EE MCQ

SSC JE Electrical 2018 with solution SET-2
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Determine the Norton’s current (in A) and Norton’s resistance (in ohms) respectively, for the given electrical circuit across the load resistance RL.

Determine the resistance RN of the network as seen from the network terminals. (Its value is the same as that of Rth).

RN = (4Ω || 8Ω) + 5Ω = (4 × 8)/(4 + 8) + 5

RN = 7.66

The value I for the current used in Norton’s Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.

According to voltage Division Rule

VAB = V1R3 ⁄ (R1 + R3)

= 24 × 8 ⁄ (4 + 8)

VAB = 16 V

Now Norton Current IN is

IN = VAB ⁄ RN

IN = 16 ⁄ 7.66

IN = 2.08

Related Question

SSC JE Electrical 2018 with solution SET-2
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What will be the equivalent resistance (in Ω) for the circuit given below?

In the given circuit the resistance R9 and R8 are Parallel with Resistance R10 therefore

(R9 + R8) || R10

= {(4 + 2) × 12} ⁄ {(4 + 2) + 12}

= (6 × 12) ⁄ (6 + 12)

RA = 4Ω

Now resistance RA and R7 is parallel with Resistance R6

∴ {(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

RB = 4Ω

Now the circuit becomes as shown in the figure

Resistance RB and R7 is parallel with Resistance R4

{(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

RC = 4Ω

Now resistance RC and R3 is parallel with Resistance R2

{(4 + 2) × 6} ⁄ {(4 + 2) + 6}

RD = 3

Now our final circuit becomes as shown in the figure

Therefore the equivalent resistance is

R1 + RD

Req = 4 + 3 = 7Ω

SSC JE Electrical 2018 with solution SET-2
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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
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What is the value of Norton resistance (in Ω) between the terminal A and B for the given Norton’s equivalent circuit?

Norton equivalent resistance for the given network is

R = (R1 || R2) + R3

R = (4 || 8) + 2 = (4 x 8) ⁄ (4 + 8) + 2 = 5.6Ω

Norton equivalent resistance = 5.6Ω

SSC JE Electrical 2018 with solution SET-2
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Which property of an electrical conductor opposes a change in the current?

An inductor is a device which temporarily stores energy in the form of the magnetic field. It is usually a coil of wire. One of the basic property of the electromagnetism is that when you have current flowing through the wire it creates a small magnetic field around it.

One current first start to flow through the inductor a magnetic field start to expand then after some time magnetic field becomes constant then we have some energy stored in the magnetic field.

Once a constant magnetic field is generated in the Inductor, it will not change any further. As magnetic flux = N x I (Turns x Current), Inductor will draw a constant current to maintain the magnetic field.

Once current stop flowing the magnetic field start to collapse and the magnetic energy turned back into electric energy.

So when the current flowing through an inductor changes, the magnetic field also changes in the inductor and emf (electromotive force) is induced in the inductor as per Faraday’s law of electromagnetic induction.

According to Lenz’s law, the direction of electromotive force(emf) opposes the change of current that created it. V= -Lx dI/dt (rate of change of current)

So inductor opposes any change of current through them.