SSC JE Electrical Engineering 2011 Paper

Question : 107 of 150

Marks: +1, -0

The V – 1 relation for the networks shown in the given box is V = 41 = 9. If now a resistor R = 2 Ω is connected across it, then the value of I will be.

– 4.5 A
– 1.5 A
1.5 A
4.5 A

Validate

Solution:

V – I Relation for the Network

V = 4I + 9

Lets voltage is zero is Network then short circuit current

I_{\SC}={-9}/4

for current is zero the open circuit voltage

V_{\OC}=0

R_{\th}={V_{\OC}}/I_{\SC}-9/{-9/4}=-4Ω

I=9/{(-4+2)}=-45A

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