UPPCL RECRUITMENT 2021
Uttar Pradesh Power Corporation Limited has rolled out the official notification for the recruitment of Assistant Engineers and Junior Engineers for a total of 286 UPPCL Vacancy 2021. The Online submission of the application form for UPPCL Recruitment 2021 starts on 12th November 2021 and the last date to apply for the same is 2nd December 2021.
UPPCL-JE’21 EE: Daily Practices Quiz
UPPCL-JE’21 EE: Daily Practices Quiz 03-Nov-2021
Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 06 min.
Q1. If the speed of a DC machine is doubled and the flux remains constant, the generated e.m.f.
(a) Is doubled
(b) Is halved
(c) Remains the same
(d) None of the above
Q2. A 15 V source is connected across a 12 Ω resistor. How much energy is used in 3 minutes?
(a) 5.6 Wh
(b) 938 Wh
(c) 0.938 Wh
(d) 56.25 Wh
Q3. When the speed at which a conductor is moved through a magnetic field is increased, the induced voltage
(a) Remains constant
(d) Reaches zero
Q4. Three lights are connected in parallel across a 120-volt source. If one light burns out,
(a) The remaining two will not glow
(b) The remaining two will glow dimmer
(c) The remaining two will grow brighter
(d) The remaining two will glow with the same brightness as before
Q5. Two capacitor C1 and C2 have C1=20μ F and C2 = 30μ F, are connected in parallel
across a 100V source. The net capacitance of the circuit is?
(a) 50 μ F
(b) 10 μ F
(c) 12 μ F
(d) 60 μ F
Q6. A 100mA meter has accuracy of +2%. Its accuracy while reading 50mA will be:
Sol. E_b α ϕN
∴E_b α N
So, if speed is doubled then generated emf will also be doubled.
Sol. P=V^2/R=(15)^2/12=225/12 W
And t=3 minutes=3/60 hours
Sol. e=BlV sinӨ
So, e α v
So, when speed is increased the induced voltage will also increase.
Sol. Three lights are connected in parallel across a 120-volt source. If one light burns out, the remaining two will glow with the same brightness as before because the voltage across other two light will not change.
= 30μF + 20μF
= 50 μF
Sol. Error = 100 mA ×(±2)/100= ±2mA
% Error = (±2mA)/50mA×100= ±4%