EE MCQ

SSC JE Electrical 2019 with solution SET-1
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Let T be the net torque developed by the rotor runs at ω rad/s. What is the mechanical power developed?

By the term, torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius r meter acted upon by a circumferential force of F Newton which causes it to rotate at N r.p.m.

The angular speed of the pulley is

ω = 2πN/60 rad/sec

Work is done by this force in one revolution

= Force × distance = F × 2πR Joule

The power developed = Work Done/Time

= (F × 2πR)/60/N

= (F × R) × (2πN)/60

The power developed = T × ω watt or P = T ω Watt

pmech = (ωT)

Related Question

SSC JE Electrical 2019 with solution SET-1
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Two inductors of 4H and 6H are connected in series. The equivalent inductance of this combination is

Total inductance when inductor are connected in series

L = L1 + L2

L = 4 + 6 =10H

SSC JE Electrical 2019 with solution SET-1
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_______ is the measuring unit of inductive susceptance

Susceptance is symbolized by the capital letter B. It is the reciprocal of AC reactance. Susceptance, like reactance, can be either capacitive or inductive. In the case of a magnetic field, the susceptance is inductive. In the case of an electric field, the susceptance is capacitive. Capacitive susceptance is symbolized BC and inductive susceptance is symbolized BL. In the case of a magnetic field, the susceptance is inductive. In the case of an electric field, the susceptance is capacitive. Inductive susceptance is assigned negative imaginary number values, and capacitive susceptance is assigned positive imaginary number values.

The formula for inductive susceptance is

BL =–1/2πfL

The unit of Inductive susceptance is Siemen or Mho.

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

SSC JE Electrical 2019 with solution SET-1
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With the current direction marked in the circuit shown, the net voltage applied is

Net voltage = −V2 + V1

Net voltage = −(V2 − V1)

SSC JE Electrical 2019 with solution SET-1
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If admittance Y= a + jb, then a =?

Admittance (Y) is the reciprocal of the impedance of a circuit. Admittance of an AC circuit is analogous to the conductance of a DC circuit. The unit of Admittance is Simen or MHO

Admittance = 1/Z simen

Y = Conductance ± J Susceptance

Or the Admittance can be written as

Y = (G ± J B) Simen

Now comparing the above equation by the given equation in the question i.e Y= a + jb

∴ a = G = Conductance