Q.1 Which loss in a transformer varies significantly with load?

(A) Hysteresis loss Transformer has two type of major Losses-

(B) eddy current loss 1. Iron Losses 2. Copper Losses
(C) copper loss Iron losses are constant while copper losses are variable with load.
(D) core loss Iron Losses(Pi) = Hysteresis losses(Ph) + Eddy current losses(Pe),
Both Ph and Pe are independent of Load Current i.e. Iron losses
are constant with changing load.
Copper losses(Pc) = Conduction losses in winding conductor which are proportional to square of Current
Related MCQs
Q.2 With the same secondary voltage and filter, which Under No load condition which of the loss is negligible?
has the most ripple? (A) Hysteresis loss
(B) Eddy current loss
a. Half-wave rectifier (C) Copper loss
b. Full-wave rectifier (D) All losses have the same magnitude
c. Bridge rectifier
d. Impossible to say

For a given input frequency, the output frequency of a full-wave

rectifier is twice that of a half-wave rectifier,which makes a full-wave
rectifier easier to filter because of the shorter time between peaks.
When filtered, the full-wave rectified voltage has a smaller ripple
than does a half-wave voltage for the same load resistance and
capacitor values. The capacitor discharges less during the shorter
interval between full wave pulses, as shown in Figure .

Q.3 For a perfectly matched transmission line,

the voltage standing wave ratio is

a. Zero
b. Unity
c. Maximum
d. Minimum

When a transmission line load impedance is same as that of the characteristic impedance, the line is said to be matched
. In a matched line, maximum transmission occurs. The reflection will be zero. The standing wave ratio
S = (1 – R)/(1 + R). For R = 0, the SWR is unity for matched line
short circuits, open circuits, and purely reactive loads having infinite VSWR,

Q.4 A 15 V source is connected across a 15 resistor. 2 2

How much energy is used in three minutes? As P=V /R = (15) /15 =15 w
OR
a. 75 Wh
I= V/R=15/15=1 and P=I2R =(1)2(15)=15 w
b. 0.75 Wh
c. 90 Wh Time = 3 minutes = 3/60 hours
d. 0.90 Wh Energy= Pt = (15 *3)/60 = 3/4 =0.75 Wh

Q.5 A crack in the magnetic path of an inductor will result A crack in the magnetic path of an inductor leads to
in collapse in the strength of magnetic field. If it looses
A. unchanged inductance the strength of its magnetic field, then obviously
B. increased inductance there could be a chance of reduction in its
C. zero inductance inductance. By Electrical4u
D. reduced inductance
Q.6 Which of the following are ill suited for energy efficient Because the favourable economics of energy-efficient motors are
motors application? based on savings in operating costs, there may be certain cases
a) Pumps which are generally economically ill-suited to energy efficient
b) Fans motors. These include highly intermittent duty or special torque
c) Punch Presses applications such as hoists and cranes, traction drives, punch
d) All the above presses, machine tools, and centrifuges.

full load efficiencies are higher by 3 to 7 % T is directly proportional to square of the voltage…So
An "energy efficient" motor produces the same shaft output Torque which can be delivered by the motor is proportional
power (HP), but uses less input power (kW) than a to square of the operating voltage. When the supply voltage
standard-efficiency motor. increases by 10%, it becomes 1.1 times and therefore torque
becomes 1.1^2 = 1.21 times i.e. 21% increase. When supply
Q.7 Reduction in supply voltage by 10% will change the voltage reduces by 10%, it becomes 0.9 times, therefore
torque of the motor by ___. torque becomes 0.9^2 times i.e. 0.81 times i.e. 19% less.
a) 38% Hence we can say that 10% change in supply voltage results
b) 19% in 20% change in torque.
c) 9.5%
d) no change

Q.8 Which of the following needs to be measured

after rewinding of motor?
a) No load current
It is common practice in industry to rewind burnt-out motors. The population of
b) winding resistance
rewound motors in some industries exceed 50 % of the total population.
c) air gap
a common problem occurs when heat is applied to strip old windings : the insulation
d) all the above
between laminations can be damaged, thereby increasing eddy current losses. A
change in the air gap may affect power factor and output torque.
The impact of rewinding on motor efficiency and power factor can be easily assessed if the no-load losses of a motor
are known before and after rewinding. Maintaining documentation of no-load losses and no-load speed from the time
of purchase of each motor can facilitate assessing this impact. For example, comparison of no load current and stator
resistance per phase of a rewound motor with the original no-load current and stator resistance at the same voltage
can be one of the indicators to assess the efficacy of rewinding.

Q.9 1 db corresponds to ___________ change in

voltage or current level
A. 40%
B. 80%
C. 20%
D. 25%

All options given in Q.9 are wrong the exact

ans in 12%
The decibel is commonly used to show the ratio of
power change (increasing or decreasing) and is
defined as the value which is ten times the Base-
10 logarithm of two power levels
dB = 10log10[P2/P1]

Related MCQs. 1 db corresponds to …………..

change in power level
50%
35%
26%
22%
The gain of an amplifier is expressed in db because ___________
A. It is a simple unit
B. Calculations become easy
C. Human ear response is logarithmic
D. None of the above
The ear responds to the loudness of sound logarithmically. A general "rule of thumb" for loudness is that the power must
be increased by about a factor of ten to sound twice as loud.

Q.10 A power amplifier has a gain of 20 dB and an input level of 2

volts. Assuming that the input and output impedances are the
same, what is the voltage level at the amplifier output?
a. 10 V
b. 20 V
c. 30 V
d. 40 V

Q.11 A certain appliance uses 350 W. If it is allowed to run

continuously for 24 days, how many kilowatt-hours of energy does
it consume? Total kWH = power * time
A. 20.16 kWh =350 (watt) *24 (days)
B. 201.6 kWh =350(watt)*24 *(24) hours
C . 2 .01 kWh =201600 watt hour
D. 8.4 kWh =201.6 kWH

Q.12 An induction motor has synchronous speed of 1500

rpm. What will be the slip when it is running at a speed of
1450 rpm?
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