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Test: Capacitors & Inductors - 1 - Electrical Engineering (EE) MCQ


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15 Questions MCQ Test - Test: Capacitors & Inductors - 1

Test: Capacitors & Inductors - 1 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Capacitors & Inductors - 1 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Capacitors & Inductors - 1 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Capacitors & Inductors - 1 below.
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Test: Capacitors & Inductors - 1 - Question 1

 An Inductor works as a ________ circuit for DC supply.

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 1

Induced voltage across an inductor is zero if the current flowing through it is constant, i.e. Inductor works as a short circuit for DC supply.

Test: Capacitors & Inductors - 1 - Question 2

A power factor of a circuit can be improved by placing which, among the following, in a circuit?

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 2

Power factor = Real power / Apparent power = kW/kVA

By adding a capacitor in a circuit, an additional kW load can be added to the system without altering the kVA.

Hence, the power factor is improved.

Test: Capacitors & Inductors - 1 - Question 3

Which among the following equations is incorrect?

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 3
  • Q is directly proportional to V.
  • The constant of proportionality in this case is C, that is, the capacitance.
  • Hence Q=CV. From the given relation we can derive all the equations except for Q = C/V.
Test: Capacitors & Inductors - 1 - Question 4

What is the value of capacitance of a capacitor which has a voltage of 4V and has 16C of charge?

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 4

Q is directly proportional to V.

The constant of proportionality in this case is C, that is, the capacitance.

Hence Q = CV

From the relation,

C = Q/V = 16/4 = 4F

Test: Capacitors & Inductors - 1 - Question 5

What is the total capacitance when two capacitors C1 and C2 are connected in series?

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 5

When capacitors are connected in series, the equivalent capacitance is:
1/Ctotal=1/C1 + 1/C2

therefore,

Ctotal = C1*C2 / (C1+C2)

Test: Capacitors & Inductors - 1 - Question 6

A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the current in the circuit.

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 6

X= 2*π*f*L = 10 ohm

Z= (R2+XL2)

Therefore the total impedance

Z = 12.2 ohm

V = IZ

therefore,

I= V/Z = 100/12.2 = 8.2A

Test: Capacitors & Inductors - 1 - Question 7

The equivalent circuit of the capacitor shown below is:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 7

Due to initial condition, at t = 0 capacitor will act as a constant voltage source (at t = 0, capacitor acts as short-circuit). Hence, option (d) is correct.

Test: Capacitors & Inductors - 1 - Question 8

The strength of current in 1 Henry inductor changes at a rate of 1 A/sec. The magnitude of energy stored in the inductor after 3 sec is:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 8

∴ Current in the inductor after 3 sec is: 

1= 3 A

Hence, energy is stored after 3 sec:

Test: Capacitors & Inductors - 1 - Question 9

 The current and voltage profile of an element vs time has been shown in given figure. The element and its value are respectively:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 9
  • Since V is not proportional to R therefore, the element can’t be a resistor.
  • At t = 5 ms, even if i ≠ 0, the element behaves as a short circuit therefore, the element can’t be a capacitor (since at t = 0 only capacitor behaves as short circuit).
  • The current at t = 0 is zero and at t = 5 ms voltage across the element is zero therefore, the element must be an inductor (at t = 0, an inductor acts as open circuit and at t =∞ it behaves as short circuit).
  • From the given voltage and current profile, we have:
Test: Capacitors & Inductors - 1 - Question 10

 Figure shown below exhibits the voltage-time profile of a source to charge a capacitor of 50 μF. The value of charging current in amperes is:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 10

From given figure:

Test: Capacitors & Inductors - 1 - Question 11

 The equivalent capacitance across the given terminals A-B is:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 11
  • The equivalent combination of C2 and C3
  • The equivalent combination of this 1 μF and  C1 = 2μF is C1+ 1 μF = 3μF
  • Hence, the equivalent capacitance between terminals A and B is
Test: Capacitors & Inductors - 1 - Question 12

 The charging time required to charge the equivalent capacitance between the given terminals a-b by a steady direct current of constant magnitude of 10 A is given by:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 12

Equivalent capacitance between terminals a-b is:

Test: Capacitors & Inductors - 1 - Question 13

 An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 13

Max. value of current:

Assuming voltage as the reference phasor:

Test: Capacitors & Inductors - 1 - Question 14

The voltage and current through a circuit element is v= 100 sin (314 t + 45°) volts and i = 10 sin (314 t - 45 ° ) amps.
The type of circuit element and its value will be respectively:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 14

The phase difference between v and i is: 

Since v leads i therefore, the circuit element is an inductor.

Test: Capacitors & Inductors - 1 - Question 15

The equivalent inductance for the inductive circuit shown below at terminal “ 1 - 2 ” is:

Detailed Solution for Test: Capacitors & Inductors - 1 - Question 15

Converting the internal star connected inductance to an equivalent delta, the circuit reduces as shown below.

Hence, equivalent circuit becomes as shown below.

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