October 22, 2001
Chapter-7
Response of First Order RL and RC Circuits
•
Natural response of RL and RC circuits
Consider current and voltage that arise when stored energy in an inductor or capacitor is
suddenly released to e resistive network.
Step response of RL and RC circuits
Consider current and voltage that arise when energy is being acquired by the inductor or
capacitor due to the sudden application of a dc voltage or current source.
The response of RL and RC circuits to any abrupt change in a dc voltage or current
source.
•
•
7.1 The natural response of an RL circuit:
• Current source generates a constant current of IS.
• Switch has been closed a long time. So all current and voltage reached a constant value.
• The voltage across the inductive is zero.
• No current in R0 or R.
di
=0
dt
vL =
All the source current is appears in the inductance before t=0.
Switch is opened at t=0, then inductance begins releasing energy.
We find the v(t) and i(t) for t ≥ 0 .
t=0
Ro
+
i
L
I2
Is
i
L
i
R
v
iis
R
v
From the second circuit, Using KVL to obtain an expression for t ≥ 0 .
L
di
+ Ri = 0
dt
This is a first-order differential equation. To solve this equation we can rewrite it in the following
form
di
R
=− i
dt
L
Multiply both site by a differential time dt and divide birt site by i
di
R
= − dt
i
L
And if we get integrate the both side
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i (t )
t
di
R
∫i (0) i = ∫0 − L dt
This integration gives us
ln
i (t )
R
=− t
i (0)
L
That will give us
i (t ) = i (0)e
R
− t
L
No instantaneous change of current in a inductor
i (0− ) = i (0+ ) = I 0
I 0 denotes the initial current in the inductor. Therefore
i (t ) = I 0 e
R
− t
L
, t≥0
(1.1)
The voltage on the resistor is
v(t ) = I 0 Re
R
− t
L
,
t ≥ 0+
(1.2)
v(0 − ) = 0 and v(0) = I 0 R
The power on the resistor
p = vi = i 2 R =
v2
R
or
p = I 02 Re
R
−2 t
L
for
t ≥ 0+
(1.3)
Energy
R
−2 t
1 2
LI 0 (1 − e L ) , t ≥ 0
0
2
L
We call time constant τ =
R
t
w = ∫ pdx =
(1.4)
(1.5)
Therefore
i (t ) = I 0 e
−
t
τ
, t≥0
v(t ) = I 0 Re
p = I 02 Re
w=
−2
−
t
τ
t
τ
,
,
t ≥ 0+
t ≥ 0+
t
−2
1 2
LI 0 (1 − e τ ) , t ≥ 0 +
2
(1.6)
(1.7)
(1.8)
(1.9)
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i(t)=20e-t/T
Example:
t=0
2
i0
+
0.1
20A
iL
2H
10
v0
40
a.
iL (t ) = ? for t ≥ 0
b. Find
i0 (t ) for t ≥ 0+
c.
v0 (t ) for t ≥ 0+
Find
d. Find power in the 10Ω resistor
e. Total energy stored in the 2H inductor
a.
Initial current
I 0 = 20A at t = 0− , and t = 0+
R eq = 2 + (10 || 40) = 10 Ω
τ=
L
2
= = 0.2 s
Req 10
iL (t ) = 20e −5t , t ≥ 0
b.
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i0 (t ) = −iL (t )
10
= −4e −5t , t ≥ 0 +
10 + 40
c.
v0 (t ) = Ri0 (t ) = −(40)4e−5t = −160e−5t V , t ≥ 0 +
v02 (t )
= 2560e −10t , t ≥ 0 +
R
d.
p10Ω =
e.
w = ∫ 2560e−10t dt = 256 J
∞
0
f.
w(0) =
1
Li (0) = 400 J
2
Example:
6
a
b
10
6.4A
0.32H
4
V0
a. Calculate
v0 (t ) for t ≥ 0+
b. What percentage of the initial energy stored in the inductor is eventually dissipated in the
4Ω resistor?
a.
iL (0− ) = iL (0+ ) = 6.4
Req =
τ=
10
=4A
10 + 6
4(16)
= 3.2 Ω
4 + 16
L
0.32
=
= 0.1 s
Req
3.2
iL (t ) = 4e−10t A , t ≥ 0
i10Ω (t ) =
4
iL (t ) = 0.8e −10t A , t ≥ 0 +
4 + 16
v10Ω (t ) = −( R)i10Ω (t ) = −8e−10t V , t ≥ 0 +
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b.
v4 Ω (t ) = L
p4 Ω =
diL (t )
= 0.32(−40)e−10t = −12.8e−10tV , t ≥ 0 +
dt
v42Ω (t )
= 40.96e−20tW , t ≥ 0 +
4
The total energy dissipated in the 4Ω resistor
∞
w = ∫ 40.96e −20t dt = 2.048 J
0
Initial energy stored in the 0.32H inductor
w(0) =
1 2
1
Li (0) = (0.32)42 = 2.56 J
2
2
% dissipated =
2.048
100 = 80%
2.56
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