Question

In the given circuit, for maximum power to be delivered to \(R_L\), its value should be \(\underline{\hspace{1cm}}\) \(\Omega\). (Round off to 2 decimal places.) 

Hint:

For AC maximum power transfer, match the load resistance to the Thevenin resistance—not the magnitude of the complex impedance.

Answer 1

Correct Answer: 1.4

Inductive reactance: \[ X_L = \omega L = 1000 \times 4\text{mH} = 4\Omega. \] Capacitive reactance: \[ X_C = \frac{1}{1000 \times 0.5\text{mF}} = 2\Omega. \] Net series reactance: \[ X_{LC} = X_L - X_C = 4 - 2 = 2\Omega. \] Parallel combination with 2Ω resistor gives: \[ R_{Th} \approx 1.41\Omega. \] Thus maximum power transfer requires: \[ R_L = R_{Th} \approx 1.41\Omega, \] which lies within 1.40–1.42 Ω.