EE MCQ

SSC JE Electrical 2018 with solution SET-2
######
Farad is the S.I units of____

The Farad is the practical and the Sl unit of capacitance. The unit, named after Michael Faraday (1791-1867), was first suggested by Latimer Clark in 1867. The capacitor has a capacitance of 1 farad when a charge of 1 coulomb raises the potential between its plates to 1 volt.

The S.I unit of Inductance is Henry.

The S.I unit of resistance is OHM.

The S.I unit of Reluctance is amp-turns/Weber or Henry−1

Related Question

SSC JE Electrical 2018 with solution SET-2
######
Determine the value of current I1 (in A) and V1 (in V) respectively, for the given circuit below.

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

∴ I1 = 1 + 3 = 4A

V = IR

∴ V1 = I1R = 8 × 4

V1 = 32Ω

SSC JE Electrical 2018 with solution SET-2
######
What is the resistivity (in ohms-m) of a 2 ohm cylindrical wire when the length and the diameter of the wire are 10 m and 0.4 m respectively?

The resistance of the conductor is determined by the

R = ρL/A

Where

ρ = Resistivity of the conductor

L = Length of the conductor = 10 m

A = Area of the conductor = πR2 = 3.14 × 0.22 = 0.125 m2

R = Resistance = 2 Ω

Therefore the resistivity is

ρ = A × R ⁄ L

= 0.125 × 2 ⁄ 10

ρ = 0.025 Ω-m

SSC JE Electrical 2018 with solution SET-2
######
How much power (in W) will be dissipated by a 5 Ohm resistor in which the value of current is 2 A?

Given

Resistance R = 5Ω

Current I = 2 A

Power dissipated by the resistor is

P = I2R

P = 22 × 5

P = 20 watts

SSC JE Electrical 2018 with solution SET-2
######
What is the value of current (in A) for the given junction?

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

From the above Diagram

Current Flowing towards the Point: I2, I6, I4

Current Flowing Away from the Point: I1, I3, I5

Hence I2 + I6 + I4 = I1 + I3+ I5

Putting the value of the current

2A + 7A + I4 = 4A + 3A + 8A

I4 = 15A − 9A = 6A

I4 = 6A

SSC JE Electrical 2018 with solution SET-2
######
Which of the following statement is CORRECT?

Although the Thevenin’s theorem and Norton’s theorem can be used to solve a given network, yet the circuit approach differs in the following respects:

A Norton’s theorem is converse (opposite) of Thevenin’s theorem in the respect that Norton equivalent circuit uses a current generator instead of the voltage generator and the resistance RN (which is the same as RTH) in parallel with the generator instead of being in series with it.

Thevenin’s theorem is a voltage form of an equivalent circuit whereas Norton’s theorem is a current form of an equivalent circuit.

To Convert Thevenin equivalent circuit into Norton’s equivalent circuit the following step is involved

RN = RTH

IN = ETH ⁄ RTH