TEST 2 SOLUTION

ELECTRIC CIRCUIT

Find Rab

60uf

40uf26uf100uf

50uf

100uf

25uf

150uf

25uf

200uf

50uf

Rab

a

b

Find Rab

24uf26uf100uf

50uf

100uf

25uf

150uf

25uf

200uf

50uf

Rab

a

b

Find Rab

50uf100uf

50uf

100uf

25uf

150uf

25uf

200uf

50uf

Rab

a

b

Find Rab

25uf100uf

100uf

25uf

150uf

25uf

200uf

50uf

Rab

a

b

Find Rab

25uf25uf

100uf

150uf

25uf

200uf

50uf

Rab

a

b

100uf

Find Rab

150uf

100uf

150uf

25uf

200uf

50uf

Rab

a

b

Find Rab

75uf

100uf

25uf

200uf

50uf

Rab

a

b

Find Rab

75uf

100uf

25uf

200uf

50uf

Rab

a

b

Find Rab

200uf

200uf

50uf

Rab

a

b

Find Rab

33.33ufRab

a

b

Find Rab

60uf

40uf26uf100uf

50uf

100uf

25uf

150uf

25uf

200uf

50uf

Rab

a

b

c

d e f

200ufa bc d25uf 50uf

e25uf 150uf

100uf

f50uf26uf

60uf 40uf

100uf

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

RLa b

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

a b

Remove the RL First thing to notice is that this circuit contained a dependent source, thus ‘normal’ method can’t be used. (because we need source to calculate the dependent source)

Therefore, turn off (shorted) the independent voltage source and take out (open) the independent current source.

We will put a 1A current source at terminal a-b. Then use linearity,

0

0

ivRth

Obtain the Thevenin equivalent circuit for terminal a-b

RTH: Obtain V1, V2 and V3.

+-

4 Ω 3 Ω

2 Ω

6 Ω

1 A

ix30ix

V1 V2 V3

5 Ω

a b

At V1,

042

1 211

VVV

Straightaway we get,

412 VVix

024 211 VVV

034 21 VV

Obtain the Thevenin equivalent circuit for terminal a-b

RTH: Obtain V1, V2 and V3.

+-

4 Ω 3 Ω

2 Ω

6 Ω

1 A

ix30ix

V1 V2 V3

5 Ω

a bAt V2,

0634

23212

VVVVV

0122

1244

1233 23212

VVVVV

0439 312 VVV

Obtain the Thevenin equivalent circuit for terminal a-b

RTH: Obtain V1, V2 and V3.

+-

4 Ω 3 Ω

2 Ω

6 Ω

1 A

ix30ix

V1 V2 V3

5 Ω

a bAt V3,

0530

31 323

xiVVV

0905815 23 xiVV

04

905815 2123

VVVV

04545101630 2123 VVVV

30354516 213 VVV

Obtain the Thevenin equivalent circuit for terminal a-b

Therefore RTH

VV 16.11

VV 51.02

VV 27.03 +-

4 Ω 3 Ω

2 Ω

6 Ω

1 A

ix30ix

V1 V2 V3

5 Ω

a b

89.01

13

0

0 VVivRth

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

a b

Obtain VTH: Find V1, V2 and V3

Straightaway we get,

412 VVix

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

a b

Obtain VTH: Find V1, V2 and V3

02

01284

4 121

VVV

02128

44 121

VVV

0256216 121 VVV

2403 21 VV

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

a b

Obtain VTH: Find V1, V2 and V3

0364

32212

VVVVV

012

44122

1233 32212

VVVVV

0639 212 VVV

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

a b

Obtain VTH: Find V1, V2 and V3

0530

34 323

xiVVV

09035560 323 xiVVV

0905860 23 xiVV

04

905860 2123

VVVV

120163545 321 VVV

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

a b

Obtain VTH: Find V1, V2 and V3

VV 1101

VV 902

VV 1203

VVVVVVab 1012011013

Obtain the Thevenin equivalent circuit for terminal a-b

+-

+-

4 Ω 3 Ω

2 Ω

6 Ω128 V

4 A

ix30ix

V1 V2 V3

5 Ω

a b

Obtain RTH: Find V1, V2 and V3

VV 1101

VV 902

VV 1203

VVVVVVab 1012011013

Obtain the Thevenin equivalent circuit for terminal a-b

+-

0.89 Ω

10V

a

b

Finally

Calculate the maximum power transfer

+-

0.89 Ω

10V

a

b

RL

Finally

WRVP

TH

TH 09.28)89.0(4

)10(4

max22

The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)

10Ω

2Ω

2H0.1Ω20 A

40Ω

t=0

iL+

-v0

The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)

10Ω

2Ω

0.1Ω20 A

40ΩiL+

-v0

t<0

Because of the inductor shorted, iL(0)=20A

The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)

t>0

10Ω

2Ω

2H 40ΩiL+

-v0

1082)40||10(2eqR

2.0102

eqRL

The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)

t>0 2.0

Aeiti t

L 0)(

Ae t520

The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)

t>0Aeti t

L520)(

10Ω

2Ω

2H 40ΩiL+

-v0

Using current divider

)(1040

10)(40 titi L

Aeti t540 20)2.0()(

Aeti t540 4)(

Vetitv tL

50 160)(40)( Therefore:

The voltage pulse across a 0.5uF capacitor described by following equation

sstVetstVst

v(t))(t

14104

00

1

Solutions

• Current, i from equation

• Power, p from equation

s1 A,24105.0s 1s 0 A,24105.0s 0 A,00105.0

116

6

6

teett

itt

dtdvCi

s W,

s s W,s W,

18241082400

1211 teeetttt

pttt

dtdvCvvip

Solutions (cont.)

• Energy, w from equation

s1 J,4165.021

s 1s 0 J,4165.021

s 0 J,0

1212

22

tee

ttt

t

w

tt

Solutions (cont.)

• Energy is being stored in the capacitor whenever the power is positive. Hence the energy is being stored in the interval 0 – 1 s.

• Energy is being delivered by the capacitor whenever the power is negative. Hence the energy is being delivered for all t > 1 s.