 # Quiz: Electrical Engineering 18 April 2020

Quiz: Electrical Engineering
Exam: NLC
Topic: Basic electrical engineering
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. The relation between inductive reactance and frequency is
(a) directly proportional to
(b) inversely proportional to
(c) independent of
(d) none of the above
L1Difficulty 1
QTags Basic electrical engineering
QCreator Balwant Kumar

Q2. A 500 V sine wave generator appears across a 10 kΩ resistor. What is the instantaneous current in the resistor at a phase angle of 35°?
(a) 9.4 mA
(b) 14.6 mA
(c) 84 mA
(d) 40.6 mA
L1Difficulty 2
QTags Basic electrical engineering
QCreator Balwant Kumar

Q3. A 2 mH inductor carries a sinusoidal current of 5 A r.m.s. at a frequency of 50 Hz. The average power dissipated by the inductor is
(a) 0 W
(b) 0.25 W
(c) 0.5 W
(d) 1 W
L1Difficulty 1
QTags Basic electrical engineering
QCreator Balwant Kumar

Q4. An alternating voltage v=200 sin 314t is applied to a device which offers an ohmic resistance of 20 Ω to the flow of current in one direction while entirely preventing the flow of current in the opposite direction. The average value of current is
(a) 4.8 A
(b) 3.18 A
(c) 1.87 A
(d) 1.69 A
L1Difficulty 3
QTags Basic electrical engineering
QCreator Balwant Kumar

Q5. An R-L series a.c. circuit has 3 V across resistor and 4 V across the inductor. The supply voltage is …….. .
(a) 45 V
(b) 5 V
(c) 13 V
(d) √185 V
L1Difficulty 2
QTags Basic electrical engineering
QCreator Balwant Kumar

Q6. An a.c. series circuit has R = 6 Ω, XL = 22 Ω and XC = 14 Ω. The circuit power factor will be……..
(a) 0.75 lagging
(c) 0.6 lagging
L1Difficulty 2
QTags Basic electrical engineering
QCreator Balwant Kumar

Q7. The nature of total impedance of a series resonant circuit is
(a) purely inductive
(b) purely capacitive
(c) purely resistive
(d) none of the above
L1Difficulty 1
QTags Basic electrical engineering
QCreator Balwant Kumar

Q8. If a phasor is multiplied by j, then ……….
(a) only its magnitude changes
(b) only its direction changes
(c) both magnitude and direction change
(d) none of the above
L1Difficulty 1
QTags Basic electrical engineering
QCreator Balwant Kumar

Q9. In polar form, capacitive reactance will be written as
(a) 1/ωC∠90°
(b) 1/ωC∠-90°
(c) 1/ωC∠0°
(d) none of the above
L1Difficulty 2
QTags Basic electrical engineering
QCreator Balwant Kumar

Q10. In a parallel resonant LC circuit, the line current is
(a) equal to IL or IC
(b) much greater than IL or IC
(c) much less than IL or IC
(d) none of the above
L1Difficulty 1
QTags Basic electrical engineering
QCreator Balwant Kumar

Solution

S1. Ans.(a)
Sol. XL = 2π f L. Therefore, XL ∝f

S2. Ans.(d)
Sol. V_m=√2×V_(r.m.s.)=√2×500 volts ;I_m=V_m/R=(√2×500)/(10×10^3 ) A
Now, i=I_m sinθ=(√2×500)/(10×10^3 ) sin35°=40.6×10^(-3) A=40.6 mA

S3. Ans.(a)
Sol. The average power dissipated by an inductor is zero.

S4. Ans.(b)
Sol. The device is doing half-wave rectification. For half-wave rectified sinusoidal current, average value is Iav = Im/π = 10/π = 3.18 A (∵ Im = 200/20 = 10 A).

S5. Ans.(b)
Sol. Supply voltage = √(V_R^2+V_L^2 )=√((3)^2+(4)²)=5 V

S6. Ans.(c)
Sol. Net reactance, X = XL – XC = 20 – 12 = 8 Ω (inductive)
Z=√(R^2+X^2 )=√(6^2+8^2 )= 10 Ω; cos ϕ =R/Z = 6/10 = 0.6 lagging

S7. Ans.(c)
Sol. The power factor of a series resonant circuit is unity. Therefore, the total impedance of a series resonant circuit is purely resistive

S8. Ans.(b)
Sol.

S9. Ans.(b)
Sol.
=1/ωC∠-90° or 1/jωC∠90°

S10. Ans.(c)
Sol. In a parallel LC circuit, the current IC or IL at resonance is much greater than the line current but IC and IL are in phase opposition. Therefore, line current is much less than IL or IC.