Question

What is the short-circuit impedance at port 1 for the given two-port network?

• A. \(3.33~\Omega\)
• B. \(4~\Omega\)
• C. \(14~\Omega\)
• D. \(23.33~\Omega\)

Hint:

For \(Z\)-parameters: \[ Z_{11} = \left.\frac{V_1}{I_1}\right|_{I_2=0} \] (port 2 open-circuited), whereas a short-circuit condition implies \(V_2=0\). Always check carefully whether the question is asking for a Z-parameter or simply the impedance seen with a shorted port.

Answer 1

The Correct Option is A

Concept: The short-circuit input impedance at port 1 is defined as \[ Z_{11} = \left.\frac{V_1}{I_1}\right|_{V_2=0}. \] The condition \(V_2=0\) means that port 2 is short-circuited. Therefore, we first short-circuit port 2 and then determine the equivalent impedance seen from port 1.

Step 1: Short-circuit port 2.
When port 2 is short-circuited, \[ V_2=0. \] Thus terminal \(2\) becomes directly connected to terminal \(2'\) (ground). As a result, the \(20\Omega\) resistor is connected between the central node and ground. The circuit now contains:
• \(5\Omega\) from the middle node to ground.
• \(20\Omega\) from the middle node to ground.
• \(10\Omega\) between port 1 and the middle node.

Step 2: Combine the \(5\Omega\) and \(20\Omega\) resistors.
The \(5\Omega\) and \(20\Omega\) resistors are connected in parallel. Hence, \[ R_p = \frac{5\times20}{5+20}. \] \[ = \frac{100}{25}. \] \[ = 4\Omega. \] Therefore, the network seen after the \(10\Omega\) resistor is equivalent to \[ 4\Omega. \]

Step 3: Determine the impedance seen from port 1.
The \(10\Omega\) resistor is in series with the equivalent \(4\Omega\). Thus, \[ Z_{11} = 10+4. \] \[ = 14\Omega. \]

Step 4: Interpret the answer according to two-port terminology.
For the short-circuit driving-point impedance at port 1, \[ Z_{11} = \left.\frac{V_1}{I_1}\right|_{V_2=0} = 14\Omega. \] Hence, \[ \boxed{14~\Omega} \] which corresponds to option (C).