GATE Electrical Engineering (EE) 2021 Solved Papers

Question : 43 of 65

Marks: +1, -0

The input impedance,

Z_{\;\text "in "\;}( {s})

, for the network shown is:

Solution:

{KVL}

at primary side,

-V(s)+4 I_{1}(s)+6 s I_{1}(s)-s I_{2}(s)=0

{V}( {s})=(4 {s}+6 {s}) {I}_{1}( {s})- {sI}_{2}( {s})

.......(1)

{KVL}

at secondary side,

4 {s} {I}_{2}( {s})+5 {I}_{2}( {s})- {sI}_{1}( {s})=0

{I}_{2}( {s})=\;{ {s}}/{4 {s}+5} {I}_{1}( {s})

Put in (1)

V(s)=[(4+6 s)-\;{s^{2}}/{4 s+5}] I_{1}(s)

Z(s)=\;{(4 s+5)(4+6 s)-s^{2}}/{4 s+5}

=\;{16 s+24 s^{2}+20+30 s-s^{2}}/{4 s+5}

=\;{23 s^{2}+46 s+20}/{4 s+5}

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