Electrical Engineering Questions

Electrical Engineering
Questions

Table of Contents
Q.1................................................................................................................................................3
Q.2................................................................................................................................................3
Q.3................................................................................................................................................4
Q.4................................................................................................................................................5
Q.5................................................................................................................................................5
Q.6................................................................................................................................................6
Q.7................................................................................................................................................6
Q.8................................................................................................................................................7
Q.9................................................................................................................................................7
Q.10..............................................................................................................................................8
Q.11 .............................................................................................................................................8
Q.12..............................................................................................................................................9
Q.13..............................................................................................................................................9
Q.14............................................................................................................................................10
Q.15............................................................................................................................................10
REFERENCES..............................................................................................................................12

Q.1
The input voltage is mostly sinusoidal. The output voltage is desired to be nearly constant
from the rectifier. The waveform shown in diagram explains that the voltage is sinusoidal and the
current is non-sinusoidal (Anwar, Maksimović and Afridi, 2017). Under these conditions only
fundamental components of input current plays role in drawing out the mean AC power from
taken from the source. Power factor is ratio of true power and the KVA.
Power Factor (cosΦ)= (Mean AC input Real Power)/(Total Apparent Power).
When input power factor is poor, more VAR is required to improve it and more input current is
taken from the source. The power factor is cosine of the phase angle or phase shift between the
supply voltage and the current.
It is given that the displacement factor is 1.
v(t) =230√2 cos(100πt) Volt
i(t) = 14√2 cos(100πt-π/6) + √2 cos(300πt-π/2)
cosΦ= P/S
P= Vrms.Irms cos(θ1 -θ2)
here displacement factor is 1
so, P= Vrms.Irms
Vrms= volt
Irms= 4*Idc/√2π = 4*10/√2π= 9.0077Amp
P= 15/√2*9.0077= 95.541W
S= Vtrue*Itrue= 15/√2*10 =106.066 W
putting these values in cosΦ= P/S
cosΦ= 95.541/106.066 = 0.90077
Q.2
Three winding transformer is generally used for high ratings. The normal two winding
transformer has two winding connected on the core. Three winding transformers has one more
winding in addition with the primary and secondary winding on the same core. The third winding
is also called as tertiary winding (Tertiary Winding of Transformer | Three Winding

Transformer., 2021). The benefit of using the third winding is that it can reduce the unbalancing.
It can be helpful in limiting the current in the times of short circuit conditions.
Here given data:
N1=100turns , N2= 200 turns, N3= 300turns.
Z=(32+j32)ohm
Z= Z1(N1/(N2+N3))^2= (32+j32)[100/200+300]
Z= (32+j32)/25
Imin=230/[(32+j32)/25]= 5750/(32+j322 45∠ °)
Imin= 127.058 -45°∠
Q.3
RMS value of voltage can be expressed as a pulse function with the variable time. This is
the amplitude of the signal varying between 0 to DT and DT to T. The RMS, value of pulse
signal or square wave can be expressed as:
Given data:
T= 0.01 sec= 10ms
Duty cycle D= 38% = 0.38
T= 0.01 sec
Also Duty cycle D= Ton/T= 0.38
ton= 0.38*0.01
ton= 0.0038 sec or 3.8 ms
By signal system:
RMS Value= √(energy of ON time/Time period)
Energy of square wave is A^2.B=400*3.8*10^(-3)= 1.52 w
Here A= Amplitude of the wave= 20volt
B= Band width =ton= 3.8 ms or 3.8*10^(-3)
By putting the values in the above formula we get .
RMS Value= √2*(1.52 /2*10^-2)
RMS Value of the Wave form will be= 12.3288 Volt