In the given circuit- the two-port network has the impendance matrix -Z- - - 40 60- 60 120 - The value of ZL for which maximum power is transferred to the load is -

From maximum power transfer theorem Z_L nbsp;= Z_th Z_th nbsp;= Z_22 nbsp; Z_12× Z_21R_S + Z_11 For given data, Z_th nbsp;= 120 60 × 6010 + 40 ...

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Z- parameters

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Question

In the given circuit, the two-port network has the impendance matrix $[Z] = [\begin{matrix} 40 & 60 60 & 120 \end{matrix}]$ . The value of $Z_{L}$ for which maximum power is transferred to the load is $Ω$ .

48

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Solution

The correct option is A 48
From maximum power transfer theorem

$Z_{L} = Z_{t h}$

$Z_{t h} = Z_{22} - \frac{Z_{12} \times Z_{21}}{R_{S} + Z_{11}}$

For given data,

$Z_{t h} = 120 - \frac{60 \times 60}{10 + 40} = 48 Ω$

$Z_{L} = 48 Ω$

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In the given circuit, the two-port network has the impendance matrix [Z]=[406060120]. The value of ZL for which maximum power is transferred to the load is Ω. 48

The correct option is A 48From maximum power transfer theorem ZL=Zth Zth=Z22−Z12×Z21RS+Z11 For given data, Zth=120−60×6010+40=48Ω ZL=48Ω

In the given circuit, the two-port network has the impendance matrix $[Z] = [\begin{matrix} 40 & 60 60 & 120 \end{matrix}]$ . The value of $Z_{L}$ for which maximum power is transferred to the load is $Ω$ .

48

The correct option is A 48
From maximum power transfer theorem

$Z_{L} = Z_{t h}$

$Z_{t h} = Z_{22} - \frac{Z_{12} \times Z_{21}}{R_{S} + Z_{11}}$

For given data,

$Z_{t h} = 120 - \frac{60 \times 60}{10 + 40} = 48 Ω$

$Z_{L} = 48 Ω$