GATE Electronics and Communications (EC) 2020 Solved Papers

Question : 48 of 65

Marks: +1, -0

The current I in the given network is

Solution:

{i}_{2}=\;{120 ∠-90^{∘}}/{ {Z}}=\;{120 ∠-90^{∘}}/{(80-35 {j})}

=1.3742 ∠-66.37^{∘}

And

+120 ∠-30^{∘}+Z i_{4}=0

{I}_{4}=\;{-(120 ∠-30^{∘})}/{Z}

=\;{120 ∠ 150}/{ {Z}}

=\;{120 ∠ 150}/{(80-35 j)}

=1.3742 ∠ 173.62^{∘}

{i}_{4}= {i}+ {i}_{2}

∴ {i}= {i}_{4}- {i}_{2}

=(1.3742 ∠ 173.62)-1.3742 ∠-66.37^{∘}

=2.38 ∠ 143.625^{∘}

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