EE MCQ

SSC JE Electrical 2018 with solution SET-2
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The algebraic sum of the electric currents meeting at the common point is_________

Kirchhoff’s Current Law (KCI ) or Kirchhoff’s Junction Rule. This law is based on the conservation of charge and may be stated as under:

The algebraic sum of the currents meeting at a junction in an electrical circuit is zero.

An algebraic sum is one in which the sign of the quantity is taken into account. For example, consider four conductors carrying currents I1, I2, I3, & I4 and meeting at point O as shown in Fig

If we take the signs of currents flowing towards point O as positive, then currents flowing away from point O will be assigned negative sign. Thus, applying Kirchhoff’s current law to the junction O we have,

(I1) + ( I2) + (−I3) + (−I4) = 0 or(I1) + ( I2) = (−I3) + (−I4)

i.e., Sum of incoming currents = Sum of outgoing currents.

Therefore, Kirchhoff’s current law may also be stated as under:

The sum of currents flowing towards any junction in an electrical circuit is equal to the sum of currents flowing away from that junction. Kirchhoff’s current law is rightly called the junction rule.

Kirchhoff’s current law is true because electric current is merely the flow of free electrons and they cannot accumulate at any point in the circuit. This is in accordance with the law of conservation of charge. Hence, Kirchhoff’s current law is based on the law of conservation of charge.

Related Question

SSC JE Electrical 2018 with solution SET-2
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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
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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

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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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.