EE MCQ

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

Related Question

SSC JE Electrical 2019 with solution SET-1
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The trade name of the Nickel-Copper alloy, that is used as a heating element is:

Constantin‘, also known as ‘Eureka wire‘, is the trade-name for a copper-nickel alloy (approx. 60:40 ratio) formulated in the late 1800s by Edward Weston.

SSC JE Electrical 2019 with solution SET-1
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If N = linkage flux, then the linkage flux per unit current is defined as:

The inductance L of this coil is defined as the flux linkage per unit current and its unit is Henry.

L =Λ/I

Where

Λ = Flux linkage

I = current

SSC JE Electrical 2019 with solution SET-1
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For a three-layer stranded wire, the number of strands in the first, second, and third layer respectively are:

The total number of the conductor in stranded cable is given by

N = 3x2 − 3x + 1

Where

x = no. of layer

For layer 1

N1 = 3(1)2 − 3(1) + 1

N1 = 1

For layer 2

N2 = 3(2)2 − 3(2) + 1

N2 = 7

In layer 2 no. of conductor = N2 − N1 = 7 − 1 = 6

For layer 3

N3 = 3(3)2 − 3(3) + 1

N3 = 19

In layer 3 no. of conductor = N3 − N2 = 19 − 7 = 12