Quiz: Electrical Engineering

Exam: UPSSSC JE

Topic: Miscellaneous

Each question carries 1 mark.

Negative marking: 1/4 mark

Time: 10 Minute

Q1. A 500 V sine wave generator appears across a 20 kΩ resistor. What is the instantaneous current in the resistor at a phase angle of 35°?

(a) 14.4 mA

(b) 32.6 mA

(c) 72 mA

(d) 20.3 mA

Q2. A 90 mH inductor carries a sinusoidal current of 10 A r.m.s. at a frequency of 50 Hz. The average power dissipated by the inductor is

(a) 0.9 W

(b) 0.55 W

(c) 0 W

(d) 1.5 W

Q3. An alternating voltage v=100 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) 5 A

(b) 3.18 A

(c) 1.59 A

(d) 1.1 A

Q4. An R-L series a.c. circuit has 5 V across resistor and 12 V across the Inductor. The supply voltage is …….. .

(a) 75 V

(b) V

(c) 169 V

(d) 13 V

Q5. An a.c. series circuit has R = 8 Ω, XL = 30 Ω and XC = 22 Ω. The circuit power factor will be……..

(a) 0.80 lagging

(b) 0.50 leading

(c) 0.53 lagging

(d) 0.60 leading

Q6. The total impedance of a series resonant circuit is

(a) purely resistive

(b) purely capacitive

(c) purely inductive

(d) none of the above

Q7. When transmission voltage is increased, the line losses are

(a) decreased

(b) increased

(c) same

(d) none of the above

Q8. The feeder is designed mainly from the point of view of

(a) its current carrying capacity

(b) voltage drop in it

(c) operating voltage

(d) operating frequency

Q9. For d.c. system, the string efficiency is

(a) 50%

(b) 75%

(c) 85%

(d) 100%

Q10. In a string of suspension insulators, the shunt capacitance can be decreased by using

(a) longer cross-arms

(b) shorter cross-arms

(c) longer spans

(d) none of the above

Solution

S1. Ans.(d)

Sol.

Now,

S2. Ans.(c)

Sol. The average power dissipated by an inductor is zero.

S3. Ans.(c)

Sol. The device is doing half-wave rectification. For half-wave rectified sinusoidal current, average value is Iav = Im/π = 5/π = 1.59 A (∵ Im = 100/20 = 5 A).

S4. Ans.(d)

Sol. Supply voltage =

S5. Ans.(c)

Sol. Net reactance, X = XL – XC = 30 – 22 = 8 Ω (inductive)

11.31 Ω ; cos ϕ =R/Z = 6/11.31 = 0.530 lagging

S6. Ans.(a)

Sol. The power factor of a series resonant circuit is unity. Therefore, the total impedance of a series resonant circuit is purely resistive

S7. Ans.(a)

Sol. P = VI cosϕ or

As transmission voltage V increases, the line current I and hence line losses are decreased

S8. Ans.(a)

Sol. A feeder is a conductor that connects the substation (or localized generator) to the area where power is to be distributed. Generally, no taping are taken from the feeder. The feeder must carry the required current safely so that is designed form point of view of current carrying capacity.

S9. Ans.(d)

Sol. It is because insulator capacitance are ineffective for d.c.

S10. Ans.(a)

Sol. In order to reduce shunt capacitance, the distance of the conductor from tower must be increased i.e. longer cross-arms should be used.