Question

The input resistance $V_{in} / I_{in}$ for the network shown in the figure is}

• A. $1\ \Omega$
• B. $500\ \Omega$
• C. $25\ \Omega$
• D. $50\ \Omega$

Hint:

When dealing with dependent sources, never deactivate them while calculating input or output impedance.
Use test voltage/current method or direct KCL to find the ratio of $V/I$ directly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The objective of this problem is to determine the input resistance of the given electrical network with a dependent current source.
The input resistance is defined as the ratio of the input voltage to the input current:
\[ R_{in} = \frac{V_{in}}{I_{in}} \]
This network contains an independent input voltage source $V_{in}$, a parallel resistor of $50\ \Omega$, and a voltage-dependent current source of value $0.02 V_{in}$ connected in parallel.

Step 2: Key Formula or Approach:
We will apply Kirchhoff's Current Law (KCL) at the node connected to the positive terminal of the input source.
By summing the currents leaving the node, we can establish a relationship between $I_{in}$ and $V_{in}$.

Step 3: Detailed Explanation:

• Let the current entering the parallel combination from the source be $I_{in}$.

• The current flowing through the parallel $50\ \Omega$ resistor is:
\[ I_{R} = \frac{V_{in}}{50} = 0.02 V_{in} \]

• The current flowing through the dependent current source branch is given as:
\[ I_{dep} = 0.02 V_{in} \]

• Applying KCL at the top node:
\[ I_{in} = I_{R} + I_{dep} \]
\[ I_{in} = 0.02 V_{in} + 0.02 V_{in} \]
\[ I_{in} = 0.04 V_{in} \]

• Rearranging the terms to find the input impedance $R_{in}$:
\[ R_{in} = \frac{V_{in}}{I_{in}} = \frac{1}{0.04} = 25\ \Omega \]

Step 4: Final Answer:
The input resistance is found to be $25\ \Omega$, which corresponds to Option (C).