SSC JE Electrical Engineering 24 Jan 2018 Evening Paper

Question : 110 of 200

Marks: +1, -0

What will be the resistance (in Ohms) of bulb A for the circuit given below?

Solution:

I^2R_A=100

I^2R_B=10

R_A/R_B=10

R=20^2/110=40/11

R=R_A+R_B=40/11

Solving (1) & (2)

R_A= 3.3

So Resistance of Bulb ‘A’

R_A = 3.3\Ω

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