RESPONSE OF
FIRST-ORDER RC
AND RL CIRCUITS
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7.1 The Natural Response of an RC Circuit
7.2 The Natural Response of an RL Circuit
7.3 Singularity Functions
7.4 The Step Response of RC and RL Circuit
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7.1 The Natural Response of an RC Circuit
Resistive Circuit
=> RC Circuit
algebraic equations => differential equations
Same Solution Methods (a) Nodal Analysis
(b) Mesh Analysis
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7.1 The Natural Response of an RC Circuit
The solution of a linear circuit, called dynamic
response, can be decomposed into
Natural Response + Forced Response
or in the form of
Steady Response + Transient Response
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7.1 The Natural Response of an RC Circuit
The natural response is due to the initial
condition of the storage component ( C or L).
nThe forced response is resulted from external
input ( or force).
nIn this chapter, a constant input (DC input) will
be considered and the forced response is called
step response.
n
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7.1 The Natural Response of an RC Circuit
Example 1 : Two forms of the first order circuit for
natural response
Find v(t)
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Find i(t)
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7.1 The Natural Response of an RC Circuit
Example 1 : (cont.) Four forms of the first order
circuit for step response
V
R
TH
TH
A capacitor connected to
a Thevenin equivalent
A capacitor connected to
a Norton equivalent
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7.1 The Natural Response of an RC Circuit
Example 1 : (cont.)
V
R
TH
TH
An inductor connected
to a Thevenin equivalent
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An inductor connected
to a Norton equivalent
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7.1 The Natural Response of an RC Circuit
Example 2
t≥0
nodal analysis
ic + iR = 0
C
dv 1
+ v=0
dt R
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7.1 The Natural Response of an RC Circuit
Example 2 (cont.)
characteristic root S,
1
CS + = 0
R
1
S=−
RC
∴ v (t ) = Ke
−
1
t
RC
,t ≥ 0
t≥0
v
ic
v
iR
R
From the initial condition
v (0 + ) = v (0 - ) = V0
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∴ v ( t ) = V0 e
−
t
RC
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7.1 The Natural Response of an RC Circuit
Example 2 (cont.)
τ = RC
time constant
−
t
τ
v(t) = V0 e , t ≥ 0
t =τ ,
v(t )
= 0.36788
V0
t = 3τ ,
4.979% < 5%
t = 5τ ,
0.674% < 1%
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7.1 The Natural Response of an RC Circuit
Example 2 (cont.)
It is customary to assume that the capacitor is fully
discharged after five time constants.
t
v(t ) V0 −τ
iR (t ) =
= e ,t ≥ 0
R
R
The power dissipated in R is
2t
V0 2 − τ
p (t ) = viR =
e
R
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7.1 The Natural Response of an RC Circuit
Example 2 (cont.)
The energy dissipated in R is
t
2t
−
1
τ
wR (t ) = ∫ pdt = CV0 2 (1 − e )
2
0
1
wR (∞) = CV0 2
2
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7.1 The Natural Response of an RC Circuit
Example 3 : Find vc , vx , and ix for t > 0
This circuit contains only one energy storage element.
Step 1. Use Thevenin theorem to find the equivalent RTH
looking into a-b terminals.
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7.1 The Natural Response of an RC Circuit
Example 3 : Find vc , vx , and ix for t > 0
Step 2. Find vc(t)
Step 3. Replace vc(t) as a voltage source in the original
circuit and solve the resistive circuit.
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7.1 The Natural Response of an RC Circuit
Step 1. RTH = (8 + 12) P 5 = 4Ω
Step 2. vc (0) = 15V
0.1
dvc vc
+ =0
dt 4
τ = RC = 0.4 s
vc (t) = 15e
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−2.5 t
V, t ≥ 0
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7.1 The Natural Response of an RC Circuit
Step 3.
By using voltage divider principle
vx (t ) =
12
vc (t )
8 + 12
= 9× e
ix (t ) =
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−2.5 t
V ,t ≥ 0
−2.5 t
vx (t )
= 0.75e A
12
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7.2 The Natural Response of an RL Circuit
Example 4
iL
Remember that vc(t) and iL(t) are continuous functions
for bounded inputs.
V
t ≤ 0 , i = S = i (0− ) @ I0
RS
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7.2 The Natural Response of an RL Circuit
t≥0
mesh analysis
v L + vR = 0
L
di
+ Ri = 0
dt
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7.2 The Natural Response of an RL Circuit
characteristic equation
LS + R = 0
R
L
∴S = −
i (t ) = Ke
R
− t
L
,t ≥ 0
i (0 + ) = i (0 − ) = I 0
−
t
τ
∴ i (t ) = I 0 e , t ≥ 0
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τ @
L
, time constant
R
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7.2 The Natural Response of an RL Circuit
vR (t ) = Ri = I0 Re
pR = vR i = I0 Re
2
wR (t ) = ∫
t
0
wR (∞) =
−
t
τ
−2 t
τ
−2 t
1
τ
pR (t )dt = LI0 2 (1 − e )
2
1
LI 0 2
2
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7.2 The Natural Response of an RL Circuit
Example 5 : i(0)=10A , find i(t) and ix(t)
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7.2 The Natural Response of an RL Circuit
Step 1. Find the Thevenin equivalent circuit looking
into a-b terminals.
Apply is , find e. Then RTH = e / is
i = −is
e − 3i e
+ = is
4
2
e 3
e
+ is + = is
4 4
2
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7.2 The Natural Response of an RL Circuit
3
1
e = is
4
4
e 1
∴ = Ω
is 3
mesh equation
Step 2.
di 1
+ i=0
dt 3
L 0.5
τ= =
= 1.5 S
R 1/ 3
0.5
1
3
2
- t
3
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∴ i (t)=10e A , t ≥ 0
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7.2 The Natural Response of an RL Circuit
Step 3 Replace the inductor with an equivalent
current source with i(t) and solve the
resistive circuit.
i (t) = 10 e
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-2
t
3
A
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7.2 The Natural Response of an RL Circuit
For this problem, since 2Ω is in parallel with the
inductor, it is trivial to get
vL (t) = L
di
dt
-2
= 0.5
t
d
(10 e 3 )
dt
-2
= 0.5 × 10 × (
-2 3 t
)e
3
-2
-10 3 t
e
V
3
-2
v
-5 3 t
∴ ix = L =
e
V, t≥0
2Ω
3
=
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7.2 The Natural Response of an RL Circuit
Example 6 : The switch has been closed for a long
time. At t=0, it is opened. Find ix .
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7.2 The Natural Response of an RL Circuit
t=0-
i1 (0- ) =
40V
=8A
2+(12//4)
ix (0- ) = 0
12
3
= 8× = 6 A
12+4
4
+
∴ Initial codition i (0 ) = i (0- ) = 6 A
i (0- ) = i1 (0- ) ×
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7.2 The Natural Response of an RL Circuit
t ≥ 0+
Step 1 RTH = (4+12) // 16 = 8 Ω
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7.2 The Natural Response of an RL Circuit
Step 2 mesh analysis
di
+ 8i = 0
dt
i (t) = K e-4t , t ≥ 0
2
i (0+ ) = 6 A
∴ i (t) = 6 e-4t A , t ≥ 0
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7.2 The Natural Response of an RL Circuit
Step 3 Replace L with an equivalent current source,
and find ix’ solve the resistive circuit.
4+12
i (t)
12+4+16
= 3e-4t A , t ≥ 0
∴ ix (t) =
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7.3 Singularity Functions
Switching functions are convenient for describing
the switching actions in circuit analysis.
They serve as good approximations to the switching
signals.
unit step function
unit impulse function
unit ramp function
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u(t)
δ(t)
r(t)
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7.3 Singularity Functions
Definition
Example 7
0 , t<0
u(t) = 
1 , t>0
undefined at t=0
or more general
0 , t<t 0
u(t-t 0 ) = 
1 , t>t 0
V0 u(t-t0) is applied
to a-b terminals
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7.3 Singularity Functions
Definition
d
δ (t) = u(t) =
dt
, t<0
0

undefined , t=0
0
, t>0

Also known as the delta function
or
or
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0+
0-
∫ δ (t) dt = 1
t 0+
t 0-
∫ δ (t-t 0 ) dt = 1
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7.3 Singularity Functions
The unit impulse function is not physically
realizable but is a very useful mathematical tool.
1
2d
1
d
An approximation as d → 0
Another approximation as d → 0
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7.3 Singularity Functions
Shifting property, if t 0 ∈ [a , b]
∫a f(t)δ (t-t 0 )dt = f(t 0 )
b
Q ∫a f(t)δ (t-t 0 )dt = ∫a f(t 0 )δ (t-t 0 )dt
b
b
= f(t 0 )∫a δ (t-t 0 )dt
b
= f(t 0 )
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7.3 Singularity Functions
Definition
t
r(t) = ∫-∞ u(t)dt = t u(t)
or
0 ,
r(t) = 
t ,
t≤0
t≥0
Note that
d u(t)
dt
d r(t)
u(t) =
dt
δ (t) =
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,
u(t) = ∫ δ (t)dt
,
r(t) = ∫ u(t)dt
t
t
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7.3 Singularity Functions
Example 8
(a)
v(t) = 10 [u(t-2) - u(t-5)]
(b )
dv(t)
= 10δ (t-2) - 10δ (t-5)
dt
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7.3 Singularity Functions
(c )
i(t) = 10 u(t) -20 u(t-2)+10u(t-4)
(d )
v(t) = 5r (t)-5r (t-2)-10u(t-2)
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7.4 The Step Response of RC and RL Circuits
When a dc voltage (current) source is suddenly
applied to a circuit , it can be modeled as a step
function , and the resulting response is called
step response .
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7.4 The Step Response of RC and RL Circuits
Example 9
V ( 0− ) = V0 , i = C
Step 1
dv
, choose v as unknown
dt
Mesh Analysis
dv
RC + v = VS , t ≥ 0
dt
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7.4 The Step Response of RC and RL Circuits
Step 2
Solving the differential equation
( a ) homogeneous solution
RC
dv h
+ vh = 0
dt
v h ( t ) = Ke
−
t
RC
, t≥0
( b ) particular solution
RC
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dv p
dt
v p = VS
+ v p = VS
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7.4 The Step Response of RC and RL Circuits
(c) co m p lete so lu tio n + In itial co n d itio n
v ( t ) = vh + v p
= Ke
−
t
RC
+ Vs
v (0 + ) = v (0 − ) = V0
∴ K = V0 − Vs
V 0 , t < 0

v (t ) = 
t
−
τ
V s + (V 0 − V s ) e , t ≥ 0
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7.4 The Step Response of RC and RL Circuits
Vs
R
i(t)
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dv
−1 − RCt
i(t ) = C = C(Vo − Vs )( )e
dt
RC
Vs − Vo − RCt
=
e
, t ≥0
R
V
For Vo = 0 , then i(o+ ) = S
R
C is initially short circuited.
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7.4 The Step Response of RC and RL Circuits
Example 10
Before t=0 , the circuit is
under steady state . At t=0 ,
the switch is moved to B .
Find v(t) , t > 0
For t<0 , the circuit shown:
v ( 0− ) =
5
× 24V = 15V
3+5
4kΩ
3kΩ
5kΩ
3kΩ
v(0− )
5kΩ
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7.4 The Step Response of RC and RL Circuits
For t>0
mesh
analysis
4kΩ
dv
4 × 10 3 × ( 0.5 × 10 −3 )
+ v = 30V
dt
v ( t ) = v p + vh
= 30 + Ke
−
t
τ
, t>0
τ = RC = 4 × 10 × 0.5 × 10 −3 = 2 second
3
v ( 0 + ) = v ( 0 − ) = 15V
∴ v ( t ) = 30 + (15 − 30 ) e
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−
t
τ
,t≥0
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7.4 The Step Response of RC and RL Circuits
Example 11
Before the switch is open at
t=0, the circuit is in steady
state.
Find i(t) & v(t) for all t.
For t<0 , then
10Ω
20Ω
1
F
4
10Ω
v ( t ) = 10V , t < 0
20Ω
10
i ( t ) = − A = −1A , t < 0
10
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7.4 The Step Response of RC and RL Circuits
For t>0
1
F
4
Thevenin equivalent circuit
20
Ω
3
1
F
4
v ( 0+ ) = v ( 0− ) = 10V
dv 1 dv
=
dt 4 dt
20 1 dv
×
+ v = 20V , t > 0
3 4 dt
ic ( t ) = C
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v ( t ) = 20 + Ke
3
− t
5
= 20 + (10 − 20 ) e
3
− t
5
,t > 0
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7.4 The Step Response of RC and RL Circuits
From the original circuit
v
dv
+C
KCL : i (t ) =
dt
20 Ω
− 0.6 t
= 1+ e
,t > 0
10V , t < 0
v (t ) = 
− 0.6 t
)V , t ≥ 0
 ( 20 − 10 e
 − 1 A, t < 0
i (t ) = 
− 0.6 t
) A, t ≥ 0
 (1 + e
Note that
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i (0 − ) = − 1 A ≠ i (0 + ) = 2 A
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7.4 The Step Response of RC and RL Circuits
Example 12
Mesh analysis
di
L + RTH i = VTH , t ≥ 0
dt
i ( 0) = 0
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7.4 The Step Response of RC and RL Circuits
Solution:
i(t ) = ih (t ) + i p (t )
characteristic equation
LS + RTH = 0
R
1
S = − TH = −
L
τ
ih (t ) = Ke
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i p (t ) =
t
−
τ
VTH
RTH
t
−
VTH
+ Ke τ
RTH
+
i(0 ) = i(0− ) = 0
Note : L is equivalent to open circuit
t
−
V
i(t ) = TH (1 − e τ ) , t ≥ 0
RTH
t
−
di
V (t ) = L = VTH e τ u(t )
dt
∴i(t ) =
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7.4 The Step Response of RC and RL Circuits
t
−
VTH
τ
i (t ) =
(1 − e ) , t ≥ 0
RTH
V (t ) = L
VTH
RTH
VTH
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t
−
di
= VTH e τ u (t )
dt
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7.4 The Step Response of RC and RL Circuits
Example 13
t＜0 , steady state
1
H
3
t＝0 , switch is opened
For t<0
i (0−) =
10
= 5A
2
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7.4 The Step Response of RC and RL Circuits
For t ≧ 0
1
H
3
Mesh Analysis
1 di
+ 5i = 10V , t ≥ 0
3 dt
s = −15
ih (t ) = Ke
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1
H
3
i (t ) = ih + i p
= Ke
t
−
τ
i p (t ) = 2 A
,τ =
−
t
τ
+ 2A , t ≥ 0
1
15
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7.4 The Step Response of RC and RL Circuits
initial condition
i (0+ ) = i (0− ) = 5 A
∴ i (t ) = 2 + 3e
−
t
τ
,t≥0
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7.4 The Step Response of RC and RL Circuits
Note that if V is chosen for solution :
V ( 0− ) = 0V
KVL : V (0+ ) = 10V − (2 + 3)5 = −15V
t
−
di 1 d
τ
or V = L =
(2 + 3e )
dt 3 dt
1 −τt
= − e , t ≥ 0+
τ
+
V (0 ) = −15V , same answer.
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7.4 The Step Response of RC and RL Circuits
Example 14 : S1 is closed at t=0
S2 is closed at t=4
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Find i(2)=？
i(5)=?
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7.4 The Step Response of RC and RL Circuits
For t < 0
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t < 0 , i=0 , open circuit
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7.4 The Step Response of RC and RL Circuits
For 0+≦ t ＜ 4S
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7.4 The Step Response of RC and RL Circuits
For 0+≦ t ＜ 4S
5
di
+ 10i = 40V ,
0+ ≤ t ≤ 4S
dt
i (0 + ) = i (0 − ) = 0 A
L 5 1
=
=
R 10 2
∴ i (t ) = 4(1 − e −2 t ) A , 0 + ≤ t ≤ 4 S
τ=
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i (4 − ) = 4(1 − e −8t ) ≈ 4 A
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7.4 The Step Response of RC and RL Circuits
For t ≧ 4+
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7.4 The Step Response of RC and RL Circuits
For t ≧ 4+
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i (4 − ) = 4 A
40 − 10
VTH =
× 2 + 10 = 20V
4+2
4
22
RTH = 4 Ω P 2 Ω + 6 Ω = + 6 =
Ω
3
3
L
5
15
τ =
=
=
S
RTH 22 / 3 22
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7.4 The Step Response of RC and RL Circuits
For t ≧ 4+
22
Ω
3
i (4 + ) = i ( 4 − ) = 4 A
di 22
+
i = 20V
dT
3
where T = t − 4
5
ih (t ) = Ke
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i p (t ) =
−
22
(t − 4 )
15
, t ≥ 4+
20
60 30
=
=
= 2.727
22 / 3 22 11
22
− (t −4)
30
+ Ke 15
, t ≥ 4+
11
30
i (4 + ) = 4 A =
+K
11
30 14
K = 4−
=
11 11 22
30 14 − 15 ( t − 4 )
∴ i (t ) =
+ e
, t ≥ 4+
11 11
i (t ) =
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7.4 The Step Response of RC and RL Circuits
In Summary

0 , t ≤ 0

i (t ) = 4(1 − e −2t ) , 0 ≤ t ≤ 4
 30 14
 + e −22(t − 4) , t ≥ 4
 11 11
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Summary
n
n
n
n
Objective 1 : Be able to determine the natural response
of both RC and RL circuits.
Objective 2 : Be able to find the step response of both
RC and RL circuits.
Objective 3 : Know and be able to use the singularity
functions.
Objective 4 : Be able to analyze circuits with sequential
switching.
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Summary
Chapter Problems : 7.7
7.14
7.23
7.32
7.47
7.63
Due within one week.
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