Quiz: Electrical Engineering 24 Nov 2020

Quiz: Electrical Engineering
Exam: UPPSC AE
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. A 4 A source has a shunt internal resistance of 65 Ω. The maximum power that can be delivered to a load is
(a) 270 W
(b) 460 W
(c) 260 W
(d) None of above

Q2. Which of the following represents the characteristics of Fuse wire?
(a) high R and low MP
(b) high R and high MP
(c) low R and low MP
(d) low R and high MP

Q3. Two electric bulbs whose resistances are in the ratio 2:1 is connected in parallel to a constant voltage source. The powers dissipated by them are in the ratio
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 4 : 1

Q4. The current in a 242 W, 220 V heater operated at 200 V is
(a) 50/7 A
(b) 6/5 A
(c) 1 A
(d) 55/9 A

Q5. The Electric stress in underground cable is
(a) Maximum at the conductor and minimum at the sheath
(b) Zero at the conductor as well as at the sheath
(c) Minimum at the conductor and maximum at the sheath
(d) None of the above

Q6. The voltage at farthest load point from supply at one end will be the least always for
(a) Interconnected system
(b) Radial system
(c) Ring system
(d) None of the above

Q7. What is the maximum safe current flow in a 42 Ω, 7 W resistor?
(a) 0.480 A
(b) 0.309 A
(c) 0.408 A
(d) 0.908 A

Q8. Capacitor is used in a 1 phase ac motor for
a) Providing starting Torque
b) Improvement if pf
c) Both (a) and (b)
d) None of the above

 

 

Q9. The Most suitable circuit breaker for interrupting short line fault without resistance switching is
a) Bulk oil circuit Breaker
b) SF6 Circuit breaker
c) Air Blast circuit breaker
d) None of the above

Q10. Voltmeter calibration is done through
a) a single generator
b) a potentiometer
c) an ampere hour meter
d) a hertz meter

Solution
S1. Ans.(c)
Sol. Source is connected across shunt resistance i.e. Norton’s equivalent circuit so
Here, I_N = 4 A and R_N = 65 Ω. For maximum power in the load, R_L=R_N = 65 Ω.
Therefore, current through load = I_N/2 = 4/2 = 2 A
∴ P_max = (2)² × R_L = (2)² × 65 = 260 W

S2. Ans.(a)
Sol. Fuse wire should have High R and Low MP

S3. Ans.(a)
Sol. Power dissipation = P_1/P_2 =((E^2/R_1 )/E^2 )/R_2 = R_2/R_1

S4. Ans.(c)
Sol. Resistance, R = V^2/P=(220)^2/242 = 200 Ω
∴Current,I=(Applied voltage)/R=200/200=1A

S5. Ans.(a)
Sol. The Electric stress in underground cable is maximum at the conductor and minimum at the sheath.
E_Max=V/(r ln⁡〖R/r〗 ) V/cm

S6. Ans.(b)
Sol. The voltage at farthest load point from supply at one end will be the least always for Radial System.

S7. Ans.(c)
Sol. P = I²R or I = √(P/R)=√(7/42) = 0.408A

S8. Ans(c)
Sol. Capacitor is used in a 1 phase ac motor for both of the mentioned reason.

S9. Ans(b)
Sol. SF6 Circuit breaker

S10. Ans(b)
Sol. A Potentiometer

Leave a comment

Your email address will not be published. Required fields are marked *