Networks Chapter 1 N etwork Elements and Basic Laws
1.3
Chapter 2 Network Theorems
1.21
Chapter 3
Transient Analysis (Ac and Dc) 1.40
Chapter 4 Two-Port Networks
1.73
Chapter 5 Resonance 1.93
Unit 01.indd 1
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5:10:50
Hints/Solutions | 1.3
Chapter 1 Network elemeNts aNd BasiC laws exerCises Practice Problems 1 Directions for questions 1 to 21: Select the correct alternative from the given choices. 1. The current in the 2 W resistor ‘I’ is 1Ω
1Ω
+
V I
10V –
2
1Ω
2Ω
1 (A) A 8
3 (B) A 8
+ I – 8V
1A
21 (C) A 8
23 (D) A 8
2. The voltage eo in the figure is
10 Ω
IA
16 V
12 Ω
6Ω
R1
R1 R2
(B) 4 A
3 Ω 2
(D)
8 A 3
+
Vx
(A)
25 A 3
– 2Ω
2 Vx
1Ω
1A
(B)
25 A 6
(C) 2 A
(D) 3 A
8. In the circuit shown in the figure, R2
A
2 (C) 3 + Ω (D) 3
2 Ω 3
(B)
4
(C) 1 A
+ 25 V –
B
(A)
4
4 4
I
A R2
2
4
3Ω
o+ eo o–
(A) 48 V (B) 24 V (C) 36 V (D) 28 V 3. The driving point impedance of the infinite ladder network shown in the figure is _____. Given R1 = 3 W and R2 = 2 W R1
(A) 2 A
2
7. In the circuit shown in the figure, the current I is
2Ω
8A
(A) 132 J (B) 98 J (C) 144 J (D) 168 J 6. In the circuit of the given figure, the source current ‘I’ is
2Ω
2Ω
2Ω 6Ω
21
4. The current wave formed in a pure resistor of 5 W is shown in the figure. The power dissipated in the resistor is IA
B
2Ω 6Ω
RAB =? (A)
6
6Ω
21 W 4
(B)
5 W 6
(C) 10 W
(D) 8 W
9. What should be the value of current ‘I’ to have zero current flowing through AB? 1Ω 0
2
+ – 5V
6 t sec
4
(A) 20 W (B) 45 W (C) 60 W (D) 90 W 5. Figure shows the waveform of the current passing through an inductor of resistance 1 W and inductance 2 H. The energy absorbed by the inductor in the first four seconds is _____. I
1V
A 1Ω
B
– + 1 A 1Ω
IA
(A) 2 A (B) 4 A (C) −4 A (D) −2 A 10. The two electrical sub-networks N1 and N2 are connected through three resistors, as shown in the figure. The voltage across the 2 W resistor is 8 V and the 3 W resistor is 6 V. The voltage across the 4 W resistor is +
6A N1
8Ω – 2Ω
4Ω
N2 3Ω
2s
4s
+
t in sec
(A) 24 V
Unit 01.indd 3 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 5
6Ω –
(B) −24 V
(C) 8 V
(D) −8 V
4/25/2019 10:19:26 AM 03/05/2017 15:10:53
1.4 || Part Networks 3.6 III • Unit 1 • Networks 11. In the circuit shown in the figure, the voltages V1, V2, and V3 are 3Ω
6Ω 1Ω
V1
V2
6Ω
16. The current i4 in the circuit of the figure equal to
6Ω
i0 = 7 A
1Ω V3
8 16 , ,4 3 3
16 8 (C) , ,4 3 3
17.
5∠ 0 ° A
12. In the network shown in the figure, the current in resistor R is
B 2Ω 10 V
3Ω
5A
(A) 2 A
4A
D
(B) 3 A
3Ω
10 ∠ 60° A
10 3 10 3 ∠ − 90°Amp ∠90° A (B) 2 2 (C) 5∠60° A (D) 5∠ − 60° A 18. A delta-connected network with its Wye equivalent is shown in the given figure. The resistance R1, R2, and R3 (in Ohm) are, respectively, (A)
R C
i1
For the circuit shown in the figure, the instantaneous current i1(t) is
3A
2A
−j 2 Ω
j 2Ω
16 8 (D) 4, , 3 3
A
(B) −12 A (D) None of these
(A) 12 A (C) 4 A
16 8 8 , , 3 3 3
(B)
i3 = 4 A
i 4 =?
12V
(A)
i2 = 3 A
i1 = 5 A
(C) 4 A
(D) 9 A
a
13. In the following circuit, the voltage AB is A
50 V
5Ω
40 V
4Ω
30V
3Ω
a B
5Ω b
R1 30Ω
⇒ R2
c
15 Ω
R3 c
b 2Ω
(A) 35 V
(B) 28.2 V
(C) 38.3 V
(D) 42.6 V
(A) 1.5, 3, and 9 (B) 3, 9, and 1.5 (C) 9, 3, and 1.5 (D) 3, 1.5, and 9 19. The voltage e0 in the figure 2Ω
14. A network contains linear resistors that are connected in series across an ideal voltage source. If all the resistances are halved and the voltage is doubled, then the voltage across each resistor becomes (A) doubled (B) halved (C) not changed (D) None of these 15. Obtain the equivalent capacitance of the network given
+ 10 Ω
8A
(A) 48 V 20.
5Ω
1µF
1µF
12 Ω
1µF
1µF
(B) 0.8 mF
Unit 01.indd 4 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 6
(D) 28 V
x 2Ω
I +
6Ω
3Ω
50 V y
1µF
1µF
b
(A) 1mF
(C) 36 V
–
1µF 1µF
8Ω
12 Ω e 0 −
6Ω
(B) 24 V
a 1µF
16 V
(C) 1.9 mF
(D) 2.6 mF
The current I supplied by the source 50 V is (A) 25 A (B) 13.7 A (C) 9.8 A (D) 3.66 A 21. The voltages VC1, VC2, and VC3 across the capacitors in the circuit in the given figure, under steady state are, respectively,
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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.5 3.7 2F
~
1F
(A) 80 V, 32 V, 48 V (C) 20 V, 8 V, 12 V
25 k Ω
+ − VC2 40 k Ω
1H
10 kΩ VC1 + −
2H
+ VC3 −
3F
Practice Problems 2
6. The current ‘I’ supplied by the source in the figure is
Directions for questions 1 to 19: Select the correct alternative from the given choices. 1. A network contains linear resistors that are connected in series across an ideal voltage source. If all the resistances are halved and the voltage is doubled, then the voltage across each resistor becomes (A) doubled (B) halved (C) not changed (D) None of these 2. Twelve similar conductors of 1 W resistance form a cubical framework. Then, the resistance between two adjacent corners, two opposite corners of one face, and two opposite corners of the cube are (A)
3 5 7 , , 4 6 12
(B)
7 5 3 , , 4 6 4
(C)
7 3 5 , , 12 4 6
(D)
12 4 6 , , 7 3 5
3. In the network shown in the figure, the voltage at node 2 is
10 A
1
2Ω
1Ω
4Ω
2 5A 4Ω
(A) 2 V (B) 10 V (C) 6 V (D) 4 V 4. In the network shown in the figure, the voltage AF is 3Ω
A
10 v
2Ω
20 v
2Ω
3A
5Ω F
1Ω
6A
(A) 2 A
1Ω
(B) 3 A
Unit 01.indd 5 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 7
(C) 1 A
4Ω 4Ω
6V
4Ω
3Ω
1Ω
3 2 A (D) A 2 3 7. A resistance of 10 W is connected in series with two resistances of 20 W arranged in parallel. What resistance should be shunted across this parallel combination so that the total current taken shall be 2 A with 30 V applied? (A) 5 W (B) 10 W (C) 20 W (D) 25 W 8. The resistance of a strip of conductor is RW. If the strip is elongated such that its length is doubled, the resistance of the strip is given by R (A) 4 R (B) 2 R (C) (D) R 2 9. The current i(t) through a 10 W resistor in series with an inductance is given by i(t) = 3 + 4 sin(100t + 45°) + 4 sin(300t + 60°) A. The RMS value of the current and the power dissipated in the circuit are (A) 2 A
(A)
(B) 3 A
(C)
41A, 410 W, respectively
(B) 35A, 350 W, respectively (C) 5 A, 250 W, respectively (D) 11 A, 1210 W, respectively 10. The nodal method of circuit analysis is based on (A) KVL and Ohm’s law (B) KCL and Ohm’s law (C) KCL and KVL (D) KCL, KVL, and Ohm’s law 11. In the given circuit, the voltage v(t) is 1Ω
2Ω I 3Ω
6Ω
I
(A) 4 V (B) −4 V (C) 6 V (D) 2 V 5. For the circuit shown in the figure, the current ‘I’ is given by 4Ω
(B) 80 V, 48 V, 32 V (D) 20 V, 12 V, 8 V
1Ω e at
+ v (t ) −
e bt
(A) eat − ebt (B) eat + ebt at bt (C) a e − b e (D) a eat + b ebt 12. The rms value of the voltage defined by v(t) = 5 + 5 sin (314t + p 6) is
+ 3V –
(D) zero
(A) 5 V
(B) 2.5 V
(C) 6.12 V
(D) 10 V
4/25/2019 10:19:27 AM 03/05/2017 15:10:57
1.6 || Part Networks 3.8 III • Unit 1 • Networks 13. If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is
16. + V1
+
+
a
10 V
R4
−
V −
b
R2
If V2 = 1 V in the abovementioned network, the value of V1 will be (A) 2.5 V (B) 4 V (C) 5 V (D) 8 V
(A)
(B) 0.138 V (D) 1 V
Z
(C) 3 3Z
(B) 3Z
3
8Ω
2V
+ −
E=?
25 A
30 Ω
5A
The current I in the circuit is (A) 20 A (B) −20 A (C) 16.2 A (D) −16.12 A 15.
a
− V1
+ 4V
a
b
R ab
=
9
1.5
b
c
R ac R bc
Previous Years’ Questions 1. If R1 = R2 = R4 = R and R3 = 1.1 R in the bridge circuit shown in figure, then the reading in the ideal voltmeter connected between a and b is [2005]
a
R4
+ R2
v
1.5V
(A) 0.6 V
R3
(B) 1.8
(D) 0 V
(A) 0.238 V (B) 0.138 V (C) −0.238 V (D) 1 V 2. A fully charged mobile phone with a 12 V battery is good for a 10−min talk time. Assume that, during the talk time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V, as shown in the figure. How much energy does the battery deliver during this talk time? [2009] 12V 10V 0
Unit 01.indd 6 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 8
(C) 0.9 V
v(t) b
−
2Ω
3Ω
+ −
10 V
c
The value of Rab, Rac, and Rbc are (A) 30 W, 15 W, 5 W (B) 5 W, 30 W, 15 W (C) 5 W, 30 W, 15 W (D) 15 W, 5 W, 30 W
R1
+ 5V −
(A) −16 V (B) 4 V (C) −6 V (D) 16 V 19. What is the current through resistor 2 W in the circuit given? 0.2 µF
3
Z 3
0V + 1V
−
I 12 Ω
(D)
18. In the given circuit, the value of the voltage source E is V2
14.
2 Ω V2 = 1 V −
17. If each branch of a Delta circuit has impedance 3Z , then each branch of the equivalent Wye circuit has impedance
R3
(A) 0.238 V (C) −0.238 V
+
2Ω
2Ω
− R1
2Ω
2Ω
10min
t
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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.7 3.9 (A) 220 J (C) 13.2 kJ
7. If VA − VB = 6 V, then VC − VD is
(B) 12 kJ (D) 14.4 kJ
3. In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power +−
VD
2A
Ra
RC
Rc
+ − 10V 2A
1Ω
(A) 0 W (C) 10 W
(B) 5 W (D) 100 W
5. In the following circuit, the current through the inductor is [2012]
∼ −
C2
+ 1∠ 0A
+
∼ −
C1
(A) 2.8 and 36 (B) 7 and 119 (C) 2.8 and 32 (D) 7 and 80 10. Consider the configuration shown in the figure, which is a portion of a larger electrical network [2014]
1Ω
−1 1 A (C) A 1+ j 1+ j
(D) 0A i5 i2
6. The impedance looking into nodes 1 and 2 in the given circuit is [2012] ib
R
1kΩ 9k Ω 100Ω
(A) 50 W (C) 5 kW
Unit 01.indd 7 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 9
C3
1∠ 0A 1∠ 0V 1∠ 0V
j1Ω
(B)
(A) k2 (B) k (C) 1/k (D) k 9. Three capacitors C1, C2, and C3, whose values are 10 μF, 5 μF, and 2 μF, respectively, have breakdown voltages of 10 V, 5 V, and 2 V, respectively. For the interconnection shown in the figure, the maximum safe voltage in Volts that can be applied across the combination and the corresponding total charge in μC stored in the effective capacitance across the terminals are, respectively, [2013]
j1Ω
1Ω
RB
RA
1Ω
1A
2 A 1+ j
VC
10V
(A) −5 V (B) 2 V (C) 3 V (D) 6 V 8. Consider a delta connection of resistors and its equivalent star connection as shown in the following figure. If all elements of the delta connection are scaled by a factor k, k > 0, the elements of the corresponding star equivalent will be scaled by a factor of [2013]
Rb
1Ω
1Ω
+
R
12A
1Ω
−
R
R
1Ω
R
20V
Which of the following can be the value of the current source I? [2009] (A) 10 A (B) 13 A (C) 15 A (D) 18 A 4. In the following circuit, the power supplied by the voltage source is [2010]
(A)
VB R
R
+ − 5V
+ 60V −
I
R
2Ω
VA
R
[2012]
99i b 1 2
(B) 100 W (D) 10.1 kW
R
i3 i4
R
i1
i6
For R = 1 W and currents i1 = 2 A, i4 = −1 A, and i5 = −4 A, which one of the following is true?
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1.8 || Networks 3.10 Part III • Unit 1 • Networks (A) i6 = 5 A (B) i3 = −4 A (C) Data is sufficient to conclude that the supposed currents are impossible. (D) Data is insufficient to identify the currents i2, i3, and i6 11. A Y-network has resistances of 10 W each in two of its arms, while the third arm has a resistance of 11 W. In the equivalent ∆-network, the lowest value (in W) among the three resistances is _________. [2014]
Ii
(A) voltage-controlled voltage source (B) voltage-controlled current source (C) current-controlled current source (D) current-controlled voltage source 16. The magnitude of current (in mA) through the resistor R2 in the following figure is __________ [2014] R2
12. In the following figure, the value of the current I (in Amperes) is __________ [2014]
1 kΩ
5Ω
5Ω
R1
10 mA
5V
1A
10Ω
R4
13. In the circuit shown in the figure, the value of node voltage V2 is [2014]
V2 4Ω −j 3Ω
6Ω
2R
R
Re
j 6Ω
(B) 2 + j 22 V (D) 2 − j 22 V
R
R
R
R
R
R
R
R
R R R
R
R
R
R = 300Ω R
R
R
b
19. In the given circuit, the values of V1 and V2, respectively, are [2015] 4Ω
R1 + V2 −
3Ω
7.5Ω
15. The circuit shown in the figure represents a
Unit 01.indd 8 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 10
R
a
Rab
14. For the Y-network shown in the figure, the value of R1 (in W) in the equivalent ∆-network is _____. [2014]
5Ω
R
R
R
(A) 22 + j 2 V (C) 22 − j 2 V
3 kΩ
The value of Re/R is _________ 18. In the network shown in the figure, all resistors are identical with R = 300 Ω. The resistance Rab (in Ω) of the network is ______. [2015]
+ −
4 0° A
2 mA
17. The equivalent resistance in the infinite ladder network shown in the figure is Re. [2014]
10 0° V
V1
R3 4 kΩ
2 kΩ
I + −
R
AiIi
[2014]
I 5A
4Ω
4Ω
2I
+ V1 −
(A) 5 V, 25 V (B) 10 V, 30 V (C) 15 V, 35 V (D) 0 V, 20 V 20. A dc voltage of 10 V is applied across an n-type silicon bar having a rectangular cross-section and a length of 1 cm as shown in figure. The donor doping
4/25/2019 10:19:30 AM 03/05/2017 15:11:05
M02_
concentration ND and the mobility of electrons mn and 1016 cm−3 and 1000 cm−2V−1s−1, respectively. The average time (in ms) taken by the electrons to move from one end of the bar to other end is ______. [2015] 10V
n −Si
concentration ND and the mobility of electrons mn and −1 −1 1016 cm−3 and 1000 cm−21Vcm s , respectively. The average time (in ms) taken by the electrons to move from 21. one For end the of circuit shown in end the isfigure, the Thevenin’s the bar to other ______. [2015] equivalent voltage (in volts) across terminals a-b is 10V _______. [2015] 3Ω a
21. For the circuit shown in the figure, theb Thevenin’s equivalent voltage (in volts) across terminals a-b is 22. In the circuit shown, the voltage Vx (in volts) is _______. [2015] 3Ω ________. [2015] 0.5Vx 12V
a
+
R2
R1
3 5 25. In the figure shown, the current i(in ampere) is + _________ . [2016] R3 + _
60 V 1A
5Ω 1Ω
+ 0.25V x −
8Ω
− 22. In the circuit shown, the voltage Vx (in volts) is ________. [2015] 0.5V
23. In the given circuit,x each resistor has a value equal to 1W [2016] a
10Ω + Vx 20Ω −
5A
1 W 3 38 (D) W 15 R3 (B)
0.04 Vx
5
Vx
8V ±
_
b 20Ω
Vx
5A
R2
R
1 1 W 6 5 9 (C) W 20
(A)
6Ω
1A
+
+ 0.25V x −
8Ω
23. In the given circuit, each resistor has a value equal to 1W [2016] a b
1Ω
± 8V
1Ω
1A
5Ω
b
1 W 6 9 (C) W 20 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 11 (A)
1 W 3 8 (D) W 15 (B)
8V ±
1 Ω consisting of resistance A in 26. A connection is made series with a parallel combination of resistances B and C. Three resistors of value 10 Ω ,15ΩΩ and 2 Ω are pro1Ω ± 8 Vpermutations of the given vided. Consider all possible resistors into the positions A, B, C, and iidentify the configurations with maximum possible overall resistance and the ones with 1a Ωminimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is _________. [2017] 27. Consider the network shown below with R1 = 1 W, R2 = 2 W and R2 = 3 W. The network is connected to a constant voltage source of 11 V.
Chapter 1 • Network Elements and Basic Laws | 3.11 What is the equivalent resistance across the terminals a and b?
1Ω
25. In the figure shown, the current i(in ampere) is i _________ . [2016]
R1
of electrons mn and pectively. The averrons to move from ____. [2015]
R1 R2
R1
R3
R1
R3 R2
R1
+ – 11 V
R1
The magnitude of current (in amperes, accurate to two decimal places) through the source is _______. [2018]03/05/2017 15:11:15
24. In the circuit shown in the figure, the magnitude of the current (in amperes) through R2 is __________. [2016]
re, the Thevenin’s s terminals a-b is [2015]
5:11:05
10Ω
24. In the iscircuit shown in resistance the figure,across the magnitude of thea What the equivalent the terminals current and b? (in amperes) through R2 is __________. [2016]
+
6Ω
1 cm 1A
1 1 W (B) W 3 6 9 8 (C) W (D) W Hints/Solutions 1.9 Chapter 1 20 • Network Elements and Basic Laws || 3.11 15 (A)
Vx 60 V _ 5 24. In the circuit shown in the figure, the magnitude of the 0.04 V _ current (in amperes) through R2 xis __________. [2016]
n −Si
12V
What is the equivalent resistance across the terminals a and b?
R1
R2
5
3
01.indd 9 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 11 a Unit
R3
+
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3.12 | Part III • Unit 1 • Networks
hiNts/solutioNs Practice Problems 1 1. Convert current source into its equivalent voltage source and solve by applying super position theorem. Hence, the correct option is (C). 2. Using super position theorem, considering 8 A current source only 10 Ω
+ 80 V –
eIIo o–
12 Ω
(6 ||12) = 20 V (6 ||12) + 10 + 2 Considering 16 V voltage source only
e0′ = 80 ×
2Ω o+
16 V
e0|| = 16 ×
6Ω
12 Ω
= 20 + 8 = 28 V Hence, the correct option is (D).
(R + R ) R eq
4Ω
4Ω
8V
2Ω 4Ω
4Ω
4Ω
Observing the circuit, the conducting path consists of 2 W resistance. 8 I= =4A 2 Hence, the correct option is (B). 7. Converting the current source of 1 A with a parallel resistance of 1 W into voltage source 3Ω
II
(10 + 2)|| 12 =8V 6 + (10 + 2)||12
1
2Ω
eo o–
Hence, total output voltage = e0 = e0| + e0||
3. Req = R1 +
2Ω
o+
6Ω
10 Ω
Total energy = 108 + 36 = 144 J Hence, the correct option is (C). 6. Redrawing the given circuit
2
R1 + R2 + Req
or Req . R1 + Req . R2 + Req2 = R12 + R1R2 + Req R1 + R1R2 + Req . R2 Hence, the correct option is (D).
Vx
+
+ 25V
−
2Vx
2Ω
−
+
−
1Ω 1V
Applying KVL for the loop in the above circuit 3I + 2(3I) + 2I + I − 1 = 25 V 12I = 24 24 I= =2A 12 Hence, the correct option is (C). 8. (A) Redrawing the circuit
2
4. I 2rms =
1 (3t ) 2 dt 2 ∫0
2Ω
2Ω
A
2
=
2Ω
9 t3 2 3 0
B
3 = × 8 = 12 2
6Ω
Converting Delta formed by three 6 W resistors into star
Power I2 R = 12 × 5 = 60 W Hence, the correct option is (C). 5. Total energy absorbed = Energy dissipated in resistor + Energy stored in inductor Energy dissipated in resistor =
2Ω
2
2Ω
A 2Ω 2Ω
2Ω 2Ω
2Ω
B 2Ω
2
∫ (3t ) × 1dt + 6 × 1 × 2 2
6Ω
6Ω
2Ω
A
0
= 36 + 72 = 108 J
1 1 Energy stored in inductor = LI 2 = × 2 × 36 = 36 2 2
Unit 01.indd 10 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 12
B
2Ω
2Ω
4Ω
2Ω
2Ω
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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.11 3.13 6Ω
Reducing 6Ω
2Ω A
6Ω
2Ω 2Ω B
Reducing further Req = RAB =
3Ω
10 Ω 3
V1
21 W 4
6Ω
V1 = 3 ×
Applying KCL at node ‘B’
N1
12.
B 2Ω
2A
5A 4A
3Ω
R C
N2 6V –
X
Taking cutset XX, the total current at the junction is zero, i.e., I2W + I4W + I3W = 0 4 + I4W + 2 = 0 I4W = − 6 A Voltage across 4 W resistance = − 6 × 4 = − 24 V Hence, the correct option is (B). 11. Redraw the given circuit as shown in the figure. given below.
Unit 01.indd 11 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 13
3A
A
+ 8V – 2Ω
+
16 16 = V 9 3
8 8 = V 3 3 V3 = 1 × 4 = 4 V Hence, the correct option is (C).
2VA − VA = 5 ⇒ VA = 5, VB = 5,
4Ω
12V
8A 3
V2 = 1 ×
(1)
X
1Ω
1Ω
3Ω
VA − 5 VA + 1 − VB + =1 1 1
10.
V3
6Ω
16 A 9
2VB − VA = I + 1 2VB − VB = I + 1 VB = I + 1 = 5 I=5−1=4A Hence, the correct option is (B).
V2
12V
6Ω
Hence, the correct option is (A). 9. Current through AB will be zero if VA = VB Applying KCL at node A,
V B V B −V A −1 + =I 1 1
1Ω
Total resistance seen through supply terminals is 2 W. 12 Total current = =6A 2 Current distribution among branches is indicated in the figure shown below.
Alternate method: From the figure, we can conclude that equivalent resistance will be greater than 4 W (as two 2 W resistors are in series with other section of network) and less than 6 W (the section of the network excluding two 2 W series resistors cannot be more than 2 W). 21 So, the answer is W. 4
2VA − VB = 5
1Ω
D
On applying KCL at node D Current through R is = 5 + 4 = 9 A Hence, the correct option is (D). 1 1 1 50 × + 40 × + 30 × 5 4 3 13. VAB = 1 1 1 + + 5 4 3 Hence, the correct option is (C). 14. Let source voltage Vs and n no.of resistors are connected in series voltage across each Resistor (R) is VR let all resistors are equal so VR= Vs/n if Vs is doubled then VR doubled
4/25/2019 10:19:33 AM 03/05/2017 15:11:42
1.12 || Part Networks 3.14 III • Unit 1 • Networks 15. The circuit can be reprinted as
i1′ = 5∠0°
a
with 10∠60°A alone 2µF
1µF
j2 Ω
0.5µF
c
− j2 Ω
1µF
10∠60 ° A
3Ω
i 1" 1µF
2 µF
i1|| = 10 ∠60°
0.5 µF
10 3 ∠90° 2 Hence, the correct option is (A).
b
i1 = i1|| − i1| = 10 ∠60° − 5∠0° =
Applying l → ∆
1µF
0.5µF
C1
0.25µF
30 × 5 = 3Ω 5 + 30 + 15
R2 =
15 × 5 = 1.5 Ω 5 + 30 + 5
0.5µF
C2 2µF
18. R1 =
C3
30 × 15 = 9Ω 5 + 30 + 5 Hence, the correct option is (D). R3 =
1× 2 1+ 2 +1 1×1 C3 = 4 1× 2 C2 = 1+ 4 Further simplifying C1 =
19. 2Ω
10 Ω
16 V 12 Ω
80 V
I1
a
6Ω
+ e0
−
I2
80 = I1(10 + 2 + 6) − I2 6 + 16 64 = I1(18) − I26
1µF
0.9µF
− I2(12 + 6) + I16 + 16 = 0 16 = I16 + I2(18)
b
i1 = 5 A
18 64 −6 16 228 + 384 672 I2 = = = 324 − 36 288 (18) 2 − 36
i2 = 3 A
I0 = 7 A
7+5 = 12 A
I2 = 2.33 E0 = 12 × 2.33
I3 = 4 A
I 4 = −12 A
Hence, the correct option is (B). 17. With 5∠0° A alone
= 28 V Hence, the correct option is (D). 20.
5Ω
j2 Ω 5∠0°
Unit 01.indd 12 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 14
(2)
64 18 −6 I1 16 = −6 18 I 2
Ceq = 1.9 mF. Hence, the correct option is (C). 16.
(1)
I
i 1’
3Ω
2Ω 6||3
12||8 + − 50 V
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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.13 3.15 10 k Ω
12 × 8 + 5 = 9.80 W 12 + 8 6×3 =2W 6+3 Req = 2 +
100 V
9.8 × 2 = 3.66 W 9.8 + 2
+ VC1 −
+ −
40 = 80 V 40 + 10 C3 3 VC2 = 80 × = 80 × = 48 V C 2 + C3 5
VC1 = 100 ×
50 = 13.7 A 3.66 Hence, the correct option is (B). 21. In steady state, the inductors are short circuited and the capacitors are open circuit. I=
C2 = 16 × 2 = 32 V C 2 + C3 Hence, the correct option is (B).
VC3 =
Practice Problems 2
Current in loop L1 is 4 A.
2. The resistance measured between two adjacent corners, two opposite corners of one face and two opposite cor7 3 5 ners of the cube are , , . 12 4 6 If the cube is formed by R W resistance conductors, 7 3 5 then the resistances will be R , R , R . This is a 12 4 6 standard question in which all the resistors have equal resistance. Solution method is not important. Just remembering the answer is suggested. Hence, the correct option is (C). 3. Two 4 W resistors that are in parallel can be replaced by equivalent resistance of 2 W and the simplified network is shown below after converting the current source with a parallel resistance into equivalent voltage source.
Current in loop L2 is 3 A.
1
VAF = 8 − 10 + 6 = 4 V Hence, the correct option is (A). 5. Solve the problem by application of the superposition theorem. I1 4Ω
6A
+ −
I1
20V
L1
I1 = 6 ×
10 − 10 =0A 5
1Ω
1 =2A 3
Considering 3 V source I2 4Ω
2Ω
+ 3V
3Ω –
A
10V
2Ω
L2
5Ω
2Ω
A F + − 8V − 10V + + − 6V F
Unit 01.indd 13 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 15
2Ω
4Ω
6A
1Ω
3Ω
3Ω
2Ω + 10V −
Current flowing in the circuit is = Voltage at node 2 is = 10 V Hence, the correct option is (B). 4.
2Ω
1Ω
2
2Ω 1Ω
10V
40 kΩ
1Ω
3A
Current supplied by 3 v Source = I2 = 2 ×
3 =2A 1.5
3 =1A 3+ 3
Current through 2 W resistor is 3 A. Hence, the correct option is (B).
4/25/2019 10:19:35 AM 03/05/2017 15:12:09
1.14 || Part Networks 3.16 III • Unit 1 • Networks R1 = 4R Hence, the correct option is (A).
6. Redraw the given circuit 4Ω
2
4Ω
6V
2 Power = I rms R = 25 × 10 = 250 W Hence, the correct option is (C).
4Ω
3Ω
1Ω
11. V(t) = L
di dt
d [e at + e bt ] dt = a eat + b ebt Hence, the correct option is (D). =1
4Ω 6Ω
4Ω
3Ω
4Ω
6V
2
5 12. RMS value = 5 + = 6.12 V 2 2
Hence, the correct option is (C).
1Ω
2Ω 6V 1Ω
6 = 2A 2 +1 Hence, the correct option is (A). 7. Combined resistance of parallel combination of two 20 20 W resistors = = 10 W 2 If the parallel resistance required is R W, then 10 + 10 R 30 = 10 + R 2 I=
10 R =5 10 + R ⇒ R = 10 W Hence, the correct option is (B). l 8. R ∝ a R corresponds to l, a R → l, a
a Length becomes 2l, area becomes , and new resist2 2l ance R1 ∝ a 2
i.e., R1 ∝
2
4 4 9. R.M.S value = 32 + + =5A 2 2
6Ω
4l a
Unit 01.indd 14 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 16
10 10 ×R− × 1.1 R 2R 2.1 R 5 − 5.238095 = −0.238095 Hence, the correct option is (C). 14. By superposition theorem 30 I1′ = 25 × = 15 A 30 + 20 12 I1′′ = 5 × = 1.2 A 30 + 20 I = 15 + 1.2 = 16.2 A Hence, the correct option is (C).
13.
15. Rab =
3 × 1.5 + 9 × 1.5 + 9 × 3 =5W 9
Rac =
3 × 1.5. + 9 × 1.5 + 9 × 3 = 30 W 1.5
3 × 1.5 + 9 × 1.5 + 9 × 3 = 15 W 3 Hence, the correct option is (C).
Rbc = 16.
2Ω
V2
2Ω
+ I1
Vi
I 1′
V i′
2Ω
2Ω
2Ω
−
V2 = 1 V 1 1 I1′ = + = 1 A 2 2 V1′ = 1 × 2 + 1 = 3 V
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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.15 3.17 V 1′ 3 = 2 2 3 5 I1 = + 1 = = 2.5 A 2 2 5 V1 = × 2 + 3 2 V1 = 8 V Hence, the correct option is (D). 17.
=
3Z 2
( )
Z 3 3
=
Z 3
.
Hence, the correct option is (A). 18. By KVL − 1 − E − 5 − 10 = 0 E = − 16 V Hence, the correct option is (A). 19. For dc, capacitor will not pass current \ Current through 2 W = 0 Hence, the correct option is (D).
3Z . 3Z 3Z + 3Z + 3Z
1 = 2 100 + × 2 × 10 × 60 2 = 13,200 = 13.2 kJ Hence, the correct option is (C).
Previous Years’ Questions 1. 3.
20 V
+ −
+–
I′
V
+ 60 V −
Given that 60 V source is absorbing power so I + I1 = 12 ⇒ I < 12 Hence, the correct option is (A). Given that R1 = R2 = R4 = R and R3 = 1.1 R R1 1.1 Vb = × 10 = = 5.23V R1 + R2 2.1 Va =
4. 1Ω
I+2
1Ω
0
= 2 × shaded area
Unit 01.indd 15 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 17
2A
+ − 10V
1Ω
10
2. Energy delivered in 10 minutes is E = ∫ VI dt
0
I
1A
⇒ Va − Vb = 5 − 5.23 V = − 0.23 V Hence, the correct option is (C).
10
3A 1Ω
1Ω
R1 × 10 = 5V R1 + R2
= 2 ∫ V dt
I+3
Consider I current is flowing from voltage source write loop equation. 10 = 2(I + 3) + 2(I + 2) ⇒I=0 So, power supplied by voltage source (P) = VI = 0 W. Hence, the correct option is (A). 5. The part of given network can be drawn as IL = 1 ∠ 0 ×
1 1 + j1
4/25/2019 10:19:37 AM 03/05/2017 15:12:30
1.16 || Part Networks 3.18 III • Unit 1 • Networks
1∠ 0 A
1Ω
j1Ω
Alternatively, by using superposition theorem also we can solve this problem, there are four sources, take each source at a time and calculate the current in desired branch and add all these four effects, to get the final current in inductor. Hence, the correct option is (C). 6. The circuit can be drawn as 1
Ib
Ra Rc K 2 Ra Rc = K⋅ Ra + Rb + Rc ( Ra + Rb + Rc ) K
RC =
Ra Rb K 2 Ra Rb = K⋅ Ra + Rb + Rc ( Ra + Rb + Rc ) K
\ Elements corresponding to star connection are also scaled by a factor ‘K’. Hence, the correct option is (B). 9. VC1 = Voltage across C1 = V Let ‘V’ be applied voltage
Ix
VC2 = Voltage across (C2) =
C3 2 .V = .V C 2 + C3 7
VC3 = Voltage across (C3) =
C2 5 .V = .V C 2 + C3 7
From the breakdown voltages
Vx
99I b
100
10K
RB =
VC1 max 10 Volts ⇒ Vmax1 = 10 Volts 2
To find out the impedance looking into nodes 1 and 2, connect voltage source Vx, current drawn by the circuit is Ix, V So, impedance = X IX By applying KCL at node 1. V 99Ib + Ib + Ix = X 100 −V Ib = X Vx is voltage across 10 kW. 10 k V By solving these two equations, X = 50Ω IX Hence, the correct option is (A). 7. VA − VB = 6 V, current through 2 W is 3 A, the same current leaving at VB will enter at VD. So the current flowing through 1 W resistor is 2A + 3A from VD to VC. That is, voltage VC − VD = 1 × −5 = −5 V. Hence, the correct option is (A). 8. After scaling the elements of delta configuration network is given by KRa
KRb
RA =
KRc
Rb Rc K 2 Rb Rc = K⋅ Ra + Rb + Rc ( Ra + Rb + Rc ) K
Unit 01.indd 16 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 18
VC2 max =
2 V = 5 Volts ⇒ Vmax 2 = 7.5 Volts 3
VC3 max =
5 V = 2 Volts ⇒ Vmax 3 = 2.8 Volts 7
Maximum safe voltage = minimum {Vmax1, Vmax2, Vmax3} = 2.8 V Effective capacitance = (5m//2m) + 10m = 11.42 mF Total charge (Q) = C.V = 11.42 × 2.8 mC = 32 mC Hence, the correct option is (C). 10. i5 i2 R
R i3
i4
R i1
i6
Given R = 1 W i1 = 2A, i4 = −1A, i5 = −4A According to KCL i3 + i6 = i1 i4 + i1 = i2 i2 = 1A i3 = i2 + i5
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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.17 3.19 = −3 A \ i6 = i1 − i3 =2+3=5A Hence, the correct option is (A). 11. From the given data
V1 − V2 = 10 V Ideal voltage source existing between two non-reference nodes. Sos apply super node at node V1 and V2. −4 +
V1 V1 − V2 V2 − V1 V2 V2 + + + + =0 − j3 4 4 6 j6
−4 −
V1 V2 V2 + + =0 j3 6 j6
V V2{1 + j1} = 1 + 4 j 6 j 3 V2{1 + j1} = 2V1 + j24
RR 100 ZA = R1 + R3 + 1 3 = 21 + = 32Ω R2 11
But V1 = 10 + V2 V2{1 + j1} = 20 + 2V2 + j24
RR 100 ZB = R1 + R2 + 1 2 = 20 + = 29.09Ω R3 11
V2{−1 + j1} = (20 + j24)
\ From all the above, lowest value of Z is ZB = 29.09 W Hence, the correct answer is 29.08 to 29.10. 12. 5Ω
5Ω
V1
I 5V
+ −
{20 + j 24} = (2 − j22) volts ( −1 + j1) Hence, the correct option is (D). 14. Y − ∆ conversion From the given data 5×3 R1 = 5 + 3 + 7.5 = 8 + 2 = 10 W Hence, the correct answer is 9 to 11. 15. Dependent current source depends on Ii value \ It is a current controlled current source Hence, the correct option is (C). 16. Applying source transformation to the given network it becomes V2 =
RR ZC = R2 + R3 + 2 3 = 32Ω R1
1A
10Ω
Applying nodal analysis at node V1. V1 − 5 V −1+ 1 = 0 5 15 3V1 − 15 − 15 + V1 = 0 4V1 = 30 V1 = 7.5 volts V 7.5 I= 1 = = 0.5Amp 15 15
AQ1
R + V -
V R 1k
2k 20V +-
↑ IN
R
4k
− + 8V
I
13. 3k Ω
o
10 ∠ 0 V +−
V1 o
4∠0 A
Unit 01.indd 17 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 19
4Ω
−j 3 Ω
I =
V2 6Ω
j6 Ω
28 mA = 2.8 mA 10
Hence, the correct answer is 2.79 to 2.81. 18. The given network is in balanced mode, so redraw equivalent network.
4/25/2019 10:19:39 AM 03/05/2017 15:12:43
1.18 || Part Networks 3.20 III • Unit 1 • Networks a R
R
Rab
R
R
R
R
R
R
R R
R
R
b a
2R
R
R
Vd = mE =
2R
mV 1000 × 10 = = 104 cm/sec L 1
Distance L We know drift velocity = = Time τ 1cm L −4 \ t= = = 10 sec V d 10 4 cm sec
\ t = 100 m sec Hence, the correct answer is 95 to 105. 21. 3Ω
a
b Rab 12V
\ Rab = (2R||2R)||(R||R) = (R||R||R) = R/3 But give R = 300 Ω \ Rab = 100 Ω Hence, the correct answer is from 99.5 to 100.5. 19. Apply nodal analysis to the given network. −5 +
Apply nodal analysis: V th − 12 V − 1 + th = 0 3 6 2(Vth − 12) − 6 + Vth = 0 3Vth − 24 − 6 = 0 Vth = 10 volts Hence, the correct answer is 10. 22.
0.5Vx
But I=
V1 4
V1 V1 + = 5 ⇒ V1 = 5 volts 2 2 Apply KVL in 1st loop V2 − 4 × 5 − V1 = 0 V2 = 20 + 5 = 25 Volts Hence, the correct option is (A). 20.
Vth
6Ω
b
V1 V1 + + 2I = 0 4 4 V1 + 2I = 5 2
1A
10Ω + Vx 20Ω −
5A
+ 0.25V x −
8Ω
Redraw the given network 5Vx
Vx
− + 20Ω
5A
10Ω V /4 x
8Ω
+ Vx − 4
10V
Applying nodal analysis, n − Si 1 cm
ND = 1016 cm−3 and mn = 1000 cm2/V-sec
Unit 01.indd 18 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 20
V x V x + 5V x − 0.25V x + =0 20 10 100 = Vx + 2{5.75 Vx} Vx = 8 V Hence, the correct answer is 7.95 to 8.05. −5 +
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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.19 3.21 23. Apply ∆ ↔ Y conversion to the given network.
4 − 3
1Ω
24.
+
•
5Ω
3Ω
i1
5Ω
−
−60 + i1 (5Ω) + i 2 (3Ω) + i 2 (5Ω) = 0 5i1 + 8i 2 = 60 Vx = i 2 (5Ω) i1 – i2 = −0.04Vx v i1 = x − 0.04 Vx 5 i1 = 0.16vx i2 = 0.2vx 5(0.16vx) + 8(0.2vx) = 60 Vx [0.8 + 1.6] = 60 Vx = 25 volts 25 i2 = = 5A 5 i R = 5A
1Ω 1Ω 4Ω
4Ω b •
Vx
i2
Mesh equation
1Ω
4Ω
0.04 V
60V
4 − 3 1Ω
a•
R2
+
1Ω
−
4 − 3
R1
•
• 1Ω
a
(i)
2
4 −Ω 5
4 −Ω 5
[Answer: 5A] 25. 1A
b
1A
4 −Ω 5
O 1Ω
4 8 \ Rab = // 5 5 5A
4 8 × Rab = 5 5 4 8 + 5 5
Unit 01.indd 19 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 21
8 Ω 15
V1
± 8V
4A
1Ω
4A O 1Ω
± 8V
i 1Ω
4A 1A [Answer: 6.656 mA]
Apply nodal analysis at node V1
8×4 5 Rab = × 25 12 =
4A
5Ω
V1 − 0 V1 − 8 V1 − 0 V − 8 + + + =0 1 1 1 1 Choice (D)
4 V1 = 16 ⇒ V1 = 4 volts \ i = −1 Amp
[Answer: −1 Amp]
4/25/2019 10:19:41 AM 03/05/2017 15:13:16
1.20 | Networks 26. From the given data
R1 = 1 W R2 = 2 W
Given A
R3 = 3 W B
I = ?
C
Req
Rev = A + (B||C)
The given network is symmetrical, so we can redraw the given circuit. V1 = V2 and V3 = V4 I
Let A= 10 Ω, B = 5 Ω and C = 2 Ω Rmax = 10 + (5||2) = 10 + =
5× 2 7
80 Ω 7
Let A = 2 Ω, B = 5 Ω and C = 10 Ω Rmin = 2 + (10||5) 10 × 5 =2+ 15 =
6 + 10 3
1Ω
1Ω
3Ω
1Ω
3Ω
+ 11 V –
1Ω
1Ω
Req = {(1 || 1) + {1.5 || 0.5} + 0.5} = 0.5 + 0.5 + 0.375 = I=
16 Ω Rmin = 3
1Ω
11 Ω 8
11 = 8 Amp 11 8
Hence, the correct answer is 7.9 to 8.1.
Rmax 80 3 15 = × = Rmin 7 16 7 = 2.142. Hence, the correct answer is 2.12 to 2.16. I 27. R1
1 R1
2
R3 R2
R1
R3 R2
3
R1
Unit 01.indd 20
R1
+ 11 V –
4
R1
4/25/2019 10:19:45 AM
M02_
Hints/Solutions | 1.21
chaptEr 2 NEtwork thEorEms ExErcisEs Practice Problems 1 Directions for questions 1 to 28: Select the correct alternative from the given choices. 1. The maximum power transferred to the load in the circuit is given as 0.5 W. Get the values of R and RL.
6. Find the Thevenin’s resistance associated with the circuit.
10V + −
R
3VAB +−
4K Ω
A
4KΩ
2K Ω B
RL
5V
(A) 1 kW (C) 2 kW
(A) 15 W, 10 W (B) 12.5 W, 12.5 W (C) 10 W, 15 W (D) 10 kW, 10 kW 2. Find the efficiency of the circuit given for RL = 50 W.
(B) 0.45 kW (D) 0.22 kW
Common Data for Questions 7 and 8: Select the correct alternative from the given choices.
5Ω V
10V
RL
(A) 99% (B) 91% (C) 80% (D) 87% 3. Current in the circuit is given by the equation i(t) = 10cos(20p t + 50) and the impedance of the load is given as ZL = 5 + j3. Find the average power delivered to the load. (A) 353.5 W (B) 291.5 W (C) 250 W (D) 176.7 W 4. In the following circuit, the Norton equivalent current (in A) across A - B is j5
A +
− j 50
10 Ω
− 20 Ω B
1kΩ
+ ~ – 50V
V 1k Ω
+
2V
−
Unit 01.indd 21 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 23
15 Ω
A
R
10Ω
10 Ω
30Ω
30 Ω
15 Ω
15Ω
B
7. Find Thevenin’s equivalent voltage of the circuit. (A) 100 V (B) 120 V (C) 125 V (D) 150 V 8. The resistance across A − B is 10 W. Find the current through the 10 W resistor. (A) 5.1 A (B) 6.45 A (C) 3.35 A (D) 13.9 A 9. When a resistor R is fed from an electrical network, ‘N’ consumes a power of ‘P’ W, as shown in the Figure (a). If an identical network is added as shown in Figure (b) the power consumed by R will be _____. N
R
Figure (a) N
a
~
N
V ab
Figure (b)
20mA
(A) 25 V (C) 49 V
15 Ω
+ −
(A) 19.45 + j3.24 (B) 6.48 - j1.08 (C) 12.97 - j2.16 (D) 20 + j0 5. Find the Thevenin’s equivalent voltage external to the load RL. 5K
7Ω
50 Ω 200V
20∠0°
8Ω
b
(B) 50 V (D) 45 V
(A) 2P (C)
P 2
(B) P (D) between P and 4P
4/25/2019 10:19:45 AM 29/04/2017 14:23:24
1.22 || Part Networks 3.24 III • Unit 1 • Networks Statement for Linked Answer Questions 10 and 11: Select the correct alternative from the given choices.
1
–j1 Ω
1Ω
ZL
j4 Ω
10. Find the value of ZL at which maximum power is transferred to ZL (A) (1.24 - j0.676) W (B) (1.24 + j0.676) W (C) 1.31 W (D) 1.24 W 11. The maximum power transferred is (A) 201.6 W (B) 617 W (C) 2016 W (D) 6170 W Directions for questions 12 to 15: Select the correct alternative from the given choices. 12. For the circuit shown in the figure, the Thevenin’s voltage and resistance looking into x - y are _____. 1Ω
2i
X o
o
1H
+ –
1F Po
1Ω 1A
Qo
(B) 1 + S +
(A) 1 1 S
(D)
1 S
S2 + S +1 S 2 + 2S + 1
14. The short-circuit test of a two-port p network is shown in Figure (a). The voltage across the terminals 111 in the network shown in Figure (b) will be
10
+ –
5Ω
15 Ω + 20 V –
1 1′
5A
2 N1
1A 2′
Figure (a)
Unit 01.indd 22 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 24
P
+ 5Ω
VX –
10 Ω
– 100 V x + VX
Q
85 Ω 12 (D) 500 A, 3.33 W
(A) 10 A, 3.33 W
(B) 100 A,
(C) 100 A, 3.33 W
Statement for Linked Answer Questions 16 and 17: Select the correct alternative from the given choices. 16. j8Ω
6Ω
j8 Ω
RL
110∠0° ∼
o Y
(A) 4/3 V, 2 W (B) 4 V, 2/3 W (C) 4/3 V, 2/3 W (D) 4 V, 2 W 13. The Thevenin’s equivalent impedance ZTH between the nodes P and Q in the following circuit is _____.
(C) 2 + S +
(A) 2 V (B) 5 V (C) 10 V (D) 1 V 15. The Norton’s equivalent circuit at terminals PQ has a current source and a Norton’s resistance of _____.
2A
1Ω
5V
2′
2Ω
1Ω o
10V
1
6Ω
i
+ –
N1
Figure (b)
3Ω
100 ∠ 0° ∼
2
1
∼ 90∠ 0°
In the circuit shown in figure, under the maximum power transfer condition, the value of RL is _____. (A) 5 W
(B) 20 W
(C)
25 Ω 3
(D) 6 W
17. The power absorbed by RL at maximum power transfer condition is _____. (A) 1,000 W (B) 500 W (C) 625 W (D) 2,000 W Directions for questions 18 to 22: Select the correct alternative from the given choices. 18. The Thevenin’s voltage at the terminals AB of the network shown in the figure is 1Ω + 2V –
(A) 4 V
–
–
1V
2 VX
+
(B) 2 V
+
2Ω
+ 2Ω
2Ω
VX –
(C)
3 V 2
A
B
(D)
1 V 2
19. The value of the resistance R, connected across the terminals A and B, which will absorb the maximum power, is
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M02_
4:23:26
Hints/Solutions Chapter 2 • Network Theorems || 1.23 3.25
4 kΩ
3 kΩ
(B) 0 A, 20 W (D) 0 A, 12 W
24. For the following circuit, the values of Rth and Vth are
B
A
~
(A) 2 A, 12 W (C) 0.5 A, 20 W
R 4 kΩ
6 kΩ
–2ix
100Ω
a +
0.01Vx
(A) 4 kW (B) 4.11 kW (C) 8 kW (D) 9 kW 20. For the circuit shown in figure, the Thevenin’s voltage and resistance looking into X - Y are 1Ω
x
i 2i
1Ω
±
2Ω
100 Ω
800 Ω
4 2 (A) V, 2 W (B) 4 V, W 3 3 4 2 V, W (D) 4 V, 2 W (C) 3 3 21. An AC source of RMS voltage 20 V with internal impedance ZS = (1 + 2j) W feeds a load of impedance ZL = (7 + 4j) W shown in the figure. The reactive power consumed by the load is Z s = (1 +2 j ) Ω
+ −
(A) 0 V, 100 W (C) 5 V, 100 W 4A
Ix N
5V
4Ω 3A
20Ω
(B) 4
(C) 8
R
(D) 16
10 Ω
5Ω
±
25Ω
5mH
2 sin 50t
+ ~ –
+ –
~ 2 sin100t
2µF
10 Ω
(A) Thevenin’s Theorem (B) MPTT (Maximum Power Transfer Theorem) (C) Milliman’s Theorem (D) Superposition Theorem 27. Consider the following network
4Ω
2Ω 3V + −
1Ω + 6V −
5Ω
IN, RN
b
Unit 01.indd 23 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 25
30V
26. In the circuit shown in figure, which one of the following theorem can be more conveniently used to evaluate the responses in the 10 W resistors?
9V + −
a
i1
N
Figure (a) Figure (b) The network ‘N’ contains only resistances. Use the data given in Figure (a) and find the current i in Figure (b). (A) 0 A (B) 12 A (C) -6 A (D) 6 A
23. For the following circuit, the value of iN and RN are
15i1
2A 15V
10 Ω
5Ω
(A) 2
Rth
(B) 0 V, 1200 W (D) None of the above
Z L = (7 + 4 j )
(A) 8 VAR (B) 16 VAR (C) 28 VAR (D) 35 VAR 22. The value of R (in Ohms) required for maximum power transfer in the following network is
− 25 V
b
ix
y
+
–
25. Consider the following circuits
2A
20 ∠ 0° V
Vx
300 Ω
The current I is (A) 0.23 A (C) 2.25 A
(B) -0.23 A (D) -0.5 A
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