EE MCQ

SSC JE Electrical 2019 with solution SET-1
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In the circuit shown, find the equivalent resistance between A and B.

The resistance R3 & R4 are connected in series

= (30 + 70) = 100Ω

Now three resistor i.e 100Ω, 100Ω & 50Ω is connected in parallel

1/Rp = 1/100 + 1/100 + 1/50

Rp = 100/4 = 25 Ω

Resistance 50Ω and 25Ω are connected in series.

Rtotal = (50 + 25)Ω

Rtotal = 75Ω

Related Question

SSC JE Electrical 2019 with solution SET-1
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Find the net capacitance of the combination in which ten capacitors of 10 μF are connected in parallel

When the capacitance is connected in series the total capacitance is given by

Ceq = C1 + C2 + C3………CN

Ceq = 10n

Ceq =10 × 10 = 100 μF

SSC JE Electrical 2019 with solution SET-1
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Three lamps are connected in series across a 120V supply and take a current of 1.5A. If the resistance of two of the lamps is 30Ω, what is the resistance of the third lamp?

Method 1

Supply Voltage = 120 V

Current I = 1.5 A

Total Resistance R = V/I = 120/1.5 = 80 Ω

Now resistance of 3 lamp be R1, R2, & R3

Req = R1 + R2 + R3

80 = 30 + 30 + R3

R3 = 20 Ω

Method 2

Applying KVL on the above circuit

120 = 1.5(R1 + R2) + 1.5R3

120 = 1.5(30 + 30) + 1.5R3

120 = 90 + 1.5R3

R3 = 30/1.5

R3 = 20 Ω

SSC JE Electrical 2019 with solution SET-1
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The power factor of an A.C. circuit is equal to

In electrical engineering, the power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit ” R/Z”.

SSC JE Electrical 2019 with solution SET-1
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The output of the logic circuit given below represents the gate.

In the given diagram all are NOR Gate . The final output is shown in the figure.

At stage 1 the output will be \overline A \& \overline B

At stage 2 the output will be \overline {\overline A + \overline B } = A.B

And the final output will be \overline {A.B}

Hence for input A & B the output is \overline {AB} in case of Nand gate.