When one port of a two port network is open circuited, $V_{1} = 4 V_{2}$ and $I_{2} = 0.25 V_{2}$ then, which of the following is correct?

z_{22} = 0.25

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A = 8

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h_{12} = 16

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z_{22} = 0.25

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Solution

The correct option is D $z_{22} = 0.25$
From h-parameters:
$V_{1} = h_{11} I_{1} + h_{12} V_{2}$
$I_{2} = h_{21} I_{1} + I_{22} V_{2}$
For open circuited port,
$\frac{V_{1}}{V_{2}} = 4 = h_{12}$
$and \frac{I_{2}}{V_{2}} = 0.25 = h_{22}$
From T-parameter:
$V_{1} = A V_{2} - B I_{2}$
$and I_{1} = C V_{2} - D I_{2}$
$∴ A = 4$
From z-parameter:
$V_{1} = z_{11} I_{1} + z_{12} I_{2}$
$V_{2} = z_{21} I_{1} + z_{22} I_{2}$
$z_{22} = \frac{1}{0.25} = 4$

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Similar questions

Q. When port

1

of a two port circuit is short circuited,

I_{1} = 4 I_{2}

and

V_{2} = 0.25 I_{2}

. Which of the following is true?

The ABCD matrix for a two-port network is defined by
$[\begin{matrix} V_{1} I_{1} \end{matrix}] = [\begin{matrix} A & B C & D \end{matrix}] [\begin{matrix} V_{2} - I_{2} \end{matrix}]$

The parameter B for the given two-port network(in ohms, correct to two decimal places) is