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Chapter 6Chapter 6

Capacitors and InductorsCapacitors and Inductors

電路學電路學 (( 一一 ))

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Capacitors and Inductors Capacitors and Inductors Chapter 6Chapter 6

6.1 Capacitors6.2 Series and Parallel Capacitors6.3 Inductors6.4 Series and Parallel Inductors

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6.1 Capacitors (1)6.1 Capacitors (1)• A capacitor is a passive element designed to

store energy in its electric field.

• A capacitor consists of two conducting plates separated by an insulator (or dielectric).

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6.1 Capacitors (2)6.1 Capacitors (2)• Capacitance C is the ratio of the charge q on one

plate of a capacitor to the voltage difference v between the two plates, measured in farads (F).

• Where is the permittivity of the dielectric material between the plates, A is the surface area of each plate, d is the distance between the plates.

• Unit: F, pF (10–12), nF (10–9), and

vCq d

AC

and

F (10–6)

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6.1 Capacitors (3)6.1 Capacitors (3)• If i is flowing into the +ve

terminal of C– Charging => i is +ve– Discharging => i is –ve

• The current-voltage relationship of capacitor according to above convention is

td

vdCi )(

10

0

tvtdiC

vt

t and

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6.1 Capacitors (4)6.1 Capacitors (4)• The energy, w, stored in

the capacitor is

• A capacitor is – an open circuit to dc (dv/dt = 0). – its voltage cannot change abruptly.– The ideal capacitor does not dissipate energy.

2

2

1vCw

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6.1 Capacitors (5)6.1 Capacitors (5)Example 1(a) Calculate the charge stored on a 3-pF capacitor with 20 V

across it.(b) Find the energy stored in the capacitor.

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6.1 Capacitors (6)6.1 Capacitors (6)Example 2

The current through a 100-F capacitor is

i(t) = 50 sin(120 t) mA.

Calculate the voltage across it at t =1 ms and t = 5 ms.

Take v(0) =0.

Answer:

v(1ms) = 93.14mV

v(5ms) = 1.7361V

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6.1 Capacitors (7)6.1 Capacitors (7)Example 3

An initially uncharged 1-mF capacitor has the current shown below across it.

Calculate the voltage across it at t = 2 ms and t = 5 ms.

Answer:

v(2ms) = 100 mV

v(5ms) = 500 mV

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6.1 Capacitors (8)6.1 Capacitors (8)Example 4Obtain the energy stored in each capacitor in the figure under dc conditions.

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6.1 Capacitors (9)6.1 Capacitors (9)Example 5Under dc condition, find the energy stored in the capacitors in the figure.

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6.2 Series and Parallel 6.2 Series and Parallel Capacitors (1)Capacitors (1)

• The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.

Neq CCCC ...21

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6.2 Series and Parallel6.2 Series and Parallel Capacitors (2) Capacitors (2)

• The equivalent capacitance of N series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.

Neq CCCC

1...

111

21

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6.2 Series and Parallel 6.2 Series and Parallel Capacitors (3)Capacitors (3)

Example 6

Find the equivalent capacitance seen at the terminals of the circuit in the circuit shown below:

Answer:

Ceq = 40F

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6.2 Series and Parallel 6.2 Series and Parallel Capacitors (4)Capacitors (4)

Example 7Find the voltage across each of the capacitors in the circuit shown below:

Answer:

v1 = 15 V

v2 = 10 V

v3 = 5 V

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6.2 Series and Parallel 6.2 Series and Parallel Capacitors (5)Capacitors (5)

Example 8Find the voltage across each of the capacitors in the circuit shown below:

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6.3 Inductors (1)6.3 Inductors (1)• An inductor is a passive element designed

to store energy in its magnetic field.

• An inductor consists of a coil of conducting wire.

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6.3 Inductors (2)6.3 Inductors (2)• Inductance is the property whereby an inductor

exhibits opposition to the change of current flowing through it, measured in henrys (H).

• The unit of inductors is Henry (H), mH (10–3) and H (10–6).

td

idLv

l

ANL

2

and

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6.3 Inductors (3)6.3 Inductors (3)• The current-voltage relationship of an inductor:

• The power stored by an inductor:

)()(1

00

titdtvL

it

t

2

2

1iLw

• An inductor acts like a short circuit to dc (di/dt = 0)• The current through an inductor cannot change

abruptly.

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6.3 Inductors (4)6.3 Inductors (4)Example 9 The current through a 0.1-H inductor is

i(t) = 10te-5t A Find the voltage across the inductor and the energy stored in it.

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6.3 Inductors (5)6.3 Inductors (5)Example 10The terminal voltage of a 2-H inductor is

v = 10(1-t) V

Find the current flowing through it at t = 4 s and the energy stored in it within 0 < t < 4 s. Assume i(0) = 2 A.

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6.3 Inductors (6)6.3 Inductors (6)Example 11 Consider the circuit, under dc conditions, find:

(a) i, vC, and iL(b) the energy stored in the capacitor and inductor.

Answer:

i = 2 A

vC = 10 V

wL = 4 J

wC = 50 J

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6.3 Inductors (7)6.3 Inductors (7)Example 12

Determine vc, iL, and the energy stored in the capacitor and inductor in the circuit of circuit shown below under dc conditions.

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6.4 Series and Parallel6.4 Series and Parallel Inductors (1) Inductors (1)

• The equivalent inductance of series-connected inductors is the sum of the individual inductances.

Neq LLLL ...21

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6.4 Series and Parallel 6.4 Series and Parallel Inductors (2)Inductors (2)

• The equivalent capacitance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances.

Neq LLLL

1...

111

21

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6.4 Series and Parallel6.4 Series and Parallel Inductors (3) Inductors (3)

Example 13

Calculate the equivalent inductance for the inductive ladder network in the circuit shown below:

Answer:

Leq = 25mH

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6.4 Series and Parallel6.4 Series and Parallel InductorsInductors (4) (4)

Example 14

For the circuit, i= 4(2 – e-10t) mA. If i2(0) = -1 mA, find: (a) i1(0); (b) v(t), v1(t), and v2(t); (c) i1(t) and i2(t)

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6.4 Series and Parallel 6.4 Series and Parallel Inductors (5) Inductors (5)

• Current and voltage relationship for R, L, C

+

+

+

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at t = 0−, inductance acts as a short circuit

21 s

1 2

(0 )R

i iR R

12 s L

1 2

(0 ) (0 )R

i i iR R

at t = 0+, iL cannot change instantaneously.

1L s 2

1 2

(0 ) (0 )R

i i iR R

1(0 ) 0i

+v2

−

+vL−

L

2 2 2

(0 ) 0

(0 ) (0 )

v

v i R

2 2 2

L 2

(0 ) (0 )

(0 ) (0 )

v i R

v v

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2c s

1 2

(0 )R

v vR R

at t = 0−, capacitor acts as a open circuit

at t = 0+, vc cannot change instantaneously.

2c c s

1 2

(0 ) (0 )R

v v vR R

ic

i2

c

s2

1 2

(0 ) 0

(0 )

i

vi

R R

c2

2

c 2

(0 )(0 )

(0 ) (0 )

vi

R

i i

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at t = 0−,

iL(0−) = 10/5 = 2 A vc(0−) = iL(0−) 3 = 6 V

at t = 0+,

iL(0+) = iL(0−) = 2 A vc(0+) = vc(0−) = 6 V

ic(0−) = 0vL(0−) = 0

ic(0+) = −iL(0+) = −2 A vL(0+) = vc(0+) − iL(0+) 3 = 0

ic +vL

−

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Example Find iL(0+), vc(0+), dvc(0+)/dt, and diL(0+)/dt

iL(0+) = iL(0−) = 0 vc(0+) = vc(0−) = −2 V

L L(0 ) (0 )2 A/s

di v

dt L

c c(0 ) (0 )12 V/s

dv i

dt C

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6.5 Applications (1)6.5 Applications (1)

Integrator An op amp circuit whose output is proportional to the integral of the input signal

0

1( )

t

o iv v t dtRC

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Example 15If v1 = 10 cos 2t mV and v2 = 0.5t mV, find vo in the op amp circuit. Assume the voltage across the capacitor is initially zero.

6.5 Applications (2)6.5 Applications (2)

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6.5 Applications (3)6.5 Applications (3)

Differentiator An op amp circuit whose output is proportional to the rate of change of the input signal

io

dvv RC

dt

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Example 16Sketch the output voltage for the circuit, given the input voltage in the figure. Take vo = 0 at t = 0.

6.5 Applications (4)6.5 Applications (4)

vi(V)

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6.5 Applications (5)6.5 Applications (5)

Analog Computer

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2

2 2 10sin4 with (0) 4, '(0) 1o oo

d v dvv t v v

dt dt

6.5 Applications (6)6.5 Applications (6)

2

2 10sin4 2o oo

d v dvt v

dt dt

'

010sin4 2 (0)

to o

o o

dv dvt v dt v

dt dt

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6.5 Applications (7)6.5 Applications (7)

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6.5 Applications (8)6.5 Applications (8)