Engineering Circuit Analysis
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
3.2 First-Order RL Circuits
3.3 Examples
References: Hayt-Ch5, 6; Gao-Ch5;
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
Key Words:
Transient Response of RC Circuits, Time constant
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
• Used for filtering signal by blocking certain frequencies and passing
others. e.g. low-pass filter
• Any circuit with a single energy storage element, an arbitrary number
of sources and an arbitrary number of resistors is a circuit of order 1.
• Any voltage or current in such a circuit is the solution to a 1st order
differential equation.
Ideal Linear Capacitor
i(t ) =
dv
dq
c
dt
dt
vc (t +) =v C (t )
Energy stored w   pdt   cvdv 
1 2
cv
2
A capacitor is an energy storage device  memory device.
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
+
+
vs(t)
vr(t)
-
R
+
vc(t)
C
-
-
• One capacitor and one resistor
• The source and resistor may be equivalent to a circuit with many
resistors and sources.
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
Transient Response of RC Circuits
ic 
Switch is thrown to 1
KVL around the loop: ic R  vC  E
R
C
C


E
2

1
E  vc
R
dvc
R  vc  E
dt
vC  Ae
K

t
RC
E
Initial condition vC (0) v C (0)  0
vC  E (1  e

t
RC

A  E
t
)  E (1  e  )
t
dvc E 
 e
ic  C
dt R
  RC
Called time constant
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
Time Constant   RC
5
R
C


10V

E
vC  E (1  e
dvc
dt
2

1
K
t 0

E


t

dvc E t /
 e
dt

)
 
E
dvc
dt
t 0
R=2k
5V
C=0.1F
SEL>>
0V
RC
0s
V(2)
1ms
2ms
3ms
4ms
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
Transient Response of RC Circuits
vc  ic R  0
Switch to 2
R
C


E
dvc
dt
dv
vc  RC c  0
dt
ic  C
2

1
K
vc  Ae

t
RC
Initial condition vC (0) v C (0)  E
vc  Ee  t / RC  Ee  t / 
ic  
E t / 
e
R
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
Time Constant   RC
vC (t )  Ee

t
RC
 Ee

t

5
R
C


IS 


10V
E
dvC
dt
2

1
K

t 0
 
E

E
dvC
dt
t 0
R=2k
5V
C=0.1F
SEL>>
0V
0s
1.0ms
2.0ms
3.0ms
4.0ms
V(2)
Time
Ch3 Basic RL and RC Circuits
3.1 First-Order RC Circuits
6.0V
4.0V
2.0V
0V
0s
V(2)
0.5ms
V(1)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
Time
3.5ms
4.0ms
4.5ms
5.0ms
5.5ms
6.0ms
Ch3 Basic RL and RC Circuits
3.2 First-Order RL Circuits
Key Words:
Transient Response of RL Circuits, Time constant
Ch3 Basic RL and RC Circuits
3.2 First-Order RL Circuits
Ideal Linear Inductor
The
rest
of
the
circuit
i(t)
+
L
v(t)
-
d
di (t )
v(t ) 
L
dt
dt
di
P  iv  Li
dt
t
1
i (t )   v( x)dx
L 
iL (t ) i L (t )
1
2
Energy stored: wL (t )   pdt   Lidi  Li 2 (t )
• One inductor and one resistor
• The source and resistor may be equivalent to a circuit with many
resistors and sources.
Ch3 Basic RL and RC Circuits
3.2 First-Order RL Circuits
Transient Response of RL Circuits
di
dt
KVL around the loop: iR  vL  E
vL  L
Switch to 1
R
L


E
EL
2

1
di
 iR
dt
Initial condition t  0, i(0 )  i(0 )  0


K
R
 t
E
E
L
 i  (1  e )  (1  e t / )
R
R
 vR  iR  E (1  e t / )
  L/ R
Called time constant
R
R
 t 
 t
di
d E 
E
R
vL  L  L   1  e L   L    e L  Ee t /
dt
dt  R 
R L

Ch3 Basic RL and RC Circuits
3.2 First-Order RL Circuits
Time constant
. i (t)

0
t
• Indicate how fast i (t) will drop to zero.
• It is the amount of time for i (t) to drop to zero if it is dropping at
the initial rate dit
.
dt
t 0
Ch3 Basic RL and RC Circuits
3.2 First-Order RL Circuits
Transient Response of RL Circuits
Switch to 2
R
L


E
di
L  iR  0
dt
di
R
  dt
i
L
2

1
K
 i  Ae
R
 t
L
i(t )
R
ln
 t
I0
L
Initial condition
t : 0  t
i : I 0  i  t 
it 
t
1
R
I0 idi  0  L dt 
R
it 

ln i I0    t  t0
L
i(t )  I 0
t  0, I 0 
E
R
E  RL t E t /
i e  e
R
R
R
 t
e L
Ch3 Basic RL and RC Circuits
3.2 First-Order RL Circuits
SEL>>
Transient Response of RL Circuits
Input energy to L
4.0mA
R
L
2.0mA


E
2

1
K
0A
0s
1ms
2ms
3ms
4ms
I(L1)
L export its energy , dissipated by R
4.0mA
2.0mA
SEL>>
0A
0s
1ms
2ms
3ms
4ms
Ch3 Basic RL and RC Circuits
Summary
Initial Value
（t = 0）
RL
Circuits
RC
Circuits
Source
(0 state)
i0  0
Sourcefree
(0 input)
i0 
Source
(0 state)
Sourcefree
(0 input)
Steady Value
(t  )
time
constant

E
R
L/ R
i0
L/ R
v0  0
vE
RC
v0  E
v0
RC
E
R
iL 
Ch3 Basic RL and RC Circuits
Summary
The Time Constant
•
•
•
•
For an RC circuit,  = RC
For an RL circuit,  = L/R
-1/ is the initial slope of an exponential with an initial value of 1
Also,  is the amount of time necessary for an exponential to decay
to 36.7% of its initial value
Ch3 Basic RL and RC Circuits
Summary
• How to determine initial conditions for a transient circuit.
When a sudden change occurs, only two types of quantities
will remain the same as before the change.
– IL(t), inductor current
– Vc(t), capacitor voltage
• Find these two types of the values before the change and use
them as the initial conditions of the circuit after change.
Ch3 Basic RL and RC Circuits
3.3 Examples
About Calculation for The Initial Value
i
iC
vC  0   vC  0 
iL
t=0
 R1 / / R3   2
vC  0   8V 
i(0+)
iC(0+)
iL(0+)
iL  0   iL  0 
i  0 
+
+
vC(0+)=4V
_

2
 4V
2  2
8V
 2A
2  2
vL(0+)
iL  0   2A 
-
1A
4
 1A
4  4
Ch3 Basic RL and RC Circuits
3.3 Examples (Analyzing an RC circuit or RL circuit)
Method 1
1) Thévenin Equivalent.(Draw out C or L)
Simplify the circuit
Veq , Req
2) Find Leq(Ceq), and  = Leq/Req ( = CeqReq)
3) Substituting Leq(Ceq) and  to the previous solution of differential equation
for RC (RL) circuit .
Ch3 Basic RL and RC Circuits
3.3 Examples (Analyzing an RC circuit or RL circuit)
Method 2
1) KVL around the loop  the differential equation
2) Find the homogeneous solution.
3) Find the particular solution.
4) The total solution is the sum of the particular and homogeneous solutions.
Ch3 Basic RL and RC Circuits
3.3 Examples (Analyzing an RC circuit or RL circuit)
Method 3 (step-by-step)
In general,
f (t )  f ()  Ae

t

Given f(0+)，thus A = f(0+) – f(∞)
f (t )  f ()  [ f (0)  f ()]e
Initial

t

Steady
1) Draw the circuit for t = 0- and find v(0-) or i(0-)
2) Use the continuity of the capacitor voltage, or inductor current, draw
the circuit for t = 0+ to find v(0+) or i(0+)
3) Find v(), or i() at steady state
4) Find the time constant 
–
For an RC circuit,  = RC
–
For an RL circuit,  = L/R
5) The solution is:
f (t )  f ()  [ f (0)  f ()]e t /
US
I1
Ch3 Basic RL and RC Circuits
3.3
Examples
5
 
5

_
P3.1 vC (0)= 0, Find vC (t) for t  0.
I1
I1
5
I1
vc  t 
+
 v-Uc 1 
6k
R1
t=0
5
Method 3:
i1

+
   vc  0 I vc    e
t


i2

i3

R2 3k
E
9V
_
C=1000PF
pf
S
E
vc  0   0, vc     9V 
I1
3K
 3V
6K  3K
Apply Thevenin theorem :
RTh
1 
 1



 6K 3K 
1
 2K
  RTh C  2K  1000pF  2  106 s
vc  t   3  3e

t
2106
V
Ch3 Basic RL and RC Circuits
3.3 Examples5
 

_
I1
I1
5
C=1000PF
P3.2 vC (0)= 0, Find
5 vC (t) for t  0.
I1

5
t=0
vc  0  + 0
R1=10k
+
U1
-
E
+ vC
10K
IS 
vc     6V 
 4.615V
10K  3K
6V
v
_
R2 3k
R1=20k
I1
Apply
Thevenin’s
theorem :


1
1 
30
 1
RTh  

  K
13
 10K 3K 
30
  RThC  K 1000pF  2.31106 s
13
vc  t   4.615  4.615e

t
2.31106
V