Circuit

1.4

Circuit Theorems

1. vTH , RTH = ?

(C) 1 V, 3W

2W

5 W 6

(D) -1 V,

6 W 5

4. A simple equivalent circuit of the 2 terminal 6W

6V

vTH, RTH

network shown in fig. P1.4.4 is

R

Fig. P.1.4.1

(A) 2 V, 4 W

(B) 4 V, 4 W

(C) 4 V, 5 W

(D) 2 V, 5 W

i

v

Fig. P.1.4.4

2. i N , R N = ? 2W

R

2W

R 4W

15 V

v

iN, RN

(A)

(B)

Fig. P.1.4.2

(A) 3 A,

R R

10 W 3

i

(B) 10 A, 4 W

(C) 1,5 A, 6 W

i

(D) 1.5 A, 4 W

(C)

(D)

5. i N , R N = ?

3. vTH , RTH = ?

2W

2W

3W

2A

1W

vTH, RTH

6A

6 W 5

(B) 2 V,

3W

iN RN

Fig. P.1.4.5

Fig. P.1.4.3

(A) -2 V,

4W

5 W 6

(A) 4 A, 3 W

(B) 2 A, 6 W

(C) 2 A, 9 W

(D) 4 A, 2 W

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34

Circuit Theorems

6. vTH , RTH = ?

Chap 1.4

The value of the parameter are vTH

RTH

iN

RN

(A)

4 V

2 W

2 A

2 W

(B)

4 V

2 W

2 A

3 W 1.2 W 5 W

25 W

30 W 20 W

vTH, RTH

5V 5A

Fig. P.1.4.6

(A) -100 V, 75 W

(B) 155 V, 55 W

(C) 155 V, 37 W

(D) 145 V, 75 W

(C)

8 V

1.2 W

30 A 3

(D)

8 V

5 W

8 A 5

10. v1 = ? 2W

1W

7. RTH = ? 6W

2W

8V

6W

3W

1W

+ v1

6W

18 V

–

6W

2A

RTH

Fig. P.1.4.10

5V

(A) 6 V

(B) 7 V

(C) 8 V

(D) 10 V

Fig. P.1.4.7

(A) 3 W

(B) 12 W

(C) 6 W

(D) ¥

11. i1 = ? 4 kW

i1

20 V

4 kW

6 kW

8. The Thevenin impedance across the terminals ab of the network shown in fig. P.1.4.8 is

12 V

24 V

3 kW

4 kW

a 3W

Fig. P.1.4.11 6W

2A

8W

2V 8W

b

(B) 6 W

(C) 6.16 W

(D)

(B) 0.75 mA

(C) 2 mA

(D) 1.75 mA

Statement for Q.12–13:

Fig. P.1.4.8

(A) 2 W

(A) 3 A

A circuit is given in fig. P.1.4.12–13. Find the Thevenin equivalent as given in question..

4 W 3

10 W

16 W

x

y

9. For In the the circuit shown in fig. P.1.4.9 a network 2W

8W

3W

x’

RTH 4V

40 W

5V

and its Thevenin and Norton equivalent are given

iN

2A vTH

Fig. P.1.4.9

y’

Fig. P.1.4.12–13 RN

12. As viewed from terminal x and x¢ is (A) 8 V, 6 W

(B) 5 V, 6 W

(C) 5 V, 32 W

(D) 8 V, 32 W

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1A

Chap 1.4

Circuit Theorems

13. As viewed from terminal y and y¢ is (A) 8 V, 32 W

(B) 4 V, 32 W

(C) 5 V, 6 W

(D) 7 V, 6 W

35

(C) 0 A, 20 W

(D) 0 A, -20 W

19. vTH , RTH = ? i1

6W

14. A practical DC current source provide 20 kW to a 50 W load and 20 kW to a 200 W load. The maximum

3i1

power, that can drawn from it, is (A) 22.5 kW

(B) 45 kW

(C) 30.3 kW

(D) 40 kW

RN

Fig. P1.4.19

Statement for Q.15–16: In the circuit of fig. P.1.4.15–16 when R = 0 W , the current iR equals 10 A.

(A) 0 W

(B) 1.2 W

(C) 2.4 W

(D) 3.6 W

20. vTH , RTH = ?

2W

4W

iN,

4W

2W

4V +

E

4W

2W

R

4A

iR

vTH RTH

v1

5W

0.1v1

–

Fig. P.1.4.15–16. Fig. P.1.4.20

15. The value of R, for which it absorbs maximum power, is

(A) 8 V, 5 W

(B) 8 V, 10 W (D) 4 V, 10 W

(A) 4 W

(B) 3 W

(C) 4 V, 5 W

(C) 2 W

(D) None of the above

21. RTH = ? 3W

2W

16. The maximum power will be

+

(A) 50 W

(B) 100 W

(C) 200 W

(D) value of E is required

vx 4

4V

17. Consider a 24 V battery of internal resistance

the current drawn from the battery is i . The current drawn form the battery will be i 2 when RL is equal to (A) 2 W

(B) 4 W

(C) 8 W

(D) 12 W

Fig. P.1.4.21

(A) 3 W

(B) 1.2 W

(C) 5 W

(D) 10 W

22. In the circuit shown in fig. P.1.4.22 the effective resistance faced by the voltage source is 4W

18. i N , R N = ? 5W

10 W i1 20i1

iN,

30 W

vs

RN

i

i 4

Fig. P.1.4.22 Fig. P.1.4.18

(A) 2 A, 20 W

RTH

–

r = 4 W connected to a variable resistance RL . The rate of heat dissipated in the resistor is maximum when

vx

(B) 2 A, -20 W

(A) 4 W

(B) 3 W

(C) 2 W

(D) 1 W

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36

Circuits Theorems

Chap 1.4

23. In the circuit of fig. P1.4.23 the value of RTH at

ix

terminal ab is 16 V

0.75va

RL

3W

0.9 A 2W

Fig. P.1.4.26–27

8W

a

–

26. The value of RL will be

va +

4W

9V

b

Fig. P.1.4.23

(A) -3 W 8 (C) - W 3

(B) 3 W

(C) 1 W

(D) None of the above

27. The maximum power is

9 W 8

(B)

(A) 2 W

(D) None of the above

(A) 0.75 W

(B) 1.5 W

(C) 2.25 W

(D) 1.125 W

28. RTH = ?

24. RTH = ?

-2ix

200 W –

va 100

va +

100 W

50 W

RTH

100 W

300 W ix

Fig. P.1.4.24

(A) ¥ (C)

(B) 0

3 W 125

(D)

125 W 3

maximum power if RL is equal to

800 W

(C) 200 W

(D) 272.8 W

29. Consider the circuits shown in fig. P.1.4.29 ia

2W 6W

6W 2W

2W

RL

3i

RTH

–

(B) 136.4 W

100 W

200 W

vx

(A) 100 W

i 6V

+

Fig. P.1.4.28

25. In the circuit of fig. P.1.4.25, the RL will absorb

40 W

100 W

0.01vx

12 V 8V

12 V

Fig. P.1.4.25

6W

(A)

400 W 3

(B)

2 kW 9

(C)

800 W 3

(D)

4 kW 9

ib

2W 6W

6W

2W

2W

Statement for Q.26–27: In the circuit shown in fig. P1.4.26–27 the

18 V

6W

3A

maximum power transfer condition is met for the load RL .

Fig. P.1.4.29a & b

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12 V

Chap 1.4

Circuit Theorems

37

33. If vs1 = 6 V and vs 2 = - 6 V then the value of va is

The relation between ia and ib is (A) ib = ia + 6

(B) ib = ia + 2

(A) 4 V

(B) -4 V

(C) ib = 15 . ia

(D) ib = ia

(C) 6 V

(D) -6 V

30. Req = ?

34. A network N feeds a resistance R as shown in fig. 12 W

P1.4.34. Let the power consumed by R be P. If an

4W

identical network is added as shown in figure, the Req

6W

power consumed by R will be

2W

18 W

6W 9W R

N

N

N

R

Fig. P.1.4.30

(A) 18 W (C)

(B)

36 W 13

72 W 13

Fig. P.1.4.34

(D) 9 W

31. In the lattice network the value of RL for the maximum power transfer to it is

(A) equal to P

(B) less than P

(C) between P and 4P

(D) more than 4P

35. A certain network consists of a large number of ideal linear resistors, one of which is R and two constant ideal source. The power consumed by R is P1

7W

when only the first source is active, and P2 when only

6

the second source is active. If both sources are active

W

simultaneously, then the power consumed by R is

5

W

RL

9W

(A) P1 ± P2

(B)

(C) ( P1 ± P2 ) 2

(D) ( P1 ± P2 ) 2

P1 ± P2

Fig. P.1.4.31

(A) 6.67 W

(B) 9 W

36. A battery has a short-circuit current of 30 A and an

(C) 6.52 W

(D) 8 W

open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 W, the power dissipated by the bulb is

Statement for Q.32–33: A circuit is shown in fig. P.1.4.32–33.

(A) 80 W

(B) 1800 W

(C) 112.5 W

(D) 228 W

12 W 1W

3W

3W

37.

1W

va

1W

following

results

were

obtained

from

measurements taken between the two terminal of a

+ vs1

The

vs2

resistive network

–

Fig. P.1.4.32–33

32. If vs1 = vs 2 = 6 V then the value of va is

Terminal voltage

12 V

0V

Terminal current

0A

1.5 A

The Thevenin resistance of the network is

(A) 3 V

(B) 4 V

(A) 16 W

(B) 8 W

(C) 6 V

(D) 5 V

(C) 0

(D) ¥

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38

Circuit Theorems

Chap 1.4

38. A DC voltmeter with a sensitivity of 20 kW/V is

Solutions

used to find the Thevenin equivalent of a linear network. Reading on two scales are as follows

1. (B) vTH =

(a) 0 - 10 V scale : 4 V (b) 0 -15 V scale : 5 V The

Thevenin

voltage

the

Thevenin

2. (A) 2W

resistance of the network is 16 1 (A) V, MW 3 15

32 V, 3

200 kW 3

(D) 36 V,

200 kW 3

(B)

2 MW 15

(C) 18 V,

and

( 6)( 6) = 4 V, RTH = ( 3||6) + 2 = 4 W 3+ 6

isc

Fig. S.1.4.2

15 10 2 =6W R N = 2 ||4 + 2 = W, v1 = 1 1 1 3 + + 2 2 4 v isc = i N = 1 = 3 A 2

+ RL

4W

15 V

39. Consider the network shown in fig. P.1.4.39.

Linear Network

2W

v1

vab –

Fig. P.1.4.39

The power absorbed by load resistance RL is

3. (C) vTH =

shown in table : RL

10 kW

30 kW

P

3.6 MW

4.8 MW

(2)( 3)(1) 5 = 1 V, RTH = 1||5 = W 3+ 3 6

4. (B) After killing all source equivalent resistance is R Open circuit voltage = v1

The value of RL , that would absorb maximum

5. (D) isc =

6´ 4 = 4 A = i N , R N = 6 ||3 = 2 W 4+2 2W

power, is (A) 60 kW

(B) 100 W

(C) 300 W

(D) 30 kW

isc 3W

4W

6A

40. Measurement made on terminal ab of a circuit of fig.P.1.4.40 yield the current-voltage characteristics

Fig. S1.4.5

shown in fig. P.1.4.40. The Thevenin resistance is 6. (B) RTH = 30 + 25 = 55 W, vTH = 5 + 5 ´ 30 = 155 V

i(mA) +

30 Resistive Network

20 10 -4

-3 -2

-1

0

vab –

1

2

v

a

7. (C) After killing the source, RTH = 6 W

b

6W

6W

Fig. P.1.4.40

RTH

(A) 300 W

(B) -300 W

(C) 100 W

(D) -100 W Fig. S.1.4.7 ***********

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Chap 1.4

Circuit Theorems

8. (B) After killing all source, RTH = 3||6 + 8 ||8 = 6 W

6W

8W 8W

If we Thevenized the left side of xx¢ and source transformed right side of yy¢ 4 8 + 8 24 RTH = 8 ||(16 + 8) = 6 W = 5 V, vxx ¢ = vTH = 1 1 + 8 24

a

3W

39

b

Fig. S1.4.8

13. (D) v yy¢ = vTH

9. (D) voc = 2 ´ 2 + 4 = 8 V = vTH RTH = 2 + 3 = 5 W = R N ,

iN =

vTH 8 = A RTH 5

4 8 + 24 8 = 7 V, R = ( 8 + 16)||8 = 6 W = TH 1 1 + 24 8

14. (A)

10. (A) By changing the LHS and RHS in Thevenin i

equivalent 1W

1W

2W

1W

Fig. S1.4.14

+ 6W

4V

RL

r

v1

12 V

–

2

2

æ ir ö æ ir ö ÷ 50 = 20 k, ç ÷ 200 = 20 k ç è r + 50 ø è r + 200 ø Þ

( r + 200) 2 = 4( r + 50) 2

Fig. S1.4.10

Pmax

i = 30 A,

4 12 + + +2 =6 V 1 1 1 v1 = 1 1 1 + + 1+1 6 1+2

r = 100 W

( 30) 2 ´ 100 = = 22.5 kW 4

15. (C) Thevenized the circuit across R, RTH = 2 W 4W

2W

2W

11. (B) By changing the LHS and RHS in Thevenin 2W

4W

equivalent i1

2 kW

4 kW

20 V

2 kW

Fig. S1.4.15 6V

8V

17. (D) RL = r = 4 W, i =

Fig. S1.4.11

i1 =

20 - 6 - 8 = 0.75 mA 2k + 4k + 2k

24 3 = RL¢ + 4 2

12. (B) 8W

16 W

x

2

æ 10 ö ÷ ´ 2 = 50 W 16. (A) isc = 10 A, RTH = 2 W, Pmax = ç è 2 ø

y

Þ

24 =3 A 4+4

R¢L = 12 W

18. (C) i N = 0,

8W

1- i1 10 W

5W i1

4V

+

8V

20i1

30 W

1A

vtest –

x’

y’

Fig. S1.4.18

Fig. S1.4.12

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40

Circuit Theorems

20 i1 = 30 i1 - 10(1 - i1 )

Þ

i1 = 0.5 A

22. (B) vs = 4 ´

vtest = 5 ´ 1 + 30 ´ 0.5 = 20 V v R N = test = 20 W 1

3i 4

vs = 3W i

Þ

voc voc - 9 + + 0.75 va = 0 4 8

23. (C) voc = vab = -va ,

19. (B) Circuit does not contains any independent source, vTH = 0 i1

6W

+ 4W

3i1

Chap 1.4

1A

vtest

2 voc + voc - 9 + 6( -voc ) = 0 , voc = - 3 V If terminal ab is short circuited, va = 0 9 v -3 -8 A and RTH = oc = isc = = W 8 isc 9 8 3 24. (D) Using source transform i1

–

100 W

200 W – +

Fig. S1.4.19

va

va

1A

vtest –

Applying 1 A at terminal, i1 = -1 A vtest vtest - 3( -1) . V + = 1 Þ vtest = 12 4 6 v RTH = test = 12 . W 1

+

Fig. S1.4.24

va = 100 i1 + 200 i1 + 50( i1 + 1) va = 100 i1 - va

Þ

va = 50 i1

50 i1 = 300 i1 + 50 i1 + 50

20. (B) 4V isc

Þ

i1 = -

1 A 6

æ 1 ö 125 vtest = 50ç 1 - ÷ = W 6ø 3 è 25. (C)

5W

0.1v1

50 W

100 W

2i 40 W

+

i

Fig. S1.4.20 200 W

6V

v1 = 4 + 5 ´ 0.1v1

Þ

3i

v1 = 8 V –

v1 = voc = vTH Fig. S1.4.25a

For isc , v1 = 0 4 v isc = A, RTH = oc = 10 W 5 isc 21. (D) vx = 2

vx +4 4

6 = 200 i - 40 ´ 2 i

Þ

i=

1 A 20

voc = 100 ´ 3i + 200 ´ i = 25 V Þ

vx = 8 V = voc

40 W

3W

2W

100 W

v1 i

isc 4V

voc

6V

vx 4

200 W

isc 3i1

Fig. S1.4.25b Fig. S1.4.21

If terminal is short circuited, vx = 0 4 v 8 isc = = 0.8 A, RTH = oc = = 10 W 2+3 isc 0.8

6 15 15 3 40 V, i = A = v1 = = 1 1 1 4 4 ´ 200 160 + + 40 200 100

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Chap 1.4

isc =

Circuit Theorems

16 3´ 3 3 v 25 800 A, RTH = oc = + = = W 4 ´ 100 160 32 isc 3 32 3

41

30. (D) Changing the D to Y 12 W

26. (B) ix + 0.9 = 10 ix

Þ

2 W 3

2W

ix = 0.1 A 10ix

2W

1W

Req

18 W

+

6W 9W

16 V

voc

3W

0.9 A

Fig. S1.4.30

–

Fig. S1.4.26

voc = 3 ´ 10 ix = 30 ix isc = 10 ix = 1 A, RTH

æ æ 2 öö Req = 18 ||çç 14 + 10 ||ç 6 + ÷ ÷÷ = 18 ||(14 + 4) = 9 W 3 øø è è

Þ voc = 3 V 3 = = 3W 1

31. (C) RTH = 7 ||5 + 6 ||9 = 6.52 W 7W 2

3 = 0.75 W 4´ 3

6 W

RTH W

27. (A) vTH = voc = 3 V, RL = 3 W, Pmax =

5

28. (A) ix = 1 A , vx = vtest -2ix

9W

Fig. S1.4.31 100 W

0.01vx 100 W

+

300 W

1A

vtest –

For maximum power transfer RL = RTH = 6.52 W 32. (D) The given circuit has mirror symmetry. It is modified and redrawn as shown in fig. S.1.4.32a.

ix

6W

800 W

Fig. S1.4.28

1W

vtest = 1200 - 800 ix - 3vtest

4 vtest = 1200 - 800 = 400 v RTH = test = 100 W 1

Þ

1W 3W

vtest = 100 (1 - 2 ix ) + 300 (1 - 2 ix - 0.01vx ) + 800 Þ

6W

6V

3W 2W

2W

vtest = 100 V

+ va

6V

–

Fig. S.1.4.32a

29. (C) In circuit (b) transforming the 3 A source in to

Now in this circuit all straight-through connection

18 V source all source are 1.5 times of that in circuit

have been cut as shown in fig. S1.4.32b

(a). Hence ib = 15 . ia . ib

6W 1W

2W 6W

3W

6W 2W

2W

2W

18 V 18 V

va –

12 V

Fig. S.1.4.32b

6W

Fig. S1.4.29

+

va =

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6 ´ (2 + 3) =5 V 2 + 3+1

6V

42

Circuit Theorems

33. (B) Since both source have opposite polarity, hence short circuit the all straight-through connection as shown in fig. S.1.4.33 6W 1W 3W 2W

+ va

For 0 -50 V scale Rm = 50 ´ 20 k = 1 MW 4 For 4 V reading i = ´ 50 = 20 mA 10 vTH = 20mRTH + 20m ´ 200 k = 4 + 20mRTH 5 For 5 V reading i = ´ 50m = 5 mA 50

...(i)

vTH = 5m ´ RTH + 5m ´ 1M = 5 + 5mRTH

...(ii)

Solving (i) and (ii) 16 200 V, RTH = vTH = kW 3 3

6V

–

Fig. S1.4.33

va = -

39. (D) v10 k = 10 k ´ 3.6m = 6

6 ´ ( 6 ||3) = -4 V 2+1

v30 k = 30 k ´ 4.8m = 12 V 6 =

10 vTH 10 + RTH

12 =

30 vTH 30 + RTH

34. (C) Let Thevenin equivalent of both network RTH

Chap 1.4

RTH

RTH

Þ

Þ

10 vTH = 6 RTH + 60

5 vTH = 2 RTH + 60

RTH = 30 kW vTH

R

vTH

R

40. (D) At v = 0 , isc = 30 mA At i = 0, voc = - 3 V v -3 RTH = oc = = - 100 W isc 30m

Fig. S1.4.34 2

æ VTH ö ÷÷ R P = çç è RTH + R ø æ ç VTH P¢ = ç R ç ç R + TH è 2

vTH

2

ö ÷ æ VTH ÷ R = 4ç ç2R + R ÷ è TH ÷ ø

2

************

ö ÷÷ R ø

Thus P < P ¢ < 4 P 35. (C) i1 =

P1 P2 and i2 = R R

using superposition i = i1 + i2 =

P1 ± R

P2 R

i 2 R = ( P1 ± P2 ) 2 36. (C) r = P=

voc = 1. 2 W isc

24 2 ´ 2 = 112.5 W (1. 2 + 2) 2

37. (B) RTH =

38. (A) Let

voc 12 = =8W isc 15 .

1 1 = = 50 mA sensitivity 20 k

For 0 -10 V scale Rm = 10 ´ 20 k = 200 kW www.nodia.co.in