SSC JE Electrical Question paper 2017 Solved | MES Electrical
Ques 1. Two 10 ohm resistance is connected in parallel, their effective resistance will be
5
10
7
None of the above
Answer.1. 5 ohm
Explanation:
Since the resistance are connected in parallel, therefore, their effective resistance will be
R1 x R2 / R1 + R2
100/20 = 5 ohm
Ques 2. In a 3-phase half-wave rectifier, if the input phase voltage is 200 V, the PIV required for each diode will be
400 V
346 V
370 V
200 V
Answer.2. 346 V
Explanation:
Peak inverse voltage or PIV of diode = Maximum value of secondary voltage
= 200 x √3 = 346 V
Ques 3. For 100 turns as primary, 1000 V, 100 turns secondary transformer will have an output power of?
1000 V
500 V
250 V
100 V
Answer.4. 100 V
Explanation:
Transformer ratio is given as
= 1000/100 = 1000/Vs
= Vs = 100 Volt
Ques 4. Two couple coil of L1 = 0.8 H and L2 = 0.2 H have a coupling coefficient of K = 0.9. The mutual inductance M is
0.144 H
0.23 H
0.36 H
0.43 H
Answer.3. 0.36
Explanation:
Mutual inductance of two coupling coil is
M = K√L1L2
= 0.9√0.16
= 0.36 H
Ques 5. Transformer efficiency compared to other motors is
Higher
Remain same
Lower
None of these
Answer.1. Higher
Explanation:
As the Transformer is a static device, it does not have any rotating parts hence the mechanical losses are absent which increases the efficiency of the transformer.
Ques 6. A Hall effect transducer with hall coefficient KH = -1 x 10-8 is required to measure a magnetic field of 10,000 gauss. A 2 mm bismuth slab is used as the transducer with a current of 3A. The output voltage of the transducer will be
-15 × 10 -6 V
-7.5 × 10 -6 V
10 × 10 -6 V
-22.5 × 10 -6 V
Answer.1. -15 x 10 -6 V
Explanation:
Hall effect Output Voltage
= EH = KH IB / T
Where I is current
T is Thickness
B = Magnetic field = 10000 gauss = 10000 x 10-5 Wb/m2
Ques 7. Which power plant has the minimum running cost?
Coal-based
Hydel
Nuclear
None of these
Answer.2. Hydel
Explanation:
The operating cost of the hydro plant is low because of prime fuel, pressurized water comes almost free.
One of the wattmeter connections will have to be reversed
Pressure coil of the wattmeter will become ineffective
Answer.3. One of the wattmeter connections will have to be reversed
Explanation:
Some points must be noted for two wattmeter method! READ THE POINT CAREFULLY
If phase angle Φ = 0°, the readings of the two wattmeters are equal.
If phase angle Φ < 60°, the readings of both the wattmeters are positive and the total power is the sum.
If phase angle Φ = 60°, the reading P2 of the wattmeter W3 is zero.
If phase angle Φ> 60°, i.e., if the power factor of the load is less than 0.5, the reading of wattmeter W2 is negative. In such a case, the connection of either the current coil or the potential coil has to be reversed so that the pointer may deflect in a positive direction. The reading obtained after the reversal of the coil should then be taken as negative while calculating the power factor or the total power.
Now in the above question, the value of the power factor is less than 0., therefore, one of the wattmeter connections will have to be reversed.
Ques 9. Which efficiency is more in the case of cells?
Watt-hour efficiency
Ampere-hour Efficiency
Output Power efficiency
None of the these
Answer.2. Ampere-hour Efficiency
Explanation:
The amp-hour is a unit of battery energy capacity hence Ampere hour efficiency is more in case of the cell and ampere-hour efficiency is always greater than watt-hour efficiency
Ques 10. A 1000 ohms/V meter is used to measure the resistance on a 150 V scale. The meter resistance is
150 KΩ
1K
6.67 ohms
0.001 Ohms
Answer.1. 150 KΩ
Explanation:
Resistance of Voltmeter = sensitivity in ohm × voltage in volts
