The time period or periodic time T of an alternating quantity is the time taken in seconds to
complete
a. one cycle
b. alternation
c. none of the above
d. Half cycle
Answer: a
The time period of an alternating quantity is 0.02 second. Its frequency will be
Solution: f = 1/T = 1/0.02 = 50Hz
a. 25 Hz
b. 50 Hz
c. 100 Hz
d. 0.02 Hz
Answer: b
The angular frequency of an alternating quantity is a mathematical quantity obtained by
multiplying the frequency f of the alternating quantity by a factor
a. π/2
b. π
c. 2π
d. 4π
Answer: c
An alternating current is given by the expression i = 200 sin(314 +
) amperes. The
maximum value and frequency of the current are
Solution: Comparing above equation with I = Imax sin(2t ±φ), Imax=200A & 2=314
a. 200 A; 50 Hz
b. 100√2; 50 Hz
c. 200 A; 100 Hz
d. 200 A; 25 Hz
ans: a
The frequency of emf generated by a generator depends upon its
a. speed
b. number of poles
c. machine rating
d. nature of supply
e. both (a) and (b)
ans: e, since f = P*N/120
In a multipolar (P-pole) machine running at a speed of N rpm, the frequency of generated emf
is given by
a. f =
b. f =
c. f =
d. f = PN
ans:b
An ac generator running at 1000 rpm produces emf of 50 Hz. The number of poles on the
generator is
a. 2
b. 4
c. 6
d. 8
ans:c , since P = 120*f/N = 120*50/1000 = 6
8. An insulation is subjected to …………….. value of applied alternating voltage to it.
a. average
b. peak
c. rms
d. double of the peak
ans:b
9. The average value of an unsymmetrical alternating quantity is calculated over the
a. whole cycle
b. half cycle
c. unsymmetrical part of the waveform
ans:a
An alternating current is represented as i = 70.7 sin(520 +
). The frequency and rms
value of the current are
a. 82.76 Hz, 50 A
b. 41.38 Hz, 25 A
c. 41.38 Hz, 50 A
d. 82.76 Hz, 25 A
Solution: f= 520/2 = 82.76 Hz, Irms = Imax / 1.414 = 70.7/1.414 = 50A
An electric iron is rated as 230V, 500W ac. The value of 230V refers to
a. rms value
b. average value
c. peak value
d. peak to peak value
ans:a
The size (cross-sectional area) of a conductor, with regard to the heating effect, is determined
on the basis of ……….. value of current to be carried by it
a. average value
b. peak value
c. rms value
d. peak to peak value
ans:c
The length of a phasor in a phasor diagram normally represents the ………… value of the
alternating quantity
a. rms or effective
b. average
c. peak
d. none of these
ans:a
The form factor is the ratio of
a. average value to rms value
b. rms value to average value
c. peak value to average value
d. peak value to rms value
ans:b
The peak factor is the ratio of
a. average value to rms value
b. rms value to average value
c. peak value to rms value
d. peak value to average value
ans:c
The ratio of effective value to average value is called the ………… factor
a. form
b. peak
c. average
d. Q-factor
ans:a
The form factor for dc supply voltage is always
a. zero
b. unity
c. infinity
d. any value between 0 and 1
ans:b
The two quantities are said to be in phase with each other when
a. the phase difference between two quantities is zero degree or radian
b. each of them pass through zero values at the same instant and rise in the same direction
c. each of them pass through zero values at the same instant but rises in the opposite
directions
d. either (a) or (b)
ans:e
When the two quantities are in quadrature, the phase angle between them will be
a. 450 or ! "#$%#&'
b. 900 or
(
)*+,*-.
c. 1350 or
!
"#$%#&'
d. 600 or "#$%#&'
ans:b
The phase difference between the two waveforms can be compared only when they
a. have the same frequency
b. have the same peak value
c. have the same effective value
d. are sinusoidal
ans:a
If e1 = A sin(/ + 01 ) and e2 = B sin(/ + 02 ) then
a. e2 is leading e1 by an angle (θ1 –θ2)
b. e1 is lagging e2 by an angle (θ1 –θ2)
c. e2 is leading e1 by an angle (θ2 –θ1)
d. e2 is lagging e1 by an angle (θ2 –θ1)
ans:c
Two sinusoidal emfs are given as e1 = A sin(/ +
!
difference between the two quantities in degrees is
a. 750
b. 1050
c. 600
) and e2 = B sin(/ −
). The phase
d. 150
Solution: 45° – ( - 30°) = 75°
A phasor is a line which represents the
a. rms value and phase of an alternating quantity
b. average value and phase of an alternating quantity
c. magnitude and direction of an alternating quantity
d. none of the above
ans:a
If two sinusoids of the same frequency but of different amplitudes and phase difference are
added, the resultant is a
a. sinusoid of the same frequency
b. sinusoid of double the original frequency
c. sinusoid of half the original frequency
d. non-sinusoid
ans:a
Which of the following statements associated with purely resistive circuits is correct?
a. PF is unity
b. Power consumed is zero
c. Heat produced is zero
d. PF is zero
ans:a
Average power in a pure resistive circuit is equal to
a. zero
b. product of average values of current and voltage
c. product of peak values of current and voltage
d. product of rms or effective values of current and voltage
ans:d
The inductive reactance of a circuit ……………. with the increase in supply frequency
a. increases
b. decreases
c. remains unchanged
d. unpredictable
ans:a
In a pure inductive circuit, the current will
a. lag behind the voltage by 900
b. lead the voltage by 900
c. remain in phase with the voltage
d. lag or lead the applied voltage
ans:a
In a purely inductive circuit…………….power is zero
a. apparent
b. reactive
c. true
d. inductive
ans:c
If the supply frequency of a purely inductive circuit is doubled, the circuit current will be
a. half
b. doubled
c. same
d. one-fourth
Solution: XL= 2(2f) L and I = V/ XL
An instantaneous change in voltage is not possible in
a. a resistor
b. an inductor
c. a capacitor
d. a current source
ans:c
If the supply frequency of a pure capacitive circuit is tripled, the current will be
a. tripled
b. one-third
c. nine times
d. one-ninth
Soln: Xc = 1/ 2 (3)3 and I = V /Xc
In a pure capacitive circuit, the current will
a. lag behind the voltage by 900
b. lead behind the voltage by 900
c. remains in phase with voltage
d. none of the above
ans:b
In an R-L series circuit the power factor is
a. leading
b. lagging
c. zero
d. unity
ans:b
When the voltage v = Vm sin (wt – /2) applied to the pure resistance what will be the
equation of current
a. i = Im sinwt
b. i = Im sin (wt − π/2)
c. i = Im sin (wt + π/2)
d. i = Im coswt
Answer: b
When the voltage v = Vm sin (wt – /2) applied to the pure inductance what will be the
equation of current
a .i = Im sinwt
b. i = Im sin (wt − π)
c. i = Im sin (wt + π/2)
d. i = Im coswt
Answer: b
When the voltage v = Vm sin (wt – /2) applied to the pure capacitance what will be the
equation of current
a. i = Im sinwt
b. i = Im sin (wt − π/2)
c. i = Im sin (wt + π/2)
d. i = Im coswt
Answer: a
When the applied voltage varies as v = Vm sin (wt – /2), the current in the body is
i = Im sin (wt – /2), the equations of voltage & current indicate that
a. voltage and current are in phase.
b. voltage and current are out of phase
c. voltage lagging to current
d. voltage leading to current
Answer: a
The angle of phase difference between voltage and current in pure resistance is
a.90
b.60
c.0
d.180
Ans: c
The angle of phase difference between voltage and current in pure inductance is
a.90
b.60
c.0
d.180
Ans: a
A coil is connected across a 200 V, 50Hz supply and takes a current of 10 amp. Loss in the coil
is 960 W the impedance of the coil is
a.10 ohm
b. 20 ohm
c.25 ohm
d.50 ohm
Answer: b
Z = 200/10 = 20 ohm
A voltage of v= 141 sin (wt + π/3) is applied to pure resistor of 30 ohm what is the rms value of
current
a.5amp
b.3.3amp
c.2amp
d.1amp
Ans: b
Irms = Vrms/R = (141/1.414)/30 = 3.323A
A voltage of v= 141 sin (wt - π/3) is applied to pure inductor of 0.1henry what is the rms value
of current
a.5.2amp
b.3.2amp
c.2.2amp
d.1.2amp
Ans: b
Solution: Irms = Vrms/ XL = (Vmax/1.414)/ XL = (141/1.414)/2 50 ∗ 0.1 = 3.2A
The power in a single phase ac circuit is given by
a. VI
b. VI sin Ф
c. VI cos Ф
d. I2Z
e. I2XL
f. none of the above
ans:c
Power factor of an ac circuit is defined as the ratio of
a. resistance and reactance
b. resistance and impedance
c. reactance and impedance
d. reactance and resistance
ans:b
In an ac circuit the applied voltage and current drawn are represented as 6 = 78#9 sin :
and % = ;8#9 sin(: − Ф). The pf of the circuit is
a. sin Ф
b. cos Ф(=*>>,->)
c. cos Ф(?@#$%&A)
d. none of these
ans: b
In an ac circuit I cos Ф is called the
a. wattless or reactive component of current
b. quadrature component of current
c. active or useful component of current
d. none of the above
ans:c
An ac source having voltage E = 100 sin(: +
drawn from the circuit varies as I = 5 sin(: −
) is connected in an ac circuit. If the current
) the impedance of the circuit will be
a. 20 Ω
b. 16 Ω
c. 30.8 Ω
d. none of these
ans:a
Z = 100/5 = 20 Ohm
A series circuit containing passive elements has the following current and applied voltage
V = 200 sin (2000 + 50 ), i = 4 cos (2000 + 20 ). The circuit elements
a. Must be resistance and capacitance
b. Must be resistance and inductance
c. Must be inductance, capacitance and resistance
d. Could be b as well as c
Ans:d