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NHPC-JE’21 EE: Daily Practices Quiz 23-Sep-2021

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NHPC-JE’21 EE: Daily Practices Quiz

NHPC-JE’21 EE: Daily Practices Quiz 23-Sep-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 06 min.

Q1. The base load power station supplies load in a day is…
(a) Off peak hours only
(b) Peak hours
(c) Full day
(d) None of the above

Q2. One-unit electrical energy is equal to……
(a) 860 kcal
(b) 4186 kcal
(c) 9.81 kcal
(d) 36×10^5 kcal

Q3. The power dissipated in a resistor in terms of its conductance G and the voltage V across it is
(a) V²G
(b) V²/G
(c) G²V
(d) G²/V

Q4. The total charge entering the terminal of an element is given by;
q = (6t² – 12t) mC
The current at t = 3s is
(a) 12 mA
(b) 48 mA
(c) 36 mA
(d) 24 mA

Q5. Three conductance G₁ = 0.5 S, G₂ = 0.3 S and G₃ = 0.2 S are in parallel. If the total circuit current is 4 A, current in G₁ is
(a) 1.2 A
(b) 2 A
(c) 0.8 A
(d) None of the above

Q6. A Wheatstone bridge in balanced. If the galvanometer is short-circuited, the currents in the various resistors will
(a) increase
(b) decrease
(c) not change
(d) cannot say

SOLUTIONS

S1. Ans.(c)
Sol. Base load: the unvarying load which occurs almost the whole day on the station is known as base load.

S2. Ans.(a)
Sol. One-unit electrical energy=1 KWh=860 kcal=3.6×10^6 joule.

S3. Ans.(a)
Sol. Power dissipated,
P=V^2/R=V^2×1/R=V^2 G

S4. Ans.(d)
Sol. Current, i=dq/dt=d/dt (q)=d/dt (6t^2-12t)=(12t-12) mA
At t = 3s, i = 12 × 3 – 12 = 36 – 12 = 24 mA

S5. Ans.(b)
Sol. Total conductance, G_T = G₁ + G₂ + G₃ = 0.5 + 0.3 + 0.2 = 1 S
∴ Current in G₁ = 4×G1/GT =4×0.5/1=2 A

S6. Ans.(c)
Sol. Since the bridge is balanced, there is no current in the branch containing galvanometer.
Therefore, whether the branch containing galvanometer is shorted or open-circuited, there will be no change in currents or voltages anywhere in the circuit.

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