SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

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Question : 131 of 200

Marks: +1, -0

Determine the current

i_a

in the given network

- 1 A
1 A
- 4 A
4 A

Solution:

Here,

i _{ a }

is given by,

i _{ a }={ \V _{ x }-12}/{4}

...(1)
Apply

\KVL

at node

\V _{ \x }

,

{ \V _{ x }-12}/{4}+3 i _{ a }+{ \V _{ x }}/{2}=0

Put the value of

i_{a}

in above equation,

{ \V _{ x }-12}/{4}+3({ \V _{ x }-12}/{4})+{ \V _{ x }}/{2}=0

{ \V _{ x }-12}/{4}+{3 \V _{ x }-36}/{4}+{ \V _{ x }}/{2}=0

\V _{ x }-12+3 \V _{ x }-36+2 \V _{ x }=0

6 \V _{ x }=48

\V _{ x }=8 \V

Put this value of V in equation (1)

i _{ a }=(8-12) / 4=-1 \A

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