capacitors and inductors(new) - faraday

EENG223: CIRCUIT THEORY I

DC Circuits:

Capacitors and Inductors Hasan Demirel


• Introduction

• Capacitors

• Series and Parallel Capacitors

• Inductors

• Series and Parallel Inductors

Capacitors and Inductors: Introduction


• Resistors are passive elements which dissipate energy only.

• Two important passive linear circuit elements: 1. Capacitor 2. Inductor

• Capacitors and inductors do not dissipate but store energy, which can be retrieved at a later time.

• Capacitors and inductors are called storage elements.

Capacitors and Inductors: Introduction


Capacitors and Inductors MICHAEL FARADAY (1791–1867)

English chemist and physicist, discovered electromagnetic induction in 1831 which was a breakthrough in engineering because it provided a way of generating electricity. The unit of capacitance, the farad, was named in his honor.


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

Capacitors and Inductors

Plates may be aluminum foil while the dielectric may be air, ceramic, paper, or mica.

A capacitor with applied voltage v.


• Three factors affecting the value of capacitance:


1. Area(A): the larger the area, the greater the capacitance.

2. Spacing between the plates (d) : the smaller the spacing, the greater the capacitance.

3. Material permittivity (e) : the higher the permittivity, the greater the capacitance.

d

εAC


• The relation between the charge in plates and the voltage across a capacitor is given below.


Cvq

C/V1F1

v

q Linear

Nonlinear

Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F).

(1 farad = 1 coulomb/volt)



Circuit symbols for capacitors: (a) fixed capacitors, (b) variable capacitors.

(a-b)variable capacitors

(a) Polyester capacitor, (b) Ceramic capacitor, (c) Electrolytic capacitor


• Voltage Limit on a Capacitor:


The plate charge increases as the voltage increases. Also, the electric field intensity between two plates increases.

If the voltage across the capacitor is so large that the field intensity is large enough to break down the insulation of the dielectric, the capacitor is out of work. Hence, every practical capacitor has a maximum limit on its operating voltage.

Cvq According to:


• Current(i)-voltage(v) relationship of a Capacitor:


Taking the derivative of both sides

where

+

-

v

i

C


• Physical Meaning:


+

-

v

i

C

• when v is a constant voltage, then i=0; a constant voltage across a capacitor creates no current through the capacitor, the capacitor in this case is the same as an open circuit.

• If v is abruptly changed, then the current will have an infinite value that is practically impossible. Hence, a capacitor is impossible to have an abrupt change in its voltage except an infinite current is applied.


• Important Properties of capacitors:


1) A capacitor is an open circuit to dc.

2) The voltage on a capacitor cannot change abruptly.

Voltage across a capacitor: (a) allowed, (b) not allowable; an abrupt change is not possible.

4) A real, nonideal capacitor has a parallel-model leakage resistance. The leakage resistance may be as high as 100 MW and can be neglected for most practical applications. We will assume ideal capacitors.

3) Ideal capacitor does not dissipate energy. It takes power from the circuit when storing energy in its field and returns previously stored energy when delivering power to the circuit.




• A capacitor has memory.

• Capacitor voltage depends on the past history (-∞ →to) of the capacitor current. Hence, the capacitor has memory.

+

-

v C




• Energy Storing in Capacitor

+

-

v C

instantaneous power delivered to the capacitor is

energy stored in the capacitor:

note that

recall that


Ex. 6.1:(a) Calculate the charge stored on a 3-pF capacitor with 20 V across it. (b) Find the energy stored in the capacitor.


(a) since: ,Cvq

pC6020103 12 q

(a) The energy stored: pJ6004001032

1

2

1 122 Cvw

Solution


Ex. 6.2: The voltage across a 5- F capacitor is: Calculate the current through it.


(a) By definition current is:

V 6000cos10)( ttv

6105 dt

dvCi

6000105 6

)6000cos10( tdt

d

Att 6000sin3.06000sin10

dt

dvCi

Solution


Ex. 6.3: Determine the voltage across a 2-F capacitor if the current through it is :


Since:

mA6)( 3000teti

Assume that the initial capacitor voltage is zero.

Solution

3

0

3000

6106

102

1

dtev

tt

0

30003

3000

103tte

t

vidtC

v0

)0(1

,0)0(and v

V)1( 3000te


Ex. 6.4: Determine the current through a 200- F capacitor whose voltage is shown in Fig 6.9.


Solution

The voltage waveform can be described mathematically as:

otherwise0

43V 50200

31V 50100

10V 50

)(tt

tt

tt

tv


Ex. 6.4: Determine the current through a 200- F capacitor whose voltage is shown in Fig 6.9.


Solution

The Since i = C dv/dt and C = 200 F, we take the derivative of v(t) to obtain:

otherwise0

43mA10

31mA10

10mA10

otherwise0

4350

3150

1050

10200)( 6

t

t

t

t

t

t

ti

Thus the current waveform is :


Ex. 6.5: Obtain the energy stored in each capacitor in circuit below under dc condition.


Solution

Under dc condition, we replace each capacitor with an open circuit. By current division,

mA2)mA6(423

3

i

,V420001 iv

V840002 iv

mJ16)4)(102(2

1

2

1 232

111 vCw

mJ128)8)(104(2

1

2

1 232

222 vCw


Series and Parallel Capacitors


Parallel-connected N capacitors

equivalent circuit for the parallel capacitors.

Niiiii ...321




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

Niiiii ...321

dt

dvC

dt

dvC

dt

dvC

dt

dvCi N ...321

dt

dvC

dt

dvC eq

N

kK

1

Neq CCCCC ....321




Series-connected N capacitors

equivalent circuit for the series capacitors.

)(...)()()( 21 tvtvtvtv N




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

For N=2:




• Summary:

These results enable us to look the capacitor in this way: • 1/C has the equivalent effect as the resistance. The equivalent capacitor of capacitors connected in parallel or series can be obtained via this point of view, so is the Y-D connection and its transformation.


Ex. 6.6: Find the equivalent capacitance seen between terminals a and b of the circuit in the figure below.



Ex. 6.6: Find the equivalent capacitance seen between terminals a and b of the circuit in the figure below.


F4520

520

F302064

:seriesin are capacitors F5 and F20

capacitors F20 andF6 with the parallel in iscapacitor F4

capacitor. F60 the withseriesin iscapacitor F30

Solution

F20F6030

6030

eqC


Ex. 6.7: For the circuit given below, find the voltage across each capacitor.


Solution

mF10mF1

20

1

30

1

)2040(

1

eqC




Total charge

C3.0301010 3 vCq eq

This is the charge on the 20 mF and 30 mF capacitors, because they are in series with the 30 V source. ( A crude way to see this is to imagine that charge acts like current, since i = dq/dt)

V101030

3.03

2

2

C

qv

,V151020

3.03

1

1

C

qvTherefore,




Having determined v1 and v2, we can use KVL to determine v3 by

V530 213 vvv

Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage v3 and their combined capacitance is 40+20=60mF.

V51060

3.0

mF60 33

qv


Capacitors and Inductors JOSEPH HENRY (1797-1878)

American physicist, discovered inductance and constructed an electric motor.


• An inductor is made of a coil of conducting wire.


l

ANL

2

Inductance is the property whereby an inductor exhibits opposition to the change of current flowing through it, measured in henrys (H).



(H/m)104 7

0

0

2

r

l

ANL

turns.ofnumber:N

length.:l

area. sectionalcross: A

coretheoftypermeabili:

(a) Solenoidal inductor, (b) toroidal inductor, (c) chip inductor.



(a) air-core, (b) iron-core, (c) variable iron-core.

• Circuit symbols for inductors:


• Voltage(v) -current(i) relationship of an Inductor:


integrating both sides gives:

voltage-current relationship of an inductor.

+

-

v

i

L


• Physical Meaning:


• When the current through an inductor is a constant, then the voltage across the inductor is zero, same as a short circuit.

• No abrupt change of the current through an inductor is possible except an infinite voltage across the inductor is applied.

• The inductor can be used to generate a high voltage, for example, used as an igniting element.

+

-

v

i

L


• Important Properties of inductors:


1) An inductor is a shaort to dc.

2) The current through an inductor cannot change abruptly.

Current through an inductor: (a) allowed, (b) not allowable; an abrupt change is not possible.

4) A practical, nonideal inductor has a significant resistive component. This resistance is called the winding resistance. The nonideal inductor also has a winding capacitance due to the capacitive coupling between the conducting coils. We will assume ideal inductors.

3) Ideal inductor does not dissipate energy. It takes power from the circuit when storing energy in its field and returns previously stored energy when delivering power to the circuit.




• An inductor has memory.

• Inductor current depends on the past history (-∞ →to) of the inductor current. Hence, the inductor has memory.

+

-

v

i

L




• Energy Storing in Inductor

power delivered to the inductor is

energy stored in the inductor:

note that

+

-

v

i

L


Ex. 6.8: 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.


Solution

,H1.0andSince Ldt

diLv

V)51()5()10(1.0 5555 teetetedt

dv tttt

isstoredenergyThe

J5100)1.0(2

1

2

1 1021022 tt etetLiw


Ex. 6.9: Find the current through a 5-H inductor if the voltage across it is Also find the energy stored within 0 < t < 5s. Assume i(0)=0.


Solution

0,0

0,30)(

2

t

tttv

.H5and L)()(1

Since0

0 t

ttidttv

Li

A23

6 33

tt

tdtti

0

2 0305

1


Ex. 6.9: Find the current through a 5-H inductor if the voltage across it is Also find the energy stored within 0 < t < 5s. Assume i(0)=0.


Solution

0,0

0,30)(

2

t

tttv

5

0

65 kJ25.156

0

5

66060

tdttpdtw

thenisstoredenergytheand,60powerThe 5tvip

before.obtainedas

)0(2

1)5(

2

1)0()5( 2 LiLiww

kJ25.1560)52)(5(2

1 23

Alternatively


Ex. 6.10: Consider the circuit below in (a). Under dc conditions, find: (a) i, vC, and iL.

(b) the energy stored in the capacitor and inductor.


before.obtainedas

Solution


Ex. 6.10: Consider the circuit below in (a). Under dc conditions, find: (a) i, vC, and iL.

(b) the energy stored in the capacitor and inductor.


before.obtainedas

Solution:

,251

12Aii L

,J50)10)(1(2

1

2

1 22

cc Cvw

J4)2)(2(2

1

2

1 22 iL Lw

)(a :conditiondcUnder capacitor

inductor

circuitopen

circuitshort

)(b

V105 ivc


Series and Parallel Inductors


Series-connected N inductors

equivalent circuit for the sieries inductors.

Neq LLLLL ...321




Neq LLLLL ...321

• Applying KVL to the loop,

• Substituting vk = Lk di/dt results in

Nvvvvv ...321

dt

diL

dt

diL

dt

diL

dt

diLv N ...321

dt

diLLLL N )...( 321

dt

diL

dt

diL eq

N

KK

1




Paralel-connected N inductors

equivalent circuit for the parallel inductors.

Neq LLLL

1111

21




Neq LLLL

1111

21

• Using KCL,

• But

Niiiii ...321

t

t k

k

k otivdt

Li )(

10

t

t

t

t s

k

tivdtL

tivdtL

i0 0

)(1

)(1

0

2

01 t

tN

N

tivdtL 0

)(1

... 0

)(...)()(1

...11

00201

210

tititivdtLLL

N

t

tN

t

teq

N

kk

t

t

N

k k

tivdtL

tivdtL 00

)(1

)(1

01

01




• The inductor in various connection has the same effect as the resistor. Hence, the Y-Δ transformation of inductors can be similarly derived.


Ex. 6.11: Find the equivalent inductance of the circuit given below.


before.obtainedas


Ex. 6.11: Find the equivalent inductance of the circuit given below.


Solution:

10H12H,,H20:Series H42

H6427

427

H18864 eqL

: Parallel


Ex: Calculate the equivalent inductance for the inductive ladder network given in the circuit below.



Ex 6.12: For the circuit given below, If, Find:

Capacitors and Inductors .mA)2(4)( 10teti

,mA 1)0(2 i)0( (a)

1i

);(and),(),((b) 21 tvtvtv

)(and)((c) 21 titi




,mA 1)0(2 i)0( (a)

1i


)(and)((c) 21 titi

.mA4)12(4)0(mA)2(4)()(a 10 ieti t

mA5)1(4)0()0()0( 21 iii

H53212||42 eqL

mV200mV)10)(1)(4(5)( 1010 tt

eq eedt

diLtv

mV120)()()( 1012

tetvtvtv

mV80mV)10)(4(22)( 1010

1

tt eedt

ditv

isinductanceequivalentThe)(b

Solution:




,mA 1)0(2 i)0( (a)

1i


)(and)((c) 21 titi

Solution:

t t t dteidtvti0 0

10

121 mA54

120)0(

4

1)(

mA38533mA50

3 101010 ttt eet

e

t ttdteidtvti

0

10

20 22 mA112

120)0(

12

1)(

mA11mA10

101010 ttt eet

e

)()()(thatNote 21 tititi

capacitors and inductors(new) - faraday

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