SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

Question : 119 of 200

Marks: +1, -0

What will be the current (in A) through 4 ohms resistor, if a source of 4 V with internal resistance of 2 ohms is connected at x–y terminals with positive terminal at x?

0.75
0.24
1.62
1.44

Validate

Solution:

Applying KVL in loop 1 we get

-4 + 4I + 4I_1 = 0

∴ I + I_1 = 1

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

-4I_1 + 2(I - I_1) + 4 = 0

-4I_1 + 2I - 2I_1 + 4 = 0

2I - 6I_1 = -4

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

I_1= 0.75A

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