PSPCL RECRUITMENT 2021
Punjab State Power Corporation Ltd. (PSPCL), a power generating and distributing company of the Government of Punjab state, has released the notification for recruitment to the post of Clerk, Revenue Accountant, Junior Engineer (Electrical), Assistant Lineman & Assistant Sub Station Attendant Posts.
A total of 75 vacancies have been announced for the recruitment process of JE ELECTRICAL Engineering.
Exam Dates: 10 N0V-21 to 17 Nov-21
PSPCL-JE’21 EE: Daily Quiz
PSPCL-JE’21 EE: Daily Practices Quiz 22-Oct-2021
Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 06 min.
Q1. The superposition theorem applies to
(a) Current /voltage calculations
(b) Power calculation
(c) Current and power calculations
(d) Voltage and power calculations
Q2. Two electric lamps of 40 W each are connected in parallel. The power consumed by the combinations is
(a) 20 W
(b) 60 W
(c) 80 W
(d) 100 W
Q3. The equation ∇.J ⃗ = 0, is known as
(a) Poisson’s equation
(b) Laplace equation
(c) Continuity equation
(d) Maxwell equation
Q4. A d.c. motor runs at 1725 r.p.m. at full-load and 1775 r.p.m. at no-load. The speed regulation is ………
Q5. A transformer transfers electrical energy form primary to secondary usually with a change in …………
(d) Time period
Q6. A 230/2300 V transformer takes no-load current of 6.5 A and absorbs 187 W. If the resistance of the primary is 0.06 Ω, what is the core loss?
(a) 122.5 W
(b) 184.5 W
(c) 206.4 W
(d) 191.3 W
Sol. Since superposition theorem is based on linearity, it is not applicable to the effect on power (= I²R) due to each source. Note that power is a nonlinear mathematical operation. Therefore, it is applicable to current/voltage calculations
Sol. 1/RT =1/R1 +1/R2
or V^2/RT =V^2/R1 +V^2/R^2
or PT=P1+P2=40+40= 80 W
Sol. Continuity equation is
∇.J ⃗ = -(dρv)/dt
For steady current or lossless region, (dρ_v)/dt=0
So, ∇.J ⃗ = 0
Sol. Speed regulation = (N0 -NFL)/N_FL ×100=(1775-1725)/1725×100 = 2.9%
Sol. Transformers are capable of either increasing or decreasing the voltage and current levels of their supply, without modifying its frequency, or the amount of electrical power being transferred from one winding to another via the magnetic circuit.
Sol. No. load loss, W0=187 W ; Primary Cu loss = I0^2 R1=(6.5)^2×0.06=2.5 W
Iron loss = W_0-Primary Cu loss = 187 – 2.5 = 184.5 W