Chapter 15
Capacitors and Inductors
in Circuits
1
Comparison of Capacitors and Inductors
Similar Properties of a capacitors and inductors :
- Circuits containing either a capacitor and inductors carry time-varying current
rather than constant current.
- Both a capacitor and an inductor can produce electric currents due to the energy
they store, either in an electric field (capacitor) or in a magnetic field (inductor).
2
RC Circuit
A circuit with resistance and capacitance is known as an RC circuit.
An RC circuit consisting of a resistor, a capacitor, a constant source of emf, and
switches S1 and S2 is shown in the below Figure.
C
R
S1
ε
I(t)
C
+
C
-
q(t)
+
q(t)
R
S1
R
ε
I(t)
S2
S2
(a)
(b)
(c)
3
When S1 is closed and S2 open, the circuit is equivalent to a single-loop circuit
consisting of a resistor and capacitor connected across a source of emf as shown
in Figure (b). (Charging Capacitor)
Applying Kirchhoff’s Loop Rule, we have
Since I = dq/dt,
ε –R (dq/dt)– q/C = 0
0
q(t) = C · ε ( 1 - e-t/RC )
By 1st Order ODE, we get
∴ q(t) = C · ε ( 1 - e-t/τc
ε – IR – q/C =
)
where τC = RC, is the capacitive time constant of the circuit.
Since VC(t) = q(t)/C,
∴
VC(t) = ε( 1 - e-t/τc )
As the capacitor charge approaches its maximum value (and VC becomes equal to
ε), the current decreases to zero. Energy stored in the electric field of Capacitor
is
UC = ½ CVC2
Since q(t) = C · ε ( 1 - e-t/τc ) and I =
dq/dt,
∴
I (t) = ε/R· e-t/τc
4
VC
I
ε
0.63ε
ε/R
0.37ε/R
τC
t
τC
t
Time Variation of (a) the voltage across the capacitor plate and (b) the electric
current in the charging RC circuit.
5
When S1 open and S2 is closed, the circuit becomes a single-loop circuit with
only a resistor and a capacitor as shown in Figure (c).
(Discharging Capacitor)
With Kirchhoff’s Loop Rule, we obtain
∴
Since I = dq/dt,
q/C – IR = 0
q/C – R (dq/dt) = 0
By 1st Order ODE, we get
q(t) = C·εe-t/τc
and
VC(t) = εe-t/τc
where τC = RC, is the capacitive time constant of the circuit.
By q(t) = C·ε e-t/τc and I = dq/dt,
∴
I (t) = ε/R· e-t/τc
VC
Time variation of the voltage across
the capacitor plates in the discharging
RC circuit.
ε
0.37ε
τC
t
6
Example 15.1
For the below figure, given that ε = 12 V, R = 60 Ω, and C = 2.0 μ F. With S1 is
closed and S2 open,
(a)What is the capacitive time constant of the circuit?
(b)What are the charge on the capacitor plates and the current
at t = 0, t = 1.5 τ C and as t -> ∞ ?
C
Solution :
(a) Since τC = RC,
= (60 Ω)(2.0 × 10-6 F)
=1.2 × 10-4 s
(a) Since q(t) = C · ε ( 1 - e-t/τc )
and
I (t) = ε/R· e-t/τc,
therefore
At t = 0s, q = 0 C and I = ε/R = 0.20 A
At t = 1.5 τ C, q = 1.9 × 10-5 C and I = 4.5 × 10-2 A
At t -> ∞ , q = 2.4 × 10-5 C and I = 0 A
R
S1
ε
S2
7
RL Circuit
A circuit with resistance and self-inductance is known as an RL circuit.
An RL circuit consisting of a resistor, an inductor, a constant source of emf, and
switches S1 and S2 is shown in the below Figure.
L
-
R
L
+
I(t)
R
S1
ε
S1
L
+
-
q(t)
R
ε
I(t)
S2
S2
(a)
(b)
(c)
8
When S1 is closed with S2 open, the circuit is equivalent to a single-loop circuit
consisting of a resistor and an inductor connected across a source of emf as
shown in Figure (b).
VL = - L· (dI/dt)
From Faraday’s Law,
*(in accordance with Lenz’s Law, the induced emf counteracts the
increase in the current.)
Applying Kirchhoff’s loop rule, we obtain
I(t) = ε/R ( 1 - e-Rt/L )
By 1st Order ODE, we get
∴ I(t) = ε/R ( 1 - e-t/τL
ε – L(dI/dt)– IR = 0
)
where τL = L/R, is the inductive time constant of the circuit.
Since VL = - L· (dI/dt)
∴
VL(t) = -ε · e-t/τL )
The energy stored in the magnetic field of an inductor is
UL = ½ LI2
9
I
|V L |
ε/R
0.63ε/R
ε
0.37ε
t
τL
(a)
t
τL
(b)
Time Variation of (a) the electric current and (b) the magnitude of the induced
voltage across the coil in the RL circuit.
10
When S1 is opened with S2 close, the circuit becomes a single-loop circuit with
only a resistor and an inductor as shown in Figure (c).
(Discharging Capacitor)
With Kirchhoff’s Loop Rule, we obtain
IR + L· (dI/dt) = 0
By 1st Order ODE, we∴
get
I(t) = ε/Re-t/τL
and
VL(t) = - ε · e-t/τL
where τL = L/R, is the inductive time constant of the circuit.
For this set-up circuit, the energy stored in the magnetic field of inductor
decreased exponentially with time.
I
Time variation of the electric current in
the RL circuit. The induced voltage
across the coil also decays exponentially.
ε/R
0.37ε/R
τC
t
11
Freely Oscillating Circuits
Circuits that do not have a source of emf can oscillate if they start with some
energy stored in either a capacitor or an inductor.
Case 1 – LC Circuit
- An idealized circuit of zero resistance and contains an inductor and a capacitor.
Initially, energy stored in the electric field of the capacitor is
UC = ½ q02/C
which is maximum energy stored in the capacitor
When the switch is closed, the capacitor begins to discharge, producing a
current in the circuit. The current in turn creates a magnetic field in the
inductor. The net effect of this process is a transfer of energy from the
capacitor, with its diminishing electric field, to the inductor, with its increasing
magnetic field.
12
I0
I=0
I=0
+
L
q0
-
I0
-
q=0
L
C
q0
L
C
+
C
L
q=0
t = 0, T
t = T/4
t = T/2
t = 3T/4
(a)
(b)
(c)
(d)
C
q (t)
I(t)
0
T/4
T/2
I0
3T/4
q0
T
(e)
13
After the capacitor is completely discharged, the current is at its maximum
value I0, the energy in the inductor is
UL = ½ LI02
which is maximum energy stored in the inductor.
UC = UL
Due to no energy is lost,
½ q02/C = ½ LI02
∴
Because there is no energy dissipation,
∴
U = ½ q2/C + ½ LI2 = ½ q02/C = ½ LI02
The frequency of the oscillations in a resistance free LC circuit may be
found by analogy with the mass-spring system. Therefore, we could have
q(t) = q0· cos( ωt + θ ) and I(t) = -ωq0· sin( ωt + θ )
where ω = √ 1/LC , is the angular frequency of the ocillations in the circuit.
So, the frequency of the oscillations is
∴
f = 1/2π · √1/LC
14
Example 15.2
In an LC circuit, the self-inductance is 2.0 × 10-2 H and the capacitance is 8.0 × 10-6 F .
At t = 0, all of the energy is stored in the capacitor, which has charge 1.2 × 10-5 C.
(a) What is the angular frequency and frequency of the oscillations in the circuit?
(b) What is the maximum current flowing through the circuit?
(c) How long does it take the capacitor to become completely discharged?
(d) Find an equation that represents q(t)?
C
Solution :
(a) By ω = √ 1/LC,
= √1/(2.0 × 10-2 H)(8.0 × 10-6 F)
=2.5 × 103 rad/s
and f = 1/2π · √ 1/LC
= 398 Hz
(b) By
R
S1
ε
S2
½ q02/C = ½ LI02 ,
∴
I = q0√ 1/LC
= 3.0 × 10-2 A
15
(c) The capacitor becomes completely discharged in one-fourth of a cycle, or
during a time T/4, where T is the period of the oscillations.
By T = 1/f
= 1/398 Hz
= 2.5 × 10 -3 s
Therefore, the time taken for the capacitor to become fully discharged
is ( 2.5 × 10 -3 s)/4 = 6.3 × 10 -4 s
(d) The capacitor is completely charged at t = 0, so q(0) = q0.
q(t) = q0· cos( ωt + θ ),
q(0) = q0 = q0 · cos(θ)
Thus θ = 0, and
q(t) = (1.2 × 10-5C) · cos(2.5 × 103t),
By
16
Example 15.3
When the current in an R-L circuit is decaying, what fraction of the original energy stored
in the inductor has been dissipated after 2.3 time constants (t = 2.3τ)?
Solution:
The current i at any time t is:
i = I 0e
− ( R / L )t
The energy U in the inductor at any time is:
1 2
U = Li
2
1
U = LI 02 e − 2 ( R / L )t = U 0 e − 2 ( R / L ) t
2
Where, U0 = ½ LI02 is the energy at the initial time t = 0. When t = 2.3 τ = 2.3L/R, then
U = U 0 e −2 ( 2.3) = U 0 e −4.6 = 0.01U 0
That is, only 0.01 or 1 % of the energy initially stored in the inductor remains, so 99 %
has been dissipated in the resistor.
17
In-Class Exercise 15.1
When the current in an R-L circuit is decaying, what fraction of the original energy
stored in the inductor has been dissipated after 2.3 time constants (t = 2.3τ)?
Solution:
The current i at any time t is:
i = I 0e
− ( R / L )t
The energy U in the inductor at any time is:
1 2
U = Li
2
1
U = LI 02 e − 2 ( R / L )t = U 0 e − 2 ( R / L ) t
2
Where, U0 = ½ LI 02 is the energy at the initial time t = 0. When t = 2.3 τ = 2.3L/R,
then
−2 ( 2.3)
−4.6
U = U 0e
= U 0e
= 0.01U 0
That is, only 0.01 or 1 % of the energy initially stored in the inductor remains, so
99 % has been dissipated in the resistor.
18
Case 2 – RLC Circuit
- A RLC circuit with a capacitor, an inductor and resistance, is another form of
freely oscillations circuit, which is an underdamped oscillations of the
capacitor charge.
For the switch is closed in the RLC circuit, the capacitor begins to discharge
and electromagnetic energy is dissipated by the resistor at a rate I2R.
q
q0
L
R
S2
t
q0
+
C
19