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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Test  >  Network Theory (Electric Circuits)  >  Test: Dot Convention in Magnetically Coupled Circuits - Electrical Engineering (EE) MCQ

Dot Convention in Magnetically Coupled Circuits - Free MCQ Practice Test


MCQ Practice Test & Solutions: Test: Dot Convention in Magnetically Coupled Circuits (15 Questions)

You can prepare effectively for Electrical Engineering (EE) Network Theory (Electric Circuits) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Dot Convention in Magnetically Coupled Circuits". These 15 questions have been designed by the experts with the latest curriculum of Electrical Engineering (EE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 45 minutes
  • - Number of Questions: 15

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Test: Dot Convention in Magnetically Coupled Circuits - Question 1
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A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is _____________

Detailed Solution: Question 1

Q = f0 / BW
And f0 = 1/2π (LC)0.5
BW = R/L
or, Q = 1R(L/C)0.5
When R, L and C are doubled, Q’ = 50.

Test: Dot Convention in Magnetically Coupled Circuits - Question 2
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The switch in the circuit shown was on position X for a long time. The switch is then moved to position Y at time t=0. The current I(t) for t > 0 is ________

Detailed Solution: Question 2

VC (t = 0–) = 100 V
At t ≥ 0,
The discharging current 

= 

= 0.2e-1250tu(t) mA.

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Test: Dot Convention in Magnetically Coupled Circuits - Question 3
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In the circuit shown below the current I(t) for t≥0+ (assuming zero initial conditions) is ___________

Detailed Solution: Question 3

I (t) = 1.5 / 3 = 0.5
LEQ = 15 mH
REQ = 5+10 = 15Ω

I (t) A – (A – B) e-t = 0.5 – (0.5-B) e-1000t
= 0.5(0.5 – 0.375) e-1000t
= 0.5 – 0.125 e-1000t
I (t) = 0.5-0.125e-1000t.

Test: Dot Convention in Magnetically Coupled Circuits - Question 4
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For a unit step signal u (t), the response is V1 (t) = (1-e-3t) for t>0. If a signal 3u (t) + δ(t) is applied, the response will be (considering zero initial conditions)?

Detailed Solution: Question 4

For u (t) = 1, t > 0
V1 (t) = (1 - e-3t)

or, 

Ans, 

Now, for 

Response, H(s) = R(s)

Or, h (t) = 3 u (t).

Test: Dot Convention in Magnetically Coupled Circuits - Question 5
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An ideal capacitor is charged to a voltage VO and connected at t=0 across an ideal inductor L. If ω = 1 / √LC, the voltage across the capacitor at time t > 0 is ________

Detailed Solution: Question 5

Voltage across capacitor will discharge through inductor up to voltage across the capacitor becomes zero. Now, inductor will start charging capacitor.
Voltage across capacitor will be decreasing from VO and periodic and is not decaying since both L and C is ideal.
∴ Voltage across the capacitor at time t > 0 is VO cos(ωt).

Test: Dot Convention in Magnetically Coupled Circuits - Question 6
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In the circuit given below, the capacitor initially has a charge of 10 C. The current in the circuit at t = 1 sec after the switch S is closed will be ___________

Detailed Solution: Question 6

Using KVL, 

or, 

Now, 

Or, 100C – q = (100C – qo) e-t/RC

= 40e-1 = 14.7 A.

Test: Dot Convention in Magnetically Coupled Circuits - Question 7
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A rectangular voltage wave of magnitude A and duration B is applied to a series combination of resistance R and capacitance C. The voltage developed across the capacitor is ____________

Detailed Solution: Question 7

Hence, maximum voltage =
V [1 – exp (-B/RC)].

Test: Dot Convention in Magnetically Coupled Circuits - Question 8
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In the circuit given below, the input impedance ZIN of the circuit is _________

Detailed Solution: Question 8

ZIN = (-6j) || (ZA)

= 0.49 + j5.82

= 64.73 + j17.77 Ω.

Test: Dot Convention in Magnetically Coupled Circuits - Question 9
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In the circuit shown, the voltage source supplies power which is _____________

Detailed Solution: Question 9

Let the current supplied by a voltage source.
Applying KVL in outer loop,
10 – (I + 3) × (I + 1) – (I + 2) × 2 = 0
10 – 2(I + 3) – 2(I - 2) = 0
Or, I = 0
∴ Power VI = 0.

Test: Dot Convention in Magnetically Coupled Circuits - Question 10
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Initial voltage on capacitor VO as marked |VO| = 5 V, VS = 8 u (t), where u (t) is the unit step. The voltage marked V at t=0+ is _____________

Detailed Solution: Question 10

Applying voltage divider method, we get,

And,  

= 

Now, 

= 

Hence, the net current in 1Ω resistance = I1 + I′1

= 

∴ Voltage drop across 1Ω = 13/3×1
= 13/3 V.

Test: Dot Convention in Magnetically Coupled Circuits - Question 11
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In the circuit given below, for time t < 0, S1 remained closed and S2 open., S1 is initially opened and S2 is initially closed. If the voltage V2 across the capacitor C2 at t = 0 is zero, the voltage across the capacitor combination at t = 0+ will be ____________

Detailed Solution: Question 11

When S1 is closed and S2 is open,
VC1 (0–) = VC1 (0+) = 3V
When S1 is opened and S2 is closed, VC2 (0+) = VC2 (0+) = 3V.

Test: Dot Convention in Magnetically Coupled Circuits - Question 12
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In the circuit given below, the switch S1 is initially closed and S2 is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through L during t=0+ is _____________

Detailed Solution: Question 12

By KCL, we get,

Hence, 2 VL = 110
∴ VL = 55 V.

Test: Dot Convention in Magnetically Coupled Circuits - Question 13
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In the figure given below, what is the RMS value of the periodic waveform?

Detailed Solution: Question 13

The rms value for any waveform is

= 

= 

= 

= [6 + 18]1/2 = √24 = 2√ 6 A.

Test: Dot Convention in Magnetically Coupled Circuits - Question 14
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In the circuit given below, the switch is closed at time t = 0. The voltage across the inductance just at t = 0+ is ______

Detailed Solution: Question 14

A t = 0+,

∴ 

VL(0+) = 1 × 4 = 4 V.

Test: Dot Convention in Magnetically Coupled Circuits - Question 15
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Two 300 μH coils in series without mutual coupling have a total inductance of

Detailed Solution: Question 15

Series Aiding:

The equivalent inductance of series aiding connection is:

Leq = L1 + L2 + 2M

Series Opposing:

 

The equivalent inductance of series opposing connection is:

Leq = L1 + L2 – 2M

Calculation:

Given,

L1 = 300 μH

L2 = 300 μH

M = 0 μH

The equivalent inductance for series addition is,

Leq = 300  + 300  + 2 x 0  = 600 μH

Important Points
The equivalent inductance of the parallel aiding connection is:


The equivalent inductance of the parallel opposing connection is

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About this Electrical Engineering (EE) Practice Test

This test contains 15 MCQs on Dot Convention in Magnetically Coupled Circuits with detailed solutions for each question. It is part of the Network Theory (Electric Circuits) course on EduRev, designed to align with the current Electrical Engineering (EE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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