SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

Question : 115 of 200

Marks: +1, -0

What will be the value of current

I_5

(in A) for the network?

Solution:

At node A.

I_1 + I_2 + I_3 = I_4

2 + 4 + 5 = I_4

∴I_4=11A

At node B

I_4 = I_5 + I_6 + I

∴ 11 = I5 + 2 + I

∴9=I_5+I

...(i)
At node C

I = I_7 + I_8 = 4 + 3

∴I =7A

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

9 = I_5 + 7

∴I_5=2A

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