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Network Theory Multiple Choice Questions | MCQs | Quiz

Network Theory Interview Questions and Answers
Practice Network Theory questions and answers for interviews, campus placements, online tests, aptitude tests, quizzes and competitive exams.

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•   Circuit Elements
•   Voltage & Current Sources
•   Ohm's Law
•   Kirchhoffs Current Law
•   Kirchhoffs Voltage Law
•   Tree & Co-Tree
•   Tie Set Matrix
•   Tree Branch Voltages
•   Mesh Analysis
•   Supermesh Analysis
•   Nodal Analysis
•   Supernode Analysis
•   Star Delta Transformation
•   Superposition Theorem
•   Thevenins Theorem
•   Nortons Theorem
•   Reciprocity Theorem
•   Compensation Theorem
•   Power Transfer Theorem
•   Tellegens Theorem
•   Millmans Theorem
•   Sine Wave Angular Relation
•   Sine Wave Voltage
•   Resistor Phase Relation
•   Impedence Diagram
•   Series Circuits
•   Parallel Circuits
•   Instantaneous Power
•   Average Power
•   Power Factors
•   Reactive Power
•   Mesh AC Analysis
•   Nodal AC Analysis
•   Superposition AC Theorem
•   Thevenins AC Theorem
•   Nortons AC Theorem
•   Power Transfer Theorem
•   Series Resonance
•   RLC Circuits Bandwidth
•   Parallel Resonance
•   Resonant Frequency
•   Polyphase System
•   Three Phase Sources
•   Star Delta Transformation
•   Star Connected System
•   Delta Connected System
•   Balanced Circuits
•   Unbalanced Circuits
•   Power Measurement
•   R-L Circuit DC Response
•   R-C Circuit DC Response
•   R-L-C Circuit DC Response
•   Sinusoidal R-L Circuits
•   Sinusoidal R-C Circuits
•   Sinusoidal R-L-C Circuits
•   Laplace Transforms
•   Operational Transforms
•   Inverse Transforms
•   S Domain Circuit Elements
•   Transfer Function
•   Impulse Function
•   Complex Frequency
•   Series & Parallel Elements
•   One Port & Two Port
•   Poles & Zeros
•   Driving Point Functions
•   Transfer Functions
•   Open Circuit Parameters
•   Short Circuit Parameters
•   Transmission Parameters
•   Different Parameters
•   Two Port Networks
•   Network Termination
•   Image Parameters
•   Filter Networks
•   Pass & Stop Band
•   Constant K Low Pass Filter
•   M-Derived T-Section
•   Attenuators
•   Inverse Network
•   Series Equalizer
•   Shunt Equalizer
•   Hurwitz Polynomials
•   Reactive One-Ports
•   Foster One Ports Method
•   Cauer One Ports Method
•   Foster R-L Network Method
•   Cauer R-L Network Method
•   Foster R-C Network Method
•   Cauer R-C Network Method

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Network Theory Questions and Answers – Sinusoidal Response of an R-L-C Circuit

Posted on June 20, 2017 by Manish

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Sinusoidal Response of an R-L-C Circuit”.

1. The particular current obtained from the solution of i in the sinusoidal response of R-L-C circuit is?
a) ip = V/√(R2+(1/ωC+ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC+ωL)/R))
b) ip = V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
c) ip = V/√(R2+(1/ωC+ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
d) ip = V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC+ωL)/R))
View Answer

Answer: b
Explanation: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is ip = V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R)).
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2. . In the sinusoidal response of R-L-C circuit, the complementary function of the solution of i is?
a) ic = c1 e(K1+K2)t + c1 e(K1-K2)t
b) ic = c1 e(K1-K2)t + c1 e(K1-K2)t
c) ic = c1 e(K1+K2)t + c1 e(K2-K1)t
d) ic = c1 e(K1+K2)t +c1e(K1+K2)t
View Answer

Answer: a
Explanation: From the R-L circuit, we get the characteristic equation as
(D2+R/L D+1/LC)=0. The complementary function of the solution i is ic = c1 e(K1+K2)t + c1 e(K1-K2)t.

3. The complete solution of the current in the sinusoidal response of R-L-C circuit is?
a) i = c1 e(K1+K2)t + c1 e(K1-K2)t – V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
b) i = c1 e(K1+K2)t + c1 e(K1-K2)t – V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ-tan-1)⁡((1/ωC-ωL)/R))
c) i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
d) i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ-tan-1)⁡((1/ωC-ωL)/R))
View Answer

Answer: c
Explanation: The complete solution for the current becomes i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R)).

4. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.
network-theory-questions-answers-sinusoidal-response-rlc-q4
a) -38.5±j1290
b) 38.5±j1290
c) 37.5±j1290
d) -37.5±j1290
View Answer

Answer: d
Explanation: By applying Kirchhoff’s voltage law to the circuit,
network-theory-questions-answers-sinusoidal-response-rlc-q4a
On differentiating the above equation and on solving, we get roots of the characteristic equation as -37.5±j1290.

5. Find the complementary current from the information provided in the question 4.
a) ic = e-37.5t(c1cos1290t + c2sin1290t)
b) ic = e-37.5t(c1cos1290t – c2sin1290t)
c) ic = e37.5t(c1cos1290t – c2sin1290t)
d) ic = e37.5t(c1cos1290t + c2sin1290t)
View Answer

Answer: a
Explanation: The roots of the charactesistic equation are D1 = -37.5+j1290 and D2 = -37.5-j1290. The complementary current obtained is ic = e-37.5t(c1cos1290t + c2sin1290t).
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6. The particular solution from the information provided in the question 4.
a) ip = 0.6cos(500t + π/4 + 88.5⁰)
b) ip = 0.6cos(500t + π/4 + 89.5⁰)
c) ip = 0.7cos(500t + π/4 + 89.5⁰)
d) ip = 0.7cos(500t + π/4 + 88.5⁰)
View Answer

Answer: d
Explanation: Particular solution is ip = V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R)). ip = 0.7cos(500t + π/4 + 88.5⁰).

7. The complete solution of current from the information provided in the question 4.
a) i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)
b) i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)
c) i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)
d) i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)
View Answer

Answer: a
Explanation: The complete solution is the sum of the complementary function and the particular integral. So i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

8. The value of the c1 obtained in the complete solution of question 7.
a) -0.5
b) 0.5
c) 0.6
d) -0.6
View Answer

Answer: b
Explanation: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, c1 = -0.71cos (133.5⁰) = 0.49.

9. The value of the c2 obtained in the complete solution of question 7.
a) 2.3
b) -2.3
c) 1.3
d) -1.3
View Answer

Answer: c
Explanation: Differentiating the current equation, we have di/dt = e-37.5t (-1290c1sin1290t + 1290c2cos1290t) – 37.5e-37.5t(c1cos1290t+c2sin1290t) – 0.71x500sin(500t+45o+88.5o). At t = 0, di/dt = 1414. On solving, we get c2 = 1.31.

10. The complete solution of current obtained by substituting the values of c1 and c2 is?
a) i = e-37.5t(0.49cos1290t – 1.3sin1290t) + 0.7cos(500t + 133.5⁰)
b) i = e-37.5t(0.49cos1290t – 1.3sin1290t) – 0.7cos(500t + 133.5⁰)
c) i = e-37.5t(0.49cos1290t + 1.3sin1290t) – 0.7cos(500t + 133.5⁰)
d) i = e-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰)
View Answer

Answer: d
Explanation: The complete solution is the sum of the complementary function and the particular integral. So i = e-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰).
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