ChapterChapter 1515

CapacitorsCapacitors andand InductorsInductors inin CircuitsCircuits

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 S 1 and S 2 is shown in the below Figure.

C C C -+ - +

I(t) q(t) q(t) R R R

S1 ε S1 ε I(t)

S2 S2

(a) (b) (c)

3 When S 1 is closed and S 2 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 ε – IR – q/C = 0 Since I = dq/dt, ε –R (dq/dt)– q/C = 0

By 1 st Order ODE, we get q(t) = C · ε ( 1 - e-t/RC )

-t/τc ∴ q(t) = C · ε ( 1 - e ) where τC = RC , is the capacitive time constant of the circuit.

Since VC(t) = q(t)/ C,

∴ -t/τc VC(t) = ε( 1 - e )

As the capacitor charge approaches its maximum value (and V C becomes equal to ε), the current decreases to zero. Energy stored in the electric field of Capacitor is 2 UC = ½ CV C Since q(t) = C · ε ( 1 - e-t/τc ) and I = dq/dt, ∴ I(t) = ε/R· e-t/τc 4 VC I

ε ε/R 0.63ε 0.37ε/R t t τC τC

Time Variation of (a) the voltage across the capacitor plate and (b) the electric current in the charging RC circuit.

5 When S 1 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 q/C – IR = 0

Since I = dq/dt, ∴ q/C – R (dq/dt) = 0 By 1 st Order ODE, we get

-t/τc -t/τc q(t) = C ·εe and VC(t) = εe

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 ε the capacitor plates in the discharging

0.37ε RC circuit. t τ C 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 , R = (60 Ω)(2.0 × 10 -6 F) S1 ε =1.2 × 10 -4 s (a) Since q(t) = C · ε ( 1 - e-t/τc ) S2 and I(t) = ε/R· e-t/τc, therefore At t = 0s, q = 0 C and I = ε/R = 0.20 A

-5 -2 At t = 1.5 τ C, q = 1.9 × 10 C and I = 4.5 × 10 A At t -> ∞ , q = 2.4 × 10 -5 C and I = 0 A 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 S 1 and S 2 is shown in the below Figure.

L L L --+ + q(t) R R I(t) R

S1 ε S1 ε I(t)

S2 S2

(a) (b) (c)

8 When S 1 is closed with S 2 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) .

From Faraday’s Law, VL = - L· (dI/dt) *(in accordance with Lenz’s Law, the induced emf counteracts the increase in the current.)

Applying Kirchhoff’s loop rule, we obtain ε – L(dI/dt)– IR = 0

By 1 st Order ODE, we get I(t) = ε/R ( 1 - e-Rt/L )

-t/τL ∴ I(t) = ε/R ( 1 - e ) where τL = L/ R, is the inductive time constant of the circuit.

Since V L = - L· (dI/dt)

∴ -t/τL VL(t) = -ε · e )

The energy stored in the magnetic field of an inductor is

2 UL = ½ LI

9 I |VL|

ε/R ε 0.63ε/R 0.37ε t t τL τL (a) (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 S 1 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 1 st Order ODE, we∴ get

-t/τL -t/τL I(t) = ε/R e and VL(t) = - ε · e

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 ε/R the RL circuit. The induced voltage 0.37ε/R across the coil also decays exponentially. t τC

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

2 UC = ½ q0 /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 I = 0 I0 I = 0 I0

+ - L q0 C L q = 0 C L q0 C L q = 0 C - +

t = 0, T t = T/4 t = T/2 t = 3T/4

(a) (b) (c) (d)

q(t) I(t) q0 I0

0 T/4 T/2 3T/4 T

(e) 13 After the capacitor is completely discharged, the current is at its maximum value I 0, the energy in the inductor is

2 UL = ½ LI 0 which is maximum energy stored in the inductor.

Due to no energy is lost, UC = UL 2 2 ∴ ½ q0 /C = ½ LI 0 Because there is no energy dissipation,

2 2 2 2 ∴ U = ½ q /C + ½ LI = ½ q0 /C = ½ LI 0 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 = 1/2 π · √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) R S ε =2.5 × 10 3 rad/s 1

and f = 1/2 π · √ 1/LC S = 398 Hz 2

2 2 (b) By ½ q0 /C = ½ LI 0 ,

∴ 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.

By 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 × 10 3t),

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 = 2.3 τ)?

Solution: −(R / L)t The current i at any time t is: = i I0e 1 U = Li 2 The energy U in the inductor at any time is: 2

1 − − U = LI 2e (2 R / L)t = U e (2 R / L)t 2 0 0 2 τ Where, U 0 = ½ LI 0 is the energy at the initial time t = 0. When t = 2.3 = 2.3L/R, then = − )3.2(2 = − 6.4 = U U 0e U 0e 01.0 U 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 = 2.3 τ)?

Solution: −(R / L)t The current i at any time t is: = i I0e 1 U = Li 2 The energy U in the inductor at any time is: 2

1 − − U = LI 2e (2 R / L)t = U e (2 R / L)t 2 0 0 2 τ Where, U 0 = ½ LI 0 is the energy at the initial time t = 0. When t = 2.3 = 2.3L/R, then = − )3.2(2 = − 6.4 = U U 0e U 0e 01.0 U 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

L q0

R S2 t

q0 + - C

19