Download Lecture 9

INC 112 Basic Circuit Analysis
Week 8
RC Circuits
RC Circuits
i(t)
+
u(t)
AC
R
C
-
The response of RC circuits can be categorized into two parts:
• Transient Response
• Forced Response
Transient response comes from the dynamic of R,C.
Forced response comes from the voltage source.
Source-Free RC Circuits
i(t)
+
v
-
+
R
v
-
Find i(t) from R, C
C
Capacitor has some energy stored so that
The initial voltage at t=0 is V0
Initial condition
dv (t ) v(t )
C

dt
R
dv (t ) v (t )

0
dt
RC
v(0)  V0
Compare with the solution of RL circuits.
The solution of RC circuits can be obtained with the same method.
Source-free RL
di (t ) R
 i (t )  0
dt
L
i (t )  I 0 e
R
 t
L
Source-free RC
dv(t ) 1

v(t )  0
dt
RC
v(t )  V0e

1
t
RC
v(t)
V0
v(t )  V0e

1
t
RC
t
i(t)
1
v(t ) V0  RC t
i (t ) 
 e
R
R
V0/R
or
1
dv(t ) V0  RC t
i(t )  C
 e
dt
R
t
Time Constant
The product RC is time constant for RC circuits
  RC
v(t )  V0e

1
t
RC
Unit: second
 V0e

t

Forced RC Circuits
t=0
R
i(t)
V
from
+ vR(t) +
vc(t)
-
dvc (t )
ic (t )  C
dt
Use KVL, we got
C has an initial voltage of 0
C
1
vc (t )   ic (t )dt
C
V  vR (t )  vc (t )
1
V  i(t ) R   ic (t )dt
C
V  i (t ) R 
From
Differentiate both sides
1
ic (t )dt

C
di (t ) 1
 i (t )
dt
C
di (t ) 1

i (t )  0
dt
RC
0R
Solve first-order differential equation
i(t )  I 0e

1
t
RC
Where I0 is the initial current of the circuit
vc (0)  0
C has an initial voltage = 0,
But from KVL,
therefore,
So,
V  vR (t )  vc (t )
vR (0)  V
V
i (t )  e
R

and
i (0) 
V
R
1
t
RC
vR (t )  i(t ) R  Ve

1
t
RC
Natural Response
vC (t )  V  vR (t )  V  Ve

1
t
RC
 V (1  e
Force Response

1
t
RC
)
vC(t)
i(t)
V/R
1
vC (t )  V (1  e

1
t
RC
V
V  RC t
i (t )  e
R
t
t
Note: Capacitor’s voltage
cannot abruptly change
vR(t)
V
vR (t )  Ve

)
1
t
RC
t
How to Solve Problems? (RC)
• Start by finding the voltage of the capacitor first
• Assume the response that we want to find is in form of
k1  k2e

t

• Find the time constant τ (may use Thevenin’s)
• Solve for k1, k2 using initial conditions and
status at the stable point
• From the voltage, find other values that the problem ask
using KCL, KVL
Example
t=0
1M
1M
i(t)
5V
+
vc(t)
-
1uF
1V
Switch open for a long time before t=0, find and sketch i(t)
First, we start by finding vc(t)
The initial condition of C is vc(0) = 1V
The stable condition of C is vc(∞) = 3V
Assume vc(t) in form of
vc (t )  k1  k2e

t

Find the time constant after t=0 by Thevenin’s, viewing C as a load
Req  1M || 1M  500 K
Therefore, the time constant is
  RC  500 K 1F  0.5 sec
Find k1, k2 using vc(0) = 1, vc(∞) = 3
At t=0, vc(0) = 1 V
1  k1  k2
At t = ∞, vc(∞) = 3 V
3  k1  0
Therefore, k1=3, k2 = -2
vC (t )  k1  k2e
vC (t )  3  2e 2t
We can find i(t) by using Ohm’s law on the resistor
5  vc (t ) 5  (3  2e 2t )
i (t ) 

1M
1M
2  2e  2t

A  2  2e  2t A
1M

t

i(t)
4A
i(t )  2  2e 2t A
2A
t
Example
i(t)
1K
1K
t=0
5V
1K
+
vc(t)
-
3uF
The switch was opened for a long time before t=0, Find i(t)
Start with vc(t)
The initial condition of C is vc(0) = 5V
The final stable condition of C comes from voltage divider,
which is vc(∞) = 5*(1/1+0.5) = 3.33V
Assume vc(t) in form of
vc (t )  k1  k2e

t

Find the time constant after t=0 by Thevenin’s, viewing C as a load
Req  1K || 1K || 1K  333.33
Therefore, the time constant is
  RC  333.33  3F  1m sec
Find k1, k2 using vc(0) = 5, vc(∞) = 3.33
At t=0, vc(0) = 5 V
At t = ∞, vc(∞) = 3.33 V
vC (t )  k1  k2e

t

5  k1  k2
3.33  k1  0
Therefore, k1=3.33, k2 = 1.66
vC (t )  3.33  1.66e 1000t
We can find i(t) by using Ohm’s law on the resistor
5  vc (t ) 5  (3.33  1.66e 1000t )
i (t ) 

1K
1K
1.66  1.66e 1000t

A  1.66  1.66e 1000t mA
1K
i(t)
i(t )  1.66  1.66e 1000t mA
1.66mA
t
vc(t)
5V
vC (t )  3.33  1.66e 1000t
3.33V
t