6.1 the inductor is a passive element ( can not generate energy) represented graphically as a coiled...

6.1 The Inductor L

v i

div = L

dt

Is a passive element ( can not generate energy)

Represented graphically as a coiled wire

Symbolized by the letter L

Measured in henrys (H)

The voltage drop across the inductor the inductor and the current in the direction of voltage drop are related as follows

However the inductor is capable of storing energy in the magnetic field which can be released

i v

100 mH

5t

0 t < 0

10te t > 0

i =

Example 6.1

5t

0 t < 0

10te t > 0

i =

(b) At what instant of time is the current maximum ?

5t 5t10( 5te e ) A s di

=dt

5t10e (1 5t) = 0= (1 5t) 0 1

t 0.25s s

Since the maximum will occur were the derivative of the current is zero

0di

=dt

0.2

0.736

5t

0 t < 0

10te t > 0

i =

(c) Find the voltage across the terminals of the inductor ?

i v

100 mH

div = L

dt5t 5t10( 5te e ) (0.1)= 5te (1 5t) V = t > 0

0 V= t < 0

(d) Sketch the voltage across the terminals of the inductor ?

5t

0 t < 0

10te t > 0

i =

(e) Are the voltage and the current at a maximum at the same time ?

i v

100 mH

5t

0 t < 0

e (1 5t) t > 0

v =

0.2

0.736

Since the voltage across the inductor is proportional to di/dt not i

the voltage maximum is not at the same time as current maximum

5t

0 t < 0

10te t > 0

i =

(f) At what instant of time does the voltage change polarity ?

i v

100 mH

5t

0 t < 0

e (1 5t) t > 0

v =

0.2

0.736

At t = 0.2 s which corresponds to the moment when di/dt is passing through zero changing sign

5t

0 t < 0

10te t > 0

i =

(g) Is there ever an instantaneous change in voltage across the inductor ? If so at what time ?

i v

100 mH

5t

0 t < 0

e (1 5t) t > 0

v =

0.2

0.736

Yes, At t = 0 s

Note that the voltage can change instantaneously across the terminals of an inductor

Current in an Inductor in Terms of the Voltage Across the Inductor

L

v

i

div = L

dtSince

di

v = Ldt

dt dt

Multiplying both side by the differential dt

div = Ldt

dt

dt vdt = Ldi Ldi = vdtOR

Integrating both side with respect to the differentials di and dt we have

)

0 0

i( )t t

t( ti

L dx = v d Note we used x and the variables of integration, were i and t become limits on the integrals

( x and become dummy variables )

) 0

t

0

t

L t t di( ) - i( = v 0

t

0

t

1i( ) = v id (

Lt t )

10t

0 t < 0

20te t > 0( )t

v =

Example 6.2 The voltage pulse applied to the 100mH inductor. Assume i = 0 for t ≤ 0

v(t) i

(a) Sketch the voltage v(t) ?

10t

0 t < 0

20te t > 0( )t

v =v(t) i(t)

(b) Find the inductor current i(t) ?

101 20 e 0

0.1d

t

0

i(t) = 10 10= 2 1 10 e e A t > 0t tt

(c) Sketch the current i(t) ?

Note the inductor current i(t) approaches a constant vlue of 2 A as t increases.

Power and Energy in the Inductor

L

v

i

div = L

dt

0

t

0

t

1i( ) = v id (

Lt t )

p = vi

di= L i

dt

di= Li

dtdw

=dt

dw = Li di

Integrating both side with respect to the differentials dw and di were w= 0 if i= 0

0

dy 0

w i

dx = L y 12

2w = Li

The Inductor

L

v i

div = L

dt

0

t

0

t

1i( ) = v id (

Lt t )

12

2w = Lidi

p = Lidt

dvi = C

dt

0

t

0

t

1v( ) = i vd (

Ct t )

12

2w = Cv

L

v

i

dvp = Cv

dt

The Capacitor

L

v

i

div = L

dt

0

t

0

t

1i( ) = v id (

Lt t )

12

2w = Li

L

v

i

dvi = C

dt

0

t

0

t

1v( ) = i vd (

Ct t )

12

2w = Cv

dip = Li

dtdv

p = Cvdt

Inductors in Series

1 2

1 1 1 1

equ nL L L L

Therefore inductor combine like resistor

5.1.1 Capacitors in Series

5.1.1 Capacitors in Parallel

Therefore capacitor combine like conductor

Example 5.1A voltage source is applied to a 5-F capacitor as shown. Sketch the capacitor current and the stored energy as a function of time.

dv ti t C

dt

5dv t

dt

215

2 sv t

21

2w t C v t

Example 5.2A current source is applied to a 5-F capacitor. Sketch the capacitor voltage as a function of time.

1

5

t

sv t i d

The capacitor voltage is related to area Under current source. For example area at t = 1 s is

For example area at t=3 s is 10+10+5 = 25 hence V(3s)=25/5= 5V.

V(1s)=10/5= 2V.

Area of Triangle is (1)(10)=10

Example 5.3A current source is applied to a 5-H inductor as shown. Sketch

the voltage across the inductor versus time.

di tv t L

dt

21

2w t Li t

Example 5.4 A voltage source is applied to a 5-H inductor as

shown. Sketch the inductor current versus time.

1 t

i t v dL

1 t

i t v dL

EE 202-01 – Fall 2012(121)--Quiz 4

012 1 2t

( )i t

2

1( )i t ( )cv t

1 F

For the circuit shown above , if no energy initially stored in the capacitor, find the followings:

 (b) The energy stored at the capacitor at t = 1 s ?

(a) ( ) ?cv t

012 1 2t

( )i t

2

1( )i t ( )cv t

1 F

c c1t 2 ( ) 0 + ( )1v v 0 0 0

t

t d

0

c c 01( ) ( ) + ( )(a) v v

t

t

t i dc t

Solution

c c2

12 t 1 ( ) 1 + ( ) ( 2) 0 ( 2)1v v 2

t

t d t t

c c1

11 t 0 ( ) 0 + ( ) 0 1 1 1v v 1

t

t d

2

c c0

10 t 2 ( ) + ( ) 11 2v v 0

ttt d

c c2

1t 2 ( ) 0 + ( ) 0 3 31v v 2

t

t d

012 1 2t

( )i t

2

1( )i t ( )cv t

1 F

2

c c1W (1) C (1)2 v

 (b) The energy stored at the capacitor at t = 1 s ?

2c

0 t 2( 2) 2 t 1

1 1 t 0( )1 0 t 2

23 t 2

(a) v

t

tt

22 91 1 (1) 1 J2 82