Circuits with Capacitor and Inductor
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We have discussed so far circuits only with resistors. While
analyzing it, we came across with the set of algebraic
equations.
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Hereafter we will analyze circuits with inductors and
capacitors.
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When we apply KCL or KVL to the circuits (in time domain)
containing inductors and capacitors, it will produce differential
equations.
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Capacitor and Inductor do not allow either voltage or current
to jump instantaneously1 from one state to another state.
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It will be possible If infinite current or infinite voltage is available.
RC Circuit
Let us consider this circuit. The capacitor has an initial charge.
(Vc (0− ) = V0 )
t=0
R
+
−
+
Vs
Vc (t)
C
−
By KCL at t = 0+ ,
C
dVc
Vc − Vs
+
=0
dt
R
dVc
+ Vc = Vs
dt
It is a first order differential equation.
RC
Let us follow the three simple steps to solve2 the differential
equations.
1. Find the homogeneous solution.
2. Find the particular solution.
3. Find the total solution which is sum of above two.
4. Find the constants using initial conditions.
To find the homogeneous solution, set the forcing function (Vs ) to
zero.
dVcH
+ VcH = 0
RC
dt
Let us assume the solution.
VcH = Ae st
Plug into the equation,
RCAse st + Ae st = 0
2
Solving differential equations is simply a guess work
RCs + 1 = 0
1
s=−
RC
VcH = Ae (−t/RC )
Let τ = RC . where τ is a time constant.
VcH = Ae (−t/τ )
Particular solution:
RC
dVcP
+ VcP = Vs
dt
Guess any solution that satisfies this equation. In this case, it is
very simple.
VcP = Vs
Total solution:
Vc = VcH + VcP
Vc = Ae (−t/RC ) + Vs
If V0 < Vs ,
Vc
Vs
Vc (t)
To find the constant,
Vc (0− ) = Vc (0+ ) = V0
V0
At t =
0+ ,
Vc (0+ ) = A + Vs
t (s)
A = V0 − Vs
The total solution is
Vc = (V0 − V )e (−t/RC ) + Vs
The first component of equation is called transient (natural)
response. The second component is steady state (forced) response.
Example - RC Circuit
Let us consider this circuit. The capacitor has a zero initial charge.
(Vc (0− ) = 0V )
t=0
1Ω
+
−
+
3V
Vc (t)
1F
−
τ = RC = 1 sec.
Vc &i
3V
Vc (t)
Vc = 3(1 − e −t ) V
i =C
dVc
= 3e −t A
dt
i(t)
t (s)
RC Circuit
Let us consider this case.
t=0
R
+
−
+
Vs
Vc (t)
C
R
−
Before t = 0, the capacitor has been charged to
Vc (0) =
Vs
volts.
2
V
2
For t > 0
+
Vc (t)
C
−
dVc
Vc
+
=0
dt
R
dVc
RC
+ Vc = 0
dt
C
R
Since there is no forcing function, the solution Vc (t) will contain
only homogeneous solution.
Vc (t) = Ae (−t/RC )
To find A, we have to use an initial condition. At t = 0,
Vc (0) = A
∴ Vc (t) = Vc (0)e (−t/RC )
Vc
Vc (0)
Vc (t)
t (s)
RL Circuit
Let us consider the RL circuit. The current in the inductor before
t = 0 is I0 .
t=0
R
+
−
+
Vs
vL (t)
L
− iL
After t = 0,
diL
dt
L diL
Vs
+ iL =
R dt
R
It is a first order differential equation.
Vs = RiL + L
Let us follow the same steps to solve this equation.
I Homogeneous solution:
Set the forcing function to zero.
L diLH
+ iLH = 0
R dt
Assume iLH = Ae st and substitute in the above equation.
L
Ase st + Ae st = 0
R
L
s +1=0
R
R
s=−
L
iLH = Ae (−Rt/L)
Let τ =
L
. where τ is a time constant.
R
iLH = Ae (−t/τ )
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Particular solution:
L diLP
Vs
+ iLP =
R dt
R
Guess any solution that satisfies it.
∴ ILP =
I
Vs
R
Total solution:
iL = iLH + iLP
Vs
R
To find constant A, let us use an initial condition. At t = 0,
iL (t) = Ae (−t/τ ) +
iL (0) = A +
A = I0 −
Vs
R
Vs
R
∴ iL (t) = (I0 −
Vs (−t/τ ) Vs
)e
+
R
R
If I0 < I∞ ,
iL
L
Vs
where τ = . Let I∞ =
be
R
R
the current at steady state.
I∞
iL (t) = (I0 − I∞ )e (−t/τ ) + I∞
I0
First term is a natural (transient)
response. Second term is a
steady state (forced) response.
The voltage across the inductor is
vL (t) = L
diL
= R(I∞ − I0 )e (−t/τ )
dt
iL (t)
t (s)
RL Circuit
Let us consider this case.
t=0
I
R
L
iL
Before t = 0, the inductor current is iL (0) = I . After t = 0,
L
diL
+ iL R = 0
dt
Since it is a homogeneous first order differential equation, iL (t) will
contain only natural (transient) response.
The solution
iL (t) = Ae (−t/τ )
where τ =
L
. To fins A,
R
iL (0) = A
A=I
∴ iL (t) = Ie (−t/τ )
iL
I
iL (t)
t (s)
Example- RL Circuit
Find the time constant of the following circuit.
t=0
−
+
Vs
R
L
R
iL
L
RTh
where RTh is the Thevenin resistance from the inductor terminals
after voltage source is shorted.
τ=
RTh =
R
2
∴τ =
2L
R
RLC Circuit
Let us consider this circuit.
t=0
R
L
iL
+
−
+
Vs
Vc
C
−
After t = 0,
Vs = RiL + L
Since iL = C
diL
+ Vc
dt
dVc
,
dt
dVc
d 2 Vc
+ LC
+ Vc
dt
dt 2
d 2 Vc
dVc
+ RC
+ Vc = Vs
LC
2
dt
dt
It is a second order linear constant coefficient differential equation.
Vs = RC
Let us use the same steps to solve this.
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Homogeneous solution: To obtain this, make Vs = 0.
LC
d 2 VcH
dVcH
+ VcH = 0
+ RC
2
dt
dt
Assume, VcH = Ae st and substitute.
LCs 2 Ae st + RCsAe st + Ae st = 0
s2 +
1
R
s+
=0
L
LC
It is called a characteristics equation. The roots of the
characteristics equation are
s R
R 2
4
− ±
−
L
L
LC
s1,2 =
2
s1,2
Let α =
R
=− ±
2L
R
2L
2
−
1
LC
1
R
and ω0 = √ .
2L
LC
s1,2
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s
q
= −α ± α2 − ω02
Case 1: Over Damping (α > ω0 ).
The two roots are real. Let s1 = −α1 and s2 = −α2 .
VcH = A1 e −α1 t + A2 e −α2 t
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Case 2: Critical Damping (α = ω0 ).
The two roots are real and equal. Let s1 , s2 = −α.
VcH = e −αt (A1 t + A2 )
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Case 3: Under Damping (α < ω0 ).
The two roots will be complex.
q
s1,2 = −α ±  ω02 − α2
Let ωd =
q
ω02 − α2 .
s1,2 = −α ± ωd
s1 = −α + ωd , s2 = −α − ωd
VcH = A1 e (−α+ωd )t + A2 e (−α−ωd )t
A1 and A2 must be complex conjugate of each other in order
to make Vc (t) real.
∴ VcH = e −αt (B1 cos(ωd t) + B2 sin(ωd t))
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Particular Solution:
LC
dVcP
d 2 VcP
+ RC
+ VcP = Vs
dt 2
dt
Guess any solution that satisfies the above equation.
VcP = Vs
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Total solution:
1. Over Damping:
Vc (t) = A1 e −α1 t + A2 e −α2 t + Vs
2. Critical Damping:
Vc (t) = e −αt (A1 t + A2 ) + Vs
3. Under Damping:
Vc (t) = e −αt (B1 cos(ωd t) + B2 sin(ωd t)) + Vs
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To find constants:
Use initial conditions. i.e., Vc (0) and iL (0).
Vc (t)
← Under Damping
Vs
← Critical Damping
← Over Damping
t (s)
Figure: Plot of Vc (t) for different cases
LC Circuit
Let us consider this circuit.
t=0
L
iL
+
−
+
Vs
Vc
C
−
After t = 0,
Vs = L
Since iL = C
dVc
,
dt
Vs = LC
LC
diL
+ Vc
dt
d 2 Vc
+ Vc
dt 2
d 2 Vc
+ Vc = Vs
dt 2
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Homogeneous solution: To obtain this, make Vs = 0.
LC
d 2 VcH
+ VcH = 0
dt 2
Assume, VcH = Ae st and substitute.
LCs 2 Ae st + Ae st = 0
s2 +
1
=0
LC
The roots of the characteristics equation are
1
1
s1,2 = + √ , − √
LC
LC
1
Since ω0 = √ ,
LC
s1,2 = ω0 , −ω0
VcH = A1 e −(ω0 )t + A2 e (ω0 )t
In order Vc (t) to be real, A1 and A2 must be complex conjugates
of each other.
VcH = B1 cos(ω0 t) + B2 sin(ω0 t)
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Particular solution:
VcP = Vs
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Total solution:
Vc (t) = B1 cos(ω0 t) + B2 sin(ω0 t) + Vs
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To find constants: Use initial conditions.
Since there is no Resistor in the circuit, the response will oscillate
forever. It is an undamped circuit.
Vc (t)&iL (t)
Vc (t)
2Vs
Vs
iL (t)
t (s)
Figure: Plots of Vc (t) and iL (t) for LC circuit