Question

In the circuit given below, what should be the value of the resistance \(R\) for maximum dissipation of power in \(R\)?

• A. \(1.2~\text{k}\Omega\)
• B. \(2.2~\text{k}\Omega\)
• C. \(3.2~\text{k}\Omega\)
• D. \(4.2~\text{k}\Omega\)

Hint:

- Maximum power is delivered to a load when the load resistance equals the Thevenin resistance seen at its terminals.
- For an ideal voltage source, short it when computing \(R_{\text{Th}}\).

Answer 1

The Correct Option is B

Step 1: For maximum power transfer to \(R\), set \(R=R_{\text{Th}}\), where \(R_{\text{Th}}\) is the Thevenin resistance seen from the terminals of \(R\). Replace the ideal \(10~\text{V}\) source by a short circuit.
Step 2: With the source shorted, the \(2~\text{k}\Omega\) and \(3~\text{k}\Omega\) resistors are in parallel between the middle node and ground, and this combination is in series with the \(1~\text{k}\Omega\) resistor to the \(R\) terminal. Hence, \[ R_{\text{Th}} = 1~\text{k}\Omega + \left(2~\text{k}\Omega \parallel 3~\text{k}\Omega\right) = 1 + \frac{2\times 3}{2+3}~\text{k}\Omega = 1 + 1.2 = 2.2~\text{k}\Omega. \] Step 3: Therefore, \(R\) should be chosen as \(R_{\text{Th}}=2.2~\text{k}\Omega\) for maximum power dissipation.