(CV  q)  ln(CV ) 
RC
CV  q  CVe
 (1/ RC ) t
(CV  q) t
ln

CV
RC
q  CV 1  e
 t / RC

RC Circuit: Charging Capacitor
R
a
i
+
+
V
b
- -
q
C
C
Instantaneous charge q
on a charging capacitor:
q  CV 1  e
 t / RC
At time t = 0: q = CV(1 - 1); q = 0
At time t = : q = CV(1 - 0); qmax = CV
The charge q rises from zero initially to
its maximum value qmax = CV

Example 1. What is the charge on a 4-mF
capacitor charged by 12-V for a time t = RC?
Qmax
q
a R = 1400 W
Capacitor
Rise in
Charge
t
V
b
i
+
+
0.63 Q
- -
4 mF
Time, t
The time t = RC is known
as the time constant.
q  CV 1  e

 t / RC
q  CV 1  e
1


e = 2.718; e-1 = 0.63
q  CV 1  0.37
q  0.63CV
Example 1 (Cont.) What is the time constant t?
Qmax
q
Rise in
Charge
t
V
b
i
+
+
0.63 Q
a R = 1400 W
Capacitor
- -
4 mF
Time, t
The time t = RC is known
as the time constant.
t = (1400 W)(4 mF)
t = 5.60 ms
In one time constant
(5.60 ms in this
example), the charge
rises to 63% of its
maximum value (CV).
RC Circuit: Decay of Current
R
a
i
+
+
V
b
- -
q
C
C
As charge q rises, the
current i will decay.
q  CV 1  e
 t / RC
dq d
CV t / RC
 t / RC
i
  CV  CVe

e

dt dt
RC
Current decay as a
capacitor is charged:
V t / RC
i e
R

Current Decay
R
a
i
+
+
V
b
- -
q
C
C
I
i
Capacitor
Current
Decay
0.37 I
t
Time, t
Consider i when
t = 0 and t =  .
The current is a maximum
of I = V/R when t = 0.
V t / RC
i e
R
The current is zero when
t =  (because the back
emf from C is equal to V).
Example 2. What is the current i after one time
constant (t  RC)? Given R and C as before.
I
i
Capacitor
0.37 I
t
V
b
i
+
+
Current
Decay
a R = 1400 W
- -
4 mF
Time, t
The time t = RC is known
as the time constant.
V  t / RC V 1
i e
 e
R
C
e = 2.718; e-1 = 0.37
V
i  0.37  0.37imax
R
Charge and Current During the
Charging of a Capacitor.
Qmax
q
0.63 I
Capacitor
Rise in
Charge
t
Time, t
I
i
Capacitor
Current
Decay
0.37 I
t
Time, t
In a time t of one time constant, the charge q
rises to 63% of its maximum, while the current i
decays to 37% of its maximum value.
RC Circuit: Discharge
After C is fully charged, we turn switch to
b, allowing it to discharge.
a
R
a
R
- -
C
V
b
i
+
+
+
+
V
b
- -
q
C
C
Discharging capacitor. . . loop rule gives:
 E   iR;
Negative because
q
 iR of decreasing I.
C
Discharging From q0 to q:
R
a
i
dq
dt

;
q
RC
+
+
V
b
q
C
- -
C
Instantaneous charge q
on discharging capacitor:
dq
q   RCi; q   RC
dt
t dt
dq
q0 q  0 RC ;
q
t
ln q  ln q0 
RC
ln qq
q
0
t
 t 
 

RC

0
q
t
ln 
q0 RC
Discharging Capacitor
R
a
i
+
+
V
b
q
C
- -
C
q
t
ln 
q0 RC
q  q0e
 t / RC
Note qo = CV and the instantaneous current is: dq/dt.
dq d
CV t / RC
 t / RC
i
  CVe

e

dt dt
RC
V t / RC
Current i for a
i e
discharging capacitor.
C
Prob. 45. How many time constants are needed
for a capacitor to reach 99% of final charge?
a
R
q
i
+
+
V
b
C

q  qmax 1  e
C
- -
q
qmax
Let x = t/RC, then:
1
x

0.01;
e
 100
x
e
x = 4.61
 t / RC
 0.99  1  e

 t / RC
e-x = 1-0.99 or e-x = 0.01
From definition
of logarithm:
t
x
RC
ln e (100)  x
4.61 time
constants
Prob. 46. Find time constant, qmax, and time to
reach a charge of 16 mC if V = 12 V and C = 4 mF.

a 1.4 MW
V 12 V
1.8 mF
i
- -C
t = RC = (1.4 MW)(1.8 mF)
qmax = CV = (1.8 mF)(12 V);
q
qmax

+
+
bR
q  qmax 1  e
 t / RC
16m C
 t / RC

 1 e
21.6m C
t = 2.52 s
qmax = 21.6 mC
1 e
Continued . . .
t / RC
 0.741
Prob. 46. Find time constant, qmax, and time to
reach a charge of 16 mC if V = 12 V and C = 4 mF.
a 1.4 MW
V 12 V
1.8 mF
i
- -C
1
x

0.259;
e
 3.86
x
e
x = 1.35
 0.741
Let x = t/RC, then:
+
+
bR
1 e
t / RC
e
x
 1  0.741  0.259
From definition
of logarithm:
t
 1.35;
RC
Time to reach 16 mC:
lne (3.86)  x
t  (1.35)(2.52s)
t = 3.40 s
CONCLUSION: Chapter 32B
RC Circuits