A two-port network has z-parameters $z_{11}=z_{22}=10\ \Omega$, and $z_{12}=z_{21}=5\ \Omega$.
The value of $R_L$ such that maximum power is transferred to $R_L$ is $\_\_\_\_\_\_\_\_$ $\Omega$.
Port 1 of the network is driven by a 5 V source in series with a 5 $\Omega$ resistor, and the load $R_L$ is connected across port 2, as shown in the figure.

Question

Suresh said, "I did it yesterday."
Which one of the following options is the correct form of this sentence in indirect speech?

Answer 1

A reciprocal two-port network, where $z_{12}=z_{21}$, can always be redrawn as a T-shaped equivalent circuit of three plain resistors. This turns the problem into a series-parallel reduction instead of solving the z-parameter equations algebraically, which is often faster once the equivalent circuit is set up correctly.

The T-equivalent arms are:

$$ Z_a = z_{11}-z_{12} = 10-5 = 5\ \Omega $$ $$ Z_b = z_{22}-z_{21} = 10-5 = 5\ \Omega $$ $$ Z_c = z_{12} = 5\ \Omega $$

Here $Z_a$ sits in the port 1 arm, $Z_b$ sits in the port 2 arm, and $Z_c$ is the shared middle branch connecting both arms to a common reference node.

To find the resistance seen by $R_L$ at port 2, set the 5 V source to zero (short it), leaving only its series resistance $R_s=5\ \Omega$ at port 1. Looking into port 2, we see $Z_b$ in series with the parallel combination of $Z_c$ and $(Z_a+R_s)$:

$$ Z_a+R_s = 5+5 = 10\ \Omega $$ $$ (Z_a+R_s)\parallel Z_c = \frac{10\times5}{10+5} = \frac{50}{15} = 3.33\ \Omega $$ $$ Z_{out} = Z_b + (Z_a+R_s)\parallel Z_c = 5+3.33 = 8.33\ \Omega $$

By the maximum power transfer theorem, the load resistance for maximum power delivered to it equals this output resistance:

$$ R_L = 8.33\ \Omega $$

Let's summarize:

Converting the z-parameters to a T-equivalent circuit turns the problem into a plain series-parallel resistor reduction.
The 5 V source is set to zero for this calculation since only its internal resistance matters for the output resistance, its actual voltage value never enters the calculation of $R_L$.
Both this method and the direct z-parameter algebra agree on $R_L=8.33\ \Omega$, confirming option (B).

$$ \boxed{R_L = 8.33\ \Omega} $$

Answer 2

A reciprocal two-port network, where $z_{12}=z_{21}$, can always be redrawn as a T-shaped equivalent circuit of three plain resistors. This turns the problem into a series-parallel reduction instead of solving the z-parameter equations algebraically, which is often faster once the equivalent circuit is set up correctly.

The T-equivalent arms are:

$$ Z_a = z_{11}-z_{12} = 10-5 = 5\ \Omega $$ $$ Z_b = z_{22}-z_{21} = 10-5 = 5\ \Omega $$ $$ Z_c = z_{12} = 5\ \Omega $$

Here $Z_a$ sits in the port 1 arm, $Z_b$ sits in the port 2 arm, and $Z_c$ is the shared middle branch connecting both arms to a common reference node.

To find the resistance seen by $R_L$ at port 2, set the 5 V source to zero (short it), leaving only its series resistance $R_s=5\ \Omega$ at port 1. Looking into port 2, we see $Z_b$ in series with the parallel combination of $Z_c$ and $(Z_a+R_s)$:

$$ Z_a+R_s = 5+5 = 10\ \Omega $$ $$ (Z_a+R_s)\parallel Z_c = \frac{10\times5}{10+5} = \frac{50}{15} = 3.33\ \Omega $$ $$ Z_{out} = Z_b + (Z_a+R_s)\parallel Z_c = 5+3.33 = 8.33\ \Omega $$

By the maximum power transfer theorem, the load resistance for maximum power delivered to it equals this output resistance:

$$ R_L = 8.33\ \Omega $$

Let's summarize:

Converting the z-parameters to a T-equivalent circuit turns the problem into a plain series-parallel resistor reduction.
The 5 V source is set to zero for this calculation since only its internal resistance matters for the output resistance, its actual voltage value never enters the calculation of $R_L$.
Both this method and the direct z-parameter algebra agree on $R_L=8.33\ \Omega$, confirming option (B).

$$ \boxed{R_L = 8.33\ \Omega} $$

Show Hint

The Correct Option is B

Solution and Explanation

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