First Order RL Circuits
Switch S is closed When the switch S is closed for t<0, the
for t<0, and open inductor behaves as a short circuit to dc.
for t ≥ 0 The voltage across the inductor v=0,
Hence the voltage across the R is also
zero
zero.
+
i
v
V
The current component i (t )
Rg
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RL Circuits
When the switch S is open at t=0s and remains open for t>0. Since the
current through the inductor does not change instaneously.
V
i (0)
Rg
For t≥0, applying KVL, we get
di
L Ri 0
dt
di R
i0
dt L
di R
i ................((2)
d
dt L
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RL Circuits
Equation (2) can be solved as follows,
Divide both side of (2) by i and integrate both side w.r.t t , we get
1 di R
i dt dt L dt
1 R
i di L dt
R
ln i t K
L
take exp onential both side , we get
R R
tK t
i (t ) e L
e L
e K , at t 0, i (0) e K
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RL Circuits
Substituting R
t
i (0) e K in i (t ) e L
e K , we get
R
t
i (t ) i (0) e L
for t 0 s
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RL Circuits
Example : In the following circuit at t<0 , the switch is closed and at t≥0, the
switch is open. Find current through inductor iL at tt<0
0 , tt>0
0 and at tt=1ms
1ms
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RL Circuits
When t<0 switch is closed and voltage across the inductor is zero and it acts
like a short circuit.
circuit
The current iL=2A
2A
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RL Circuits
For t ≥ 0, switch is open and the current through the inductor iL (t>0) does not
change instantaneously from iL (t<0)
At t=1ms
R
We know that , t
i (t ) i (0) e L
for t 0 s
20 x 1 x 10 3
i (t ) 2 e 4
2 x 0.995 1.99 A
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RL Circuits
Example :
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
For t<0 switch is closed and capacitor behaves as a open circuit
By voltage division we get ,
V R
v for t 0
R Rg
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
At t=0 and t>0 switch is open Since the voltage across the
capacitors does not change
instantaneously, voltage v
iC iR remains same at t=0s
V R
v(0) for t 0
R Rg
Applying KCL at RC circuit , we get,
iC iR 0
dv v dv 1
C 0 v (2)
dt R dt RC
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
Equation (2) can be solved as follows,
Divide both side of (2) by v and integrate both side w.r.t t , we get
1 dv 1
v dt dt RC dt
1 1
v d
dv RC dt
1
ln v(t ) tK
RC
take exp onential both side , we get
t t
K
v(t ) e RC
e RC
e K , at t 0, v(0) e K
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
Substituting
t
v(0) e K in
i v(t ) e RC
e K , we gett
t
v(t ) v(0) e RC
for t 0 s
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
Given following circuit, Find vc and v0 for
I
I. t<0
II. t>0
III. t=1.3ms
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
For t < 0
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
120 x 1250
vC 100 V
1250 250
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
v ?
What is 0
Current through the 400 Ω resistance = 0.096A
g across the 400 Ω resistance i. e v0 = 38.4 V
Voltage
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
For t > 0 , vc , does not change instantaneously so vc =100V
Let us redraw the circuit for t ≥ 0
100V
Find voltage between these two points using voltage division rule
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
+ +
100V v’’
- -
Once you get this voltage say v’ , then get the find vo y applying voltage division
v’=32V and v0 at t=0 is 25.6 V
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
First Order RC Circuits
At t=1.3ms, we know that
t vc (t 1.3ms ) 59.5 V
vc (t ) vc (0) e RC
for t 0s
d C 4 H
vc (0) 100 V and
Req R ?
Req=R= 625Ω
Req=R=
Dr. Jagadish Nayak ,BITS Pilani, Dubai Campus
