Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 08 min.
Q1. A mho relay is used for protection of:
(a) Protection of a transformer against external fault
(b) Long Transmission Line
(c) Protection of a transformer against all the internal faults and external fault
(d) Medium Length lines
Q2. Condition for maximum power delivered in D.C. Generator is_______:
(a) Back EMF = (2 × Supply voltage)
(b) Back EMF = Supply voltage
(c) Back EMF = (Supply voltage/4)
(d) Back EMF = (Supply voltage/9)
Q4. Moving system of the induction type single phase energy meter has _________
(a) heavy aluminium disc
(b) light aluminium disc
(c) medium aluminium disc
(d) no aluminium disc
Q5. Sumpner’s test is conducted on transformers to study effect of ____________
(a) Temperature
(b) Stray losses
(c) All-day efficiency
(d) none of these
Q6. What will be the total resistance of the given circuit?
(a) 200 Ω
(b) 125 Ω
(c) 50 Ω
(d) 150 Ω
SOLUTIONS
S1. Ans.(b)
Sol. Mho relay is used for protection of long transmission line.
Important point of mho relay……
• Mho relay is a high-speed relay and is also known as admittance relay.
• Mho relay comes in the category of the distance relay protection scheme.
• Mho Relay is less affected by the power swing.
S4. Ans.(b)
Sol. Single phase induction type energy meter is also popularly known as watt-hour meter.
Aluminum disc is provided in the air gap between the series and shunt magnets. Jewel bearings support the spindle. Hence the moving system in an induction type single phase energy meter consists of light aluminum disc.
S5. Ans.(a)
Sol. Sumpner’s test is the test which is used to determine the steady temperature rise if the transformer was fully loaded continuously; this is so because under each of these tests the power loss to which the transformer is subjected is either the core-loss or copper-loss but not both.
S6. Ans.(c)
Sol. Two resistors of 50 Ω are connected in series and in parallel with 100 Ω resistor.
The equivalent resistance is
⇒ R_eq = (50 + 50) || 100
⇒ R_eq = 100 || 100
⇒ 1/R_eq =(100 × 100)/(100 + 100)
⇒ 1/R_eq =10000/200
So, R_eq = 50 Ω
