Question

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.

• A. 5.67
• B. 8.33
• C. 10.33
• D. 25.67

Hint:

Zero the 5 V source (keep its 5 \(\Omega\) series resistance) and find the resistance looking into port 2 using \(Z_{out}=z_{22}-\dfrac{z_{12}z_{21}}{z_{11}+R_s}\); set \(R_L\) equal to that.

Answer 1

The Correct Option is B

Step 1: Understand the setup.
The two-port network has \(z_{11}=z_{22}=10\ \Omega\) and \(z_{12}=z_{21}=5\ \Omega\). Port 1 is driven by a 5 V source with a 5 \(\Omega\) resistor in series, and \(R_L\) is the load on port 2. We need the value of \(R_L\) that draws the maximum possible power.

Step 2: Recall the maximum power transfer condition.
By the maximum power transfer theorem, a load draws maximum power when its resistance equals the resistance of the network as seen from its terminals, with all independent sources set to zero (the 5 V source becomes a short, leaving only its 5 \(\Omega\) series resistance \(R_s\) in the circuit). So we need the output resistance looking into port 2, \(Z_{out}\).

Step 3: Write the z-parameter equations.
\[ V_1 = z_{11}I_1+z_{12}I_2 \]
\[ V_2 = z_{21}I_1+z_{22}I_2 \]
With the source zeroed, port 1 only has the resistor \(R_s\) across it, so \(V_1=-R_sI_1\).

Step 4: Eliminate \(I_1\) to get \(Z_{out}\).
Substituting \(V_1=-R_sI_1\) into the first equation,
\[ -R_sI_1 = z_{11}I_1+z_{12}I_2 \ \Rightarrow\ I_1=\frac{-z_{12}I_2}{z_{11}+R_s} \]
Substitute into the second equation:
\[ V_2 = z_{21}I_1+z_{22}I_2 = \left(z_{22}-\frac{z_{12}z_{21}}{z_{11}+R_s}\right)I_2 \]
So
\[ Z_{out} = \frac{V_2}{I_2} = z_{22} - \frac{z_{12}z_{21}}{z_{11}+R_s} \]

Step 5: Substitute the numbers.
\[ Z_{out} = 10 - \frac{5\times5}{10+5} = 10-\frac{25}{15} = 10-1.67 = 8.33\ \Omega \]

Step 6: Apply the maximum power transfer condition.
\[ R_L = Z_{out} = 8.33\ \Omega \]
Options (A), (C) and (D) come from common slips, such as leaving the 5 \(\Omega\) source resistance out of the denominator, adding the coupling term instead of subtracting it, or an arithmetic slip in the division step, none of which apply the formula correctly.

Final Answer:
\[ \boxed{R_L = 8.33\ \Omega} \]