EE MCQ

MGVCL Exam Paper (30-07-2021 Shift 2)
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A single phase motor connected to 415 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging.

P = VIcosφ = 8715 watt

Q = VAR required to improve PF = P(tanφ₁ - tanφ₂) = 4670 VAR

Capacitive Reactance, Xc = V²/Q = 36.8790 ohm.

C = (2πfXc)⁻¹ = 86.314 μF

Related Question

MGVCL Exam Paper (30-07-2021 Shift 2)
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Which one of the following statement is TRUE?

In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors.

MGVCL Exam Paper (30-07-2021 Shift 2)
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The miniature circuit breaker (MCB) with Class D trip characteristics trips instantaneously when the current flowing through it reaches ____ the rated current.

Class B trip curve :

The MCB with class B trip characteristics trips instantaneously when the current flowing through it reaches between 3 to 5 times rated current. These MCBs are suitable for cable protection.

Class C trip curve:

MCB with class C trip characteristics trips instantaneously when the current flowing through it reaches between 5 to 10 times the rated current. Suitable Domestic and residential applications and electromagnetic starting loads with medium starting currents.

Class D trip curve:

MCB with class D trip characteristics trips instantaneously when the current flowing through it reaches between Above 10 (excluding 10) to 20 times the rated current. Suitable for inductive and motor loads with high starting currents.

Class K trip curve:

MCB with class K trip characteristics trips instantaneously when the current flowing through it reaches between 8 to 12 times the rated current. Suitable for inductive and motor loads with high inrush currents.

MGVCL Exam Paper (30-07-2021 Shift 2)
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A single phase 240 V, 1 kW heater is connected across single phase 240 V, 50 Hz supply through an SCR. For firing angle delay of 90˚, calculate the power absorbed in the heater element.

Resistance of heater, R = V²/P

R = 57.6 ohm

RMS value of voltage through controlled SCR = (Vm/2π)*(1 + cosα)

= 120 V

Power absorb by heater = Vrms²/R

= 120²/57.6

= 250 W