A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given.

Frequency of natural oscillation is given by,
fn = {((dPe/dδ)at(δo))/M)}
dPe/dδ = ((V1*V2)/X*(cosδ))
= (11/05)*cosδ
= (11/0.5)*0.5
M = (H*s)/(πf)
= 4/(50π)

fn = (1.76*(3/50π))
= 9.4 rad/sec = 1.32 Hz

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

Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL"

What is the full form of UN in 'UN Human Rights Prize'?

In circuit breaker, the term MCCB stands for

MCCB stands for Moulded Case Circuit Breaker.
MCB stands for Miniature Circuit Breaker.
ELCB stands for Earth Leakage Circuit Breaker
RCCB stands for Residual Current Circuit Breaker.

The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents.

Calculation:

Ia1 = 1/3*(Ia + a*Ib + a²*Ic)
= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)
= (546.41 + j*156.47) A

Ia2 = 1/3*(Ia + a²*Ib + a*Ic)
= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)
= (-146.41 - j*56.47) A

Iao = 1/3*(Ia + Ib + Ic)
= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)
= (100 + j*50) A