This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Basic Network Concepts”.

1. Energy per unit charge is ____________

a) Power

b) Voltage

c) Current

d) Capacitance

View Answer

Explanation: The work or energy per unit charge utilised in the process of separation of charges is known as Voltage or Potential difference. The phenomenon of transfer of charge from one point to another is termed Current. The rate at which the work is done is called Power. Charge per unit voltage is Capacitance.

2. A conductor is said to have resistance of one ohm if a potential difference of one volt across its terminals causes a current of X ampere to flow through it. X=?

a) 4

b) 2

c) 3

d) 1

View Answer

Explanation: Ohm’s law states that the potential difference (voltage) across a conductor is proportional to the current through it. The constant of proportionality is called the Resistance(R).

According to Ohm’s law, V = IR (where V is the potential difference between two points which include a resistance R).

–> I = V/R = 1V/1Ω = 1A.

3. Resistance depends on the temperature of the conductor.

a) True

b) False

View Answer

Explanation: Resistance is directly proportional to its length, inversely proportional to the area of cross section of the conductor, depends on the nature of the material and on the temperature of the conductor.

4. A 25 Ω resistor has a voltage of 150 sin377 t. Find the corresponding power.

a) 900 sin^{2} 337 t

b) 90 sin^{2} 337 t

c) 900 sin^{2} 377 t

d) 9 sin^{2} 337 t

View Answer

Explanation: Given R = 25 Ω and v = 150 sin 377 t

i = \(\frac{v}{R} = \frac{150 sin 377 t}{25}\) = 6 sin 377 t

p = vi = (150 sin 377 t)(6 sin 377 t) = 900 sin

^{2}377 t.

5. Unit of inductance is ________

a) Weber

b) Henry

c) Farad

d) Tesla

View Answer

Explanation: The unit of inductance is Henry. Weber is the unit of magnetic flux. Tesla is the unit of flux density. Farad is the unit of capacitance.

6. Inductance of an inductor is inversely proportional to its ___________

a) Number of turns

b) Area of cross section

c) Absolute permeability

d) Length

View Answer

Explanation: Inductance of an inductor, L = µN

^{2}A/l

From the above equation, Inductance of an inductor is inversely proportional to its length.

7. Energy stored in an inductor is ________

a) LI

b) LI^{2}

c) LI/2

d) LI^{2}/2

View Answer

Explanation: V = L \(\frac{di}{dt}\)

dE = Vidt = L \(\frac{di}{dt} idt\) = Lidt

E = \(\int_0^I dE = \int_0^I Lidt = \frac{1}{2} LI^2\).

8. An inductor of 3mH has a current i = 5(1 – e^{-5000t}). Find the corresponding maximum energy stored.

a) 37.5 mJ

b) 375 J

c) 37.5 kJ

d) 3.75 mJ

View Answer

Explanation: Given L = 3 mH, i = 5(1 – e

^{-5000t})

V = L \(\frac{di}{dt} = 3 × 10^{-3} \frac{d}{dt}[5(1-e^{-5000t})] = 75 e^{-5000t}\)

I = i(∞) = 5(1 – e

^{-∞}) = 5 A

E = \(\frac{1}{2}\) LI

^{2}= 0.5 × 3 × 10

^{-3}× 5

^{2}= 37.5 mJ.

9. The capacitance of a capacitor does not depend on the absolute permittivity of the medium between the plates.

a) True

b) False

View Answer

Explanation: C = Ɛ \(\frac{A}{d}\)

Where d is the distance between the plates, A is the cross-sectional area of the plates and Ɛ is absolute permittivity of the medium between the plates.

Hence, the capacitance of a capacitor depends on the absolute permittivity of the medium between the plates.

10. Which of the following is not the energy stored in a capacitor?

a) \(\frac{CV^2}{2}\)

b) \(\frac{QV}{2}\)

c) \(\frac{Q^2}{2C}\)

d) \(\frac{QC}{2}\)

View Answer

Explanation: Energy stored in a capacitor, E = \(\frac{CV^2}{2}\)

Since C = Q/V

E = \(\frac{CV^2}{2} = \frac{QV}{2} = \frac{QC}{2}\).

11. A voltage is defined by \(
v(t)=\begin{cases}

0\\

2t\\

4e^{-(t-2)}\\

\end{cases} \) for \(
\begin{cases}

t < 0 \\

0<t<2s \\

t>2s \\

\end{cases} \) respectivelyand is applied to the 10 µF capacitor. Which of the following is incorrect?

a) i = 0 for t < 0

b) i = 20µA for 0 < t < 2s

c) i = 40e^{t-2}µA for t > 2s

d) i = -40e^{t-2}µA for t > 2s

View Answer

Explanation: Using i = C \(\frac{dv}{dt}\)

For t < 0, i = 0

For 0 < t < 2s, v = 2t; i = 10 × 10

^{-6}\(\frac{d}{dt}\)(2t) = 20 µA

For t > 2s, v = 4e

^{-(t-2)}

i = 10 × 10

^{-6}\(\frac{d}{dt}\)[4e

^{-(t-2)}] = 10 × 10

^{-6}[-4e

^{-(t-2)}] = -40e

^{-(t-2)}µA.

**Sanfoundry Global Education & Learning Series – Network Theory.**

To practice all areas of Network Theory, __here is complete set of 1000+ Multiple Choice Questions and Answers__.