SSC JE Electrical Engineering 23 Jan 2018 Morning Paper

Question : 120 of 200

Marks: +1, -0

Find the current (in mA) flowing through the load resistance

R_L

Solution:

For Applying KVL in loop we get

-20+1KI+(1/2k)I_1=0

...(1)
Applying KVL in Loop (2) we get

-(1/2kI_1)-12+1k(I+I_2-I_1)=0

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

+12 - 2kI_2 + 10 = 0

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

I_1 = 5.77 A

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