EE MCQ

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

Related Question

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.

SSC JE Electrical 2018 with solution SET-2
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What will be the value of Thevenin’s voltage (in V). The Thevenin’s resistance (in Ω) and the load current (in A) respectively, across the load resistor in the given electrical circuit?

As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage ETh and internal resistance RT and this is called Thevenin equivalent circuit.

Thevenin Voltage

The Thevenin voltage used in Thevenin’s Theorem is an ideal voltage source equal to the open circuit voltage at the terminals.

In the given question, the resistance 10Ω does not affect this voltage and the resistances 30Ω and 20Ω form a voltage divider, giving

\begin{array}{l}{E_{Th}} = 100 \times \dfrac{30}{{30 + 20}}\\\\{E_{TH}} = 60V\end{array}

Thevenin’s resistance can be found by replacing 100 V source with a short-circuit.

Thevenin equivalent resistance for the given network is

R = (R1 || R2) + R3

Rth = (20 || 30) + 10 = (20 x 30) ⁄ (20 + 30) + 10 = 22Ω

Rth = 22Ω

The Load current Is calculated as

IL = ETH ⁄ (RTH + RL)

= 60 ⁄ (22 + 5) = 2.22 A

Hence the value of Thevenin voltage, Thevenin Resistance, and Load current is (60 V, 22Ω, 2.22A) respectively.

SSC JE Electrical 2018 with solution SET-2
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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
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Two wires of the same resistivity have equal length. The cross-sectional area of first wire is two times to the area of the other. What will be the resistance (in ohms) of the wire that has a large cross-sectional area, if the resistance of the other wire is 20 Ohms?

SSC JE Electrical 2018 with solution SET-2
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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