EE MCQ

SSC JE Electrical 2018 with solution SET-2
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What will be the resistance (in ohms) of bulb A for the circuit given below?

The power can be defined as

P = I2R

Let the Power dissipated by Bulb A be

P = I2RA = 100 = I2RA

And power dissipated by Bulb B be

P = I2RB = 10 = I2RB

As we Know that the current flow in the series circuit is same

RA ⁄ 100 = RB ⁄ 10

RA ⁄ RB = 10

or

RA = 10RB

Power can also be defined as the

P = V2/R

Total Power consumption = 100 Watt + 10 Watt = 110 Watt

Applied voltage = 20 Volt

∴ 110 = 202 ⁄ R

or

R = 202 ⁄ 110 = 40 ⁄ 11

Now in series connection, the equivalent resistance is the sum of the individual resistance

∴ R = RA + RB

40 ⁄ 11 = 10RB + RB

RB = 0.33 Ω

Hence RA = 10RB

= 10 × 0.33 =3.3

RA = 3.3Ω

Related Question

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

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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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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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

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.