EE MCQ

SSC JE Electrical 2019 with solution SET-1
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Find H = _____ A/m at the center of a circular coil of diameter 1 m and carrying a current of 2A

Magnetic Field Strength (H) gives the quantitative measure of strongness or weakness of the magnetic field. H = B/μo Where B = Magnetic Flux Density μo = Vacuum Permeability Magnetic Field strength at the center of circular loop carrying current I is given by B = μoI/2r B/μo = I/2r H = I/2r Where r = Radius Now Given Parameters Diameter = 1m Current = 2A ∴ Magnetic field Intensity H = (2 / 2 × 1/2) = 2 A/m

Related Question

SSC JE Electrical 2019 with solution SET-1
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Please check out below figure.

From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below. Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R Here Current I = 1/2 A = 0.5 A Voltage V = 5 V R = V/I = 0.5/5 R = 10Ω

SSC JE Electrical 2019 with solution SET-1
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Split-phase, capacitor, and shaded-pole single-phase motors classified as:

Split-phase, capacitor, and shaded-pole single-phase motors classified as single phase motor.

The single-phase motor with a single winding is not self-starting, because if the single winding is fed with a.c., it simply produces a “pulsating field” which rises and falls with the alternating current, and this type of field produces no torque in the rotor.

SSC JE Electrical 2019 with solution SET-1
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If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.

At full load copper Loss = I2R

At half load copper Loss = (I/2)2 × R = I2/4 × R

400 = I2/4 × R

I2R = 4 × 400

I2R = Full load copper Loss = 1600 W