The value of $Z_{21}$ for network shown in figure below is

- 1 Ω

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- 0.8 Ω

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0.6 Ω

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0.8 Ω

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Solution

The correct option is A $- 1 Ω$
Transforming the dependent current source into voltage source, the network is shown as,

Let $I_{3}$ be the current through $2 Ω$
Apply KVL in outer loop,
$- V_{2} + 2 V_{1} + I_{2} - I_{3} + V_{1} = 0$
$- V_{2} + 3 V_{1} + I_{2} - I_{3} = 0$ ...(i)
Also,
$- V_{1} + I_{1} + I_{2} - I_{3} + 2 V_{2} = 0$
$V_{1} = I_{1} + I_{2} - I_{3} + 2 V_{2}$ ...(ii)
From equation (i) and (ii), we get
$5 V_{2} + 3 I_{1} + 4 I_{2} - 4 I_{3} = 0$
where, $I_{3} = \frac{V_{2}}{2}$
$V_{2} = - I_{1} - \frac{4}{3} I_{2}$
Hence, $Z_{21} = {\begin{matrix} \frac{V_{2}}{I_{1}} \end{matrix} ∣ ∣ ∣}_{I_{2} = 0} = - 1 Ω$

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Similar questions

Z_{11}

Z_{21}

For the Y-network shown in the figure, the value of $R_{1}$ (in $Ω$ ) in the equivalent $△$ -network is

i

Z_{11} = 40 Ω, Z_{12} = 60 Ω, Z_{21} = 80 Ω

Z_{22} = 100 Ω

R_{L} = 20 Ω

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