1.4 Circuit Theorems

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
www.nodia.co.in
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
www.nodia.co.in
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
www.nodia.co.in
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
www.nodia.co.in
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) ¥
www.nodia.co.in
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
***********
www.nodia.co.in
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
www.nodia.co.in
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
www.nodia.co.in
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 =
www.nodia.co.in
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