SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

Question : 129 of 200

Marks: +1, -0

Determine the current in each branch

( \I _{1}, \I _{2}, \I _{3}

and

\I _{4})

of the given network.

Solution:

Let node Voltage across

12 \Ω

be V
By applying

\KCL

\I _{1}= \I _{2}+ \I _{3}+ \I _{4}

{60- \V }/{7}={ \V }/{12}+{ \V }/{6}+{ \V }/{12}

{60- \V }/{7}={ \V +2 \V + \V }/{12}

{60- \V }/{7}={4 \V }/{12}={ \V }/{3}

3 ×(60- \V )=7 \V

180-3 \V =7 \V

10 \V =180

\V =18 \Volts

\I _{1}={60- \V }/{7}={60-18}/{7}=6 \A

\I _{2}={ \V }/{12}={18}/{12}=1.5 \A

\I _{3}={ \V }/{6}={18}/{6}= 3 \A

\I _{4}={ \V }/{12}={18}/{12}=1.5 \A

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