SSC JE Electrical Engineering 26 Sep 2019 Shift 1 Paper

Question : 120 of 200

Marks: +1, -0

When only current source is active in the circuit, find the current through the 10Ω resistor.

Solution:

Applying KVL in above circuit,

-20 + 5I + 10(I -2) = 0

-20 + 5I + 10I -20 = 0

15I = 40

∴

I =40/15

∴

I = 2.66

∴ Current through ‘10Ω’ resistor

= I -2

= 2.66 -2 = 0.66A

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