Question

For the given circuit, the diode \( D \) is ideal. The power dissipated by the \( 300\ \Omega \) resistor is:

• A. 0.25 W
• B. 0.50 W
• C. 0.75 W
• D. 1 W

Hint:

For half-wave rectification: \( V_{\text{rms}} = \dfrac{V_m}{2} \). Use \( P = I^2 R \) to find power dissipated across resistors.

Answer 1

The Correct Option is C

The input is: \[ v(t) = 60 \cos(314t) \] Since the diode is ideal, it will conduct only during the positive half cycle.
Peak voltage: \( V_m = 60\ \text{V} \)
RMS value for half-wave rectified signal: \[ V_{\text{rms}} = \dfrac{V_m}{2} = 30\ \text{V} \] Total resistance: \[ R = 100 + 200 + 300 = 600\ \Omega \Rightarrow I_{\text{rms}} = \dfrac{30}{600} = 0.05\ \text{A} \] Power across \(300\ \Omega\) resistor: \[ P = I^2 R = (0.05)^2 \cdot 300 = 0.0025 \cdot 300 = 0.75\ \text{W} \]