SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

Question : 120 of 200

Marks: +1, -0

Find the value of current (in A) through the load resistance

(\R_\L )

in the given circuit diagram?

Solution:

Applying KVL in loop 1

-10 + I + 2I_1 + 5 = 0

I + 2I_1 = 5

...(1)
Applying KVL in loop 2 we get,

-5 - 2I_1 + 4I_2 + 10 = 0

∴ 4I_2 - 2I_1 = -5

...(2)
Applying KVL in loop (3) we get

-10 -4I_2 + 6 (I - I_1 -I_2) = 0

...(3)
Solving equation (1), (2) and (3) we get the required value.

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