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(B) In a half-wave rectifier,the output voltage across the resistor $R$ exists only for the positive half-cycle of the input sinusoidal voltage.The in

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(B) In a half-wave rectifier,the output voltage across the resistor $R$ exists only for the positive half-cycle of the input sinusoidal voltage.The instantaneous voltage is given by $V(t) = V_0 \sin(\omega t)$,where $V_0 = 200 \ V$.The root mean square (rms) voltage $V_{rms}$ is defined as:$V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2(t) dt}$For a half-wave rectified signal,the integration is over the interval $[0, T/2]$:$V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T/2} (V_0 \sin(\omega t))^2 dt}$Since $\omega = \frac{2\pi}{T}$,we have:$V_{rms} = V_0 \sqrt{\frac{1}{T} \int_{0}^{T/2} \sin^2(\omega t) dt} = V_0 \sqrt{\frac{1}{T} \int_{0}^{T/2} \frac{1 - \cos(2\omega t)}{2} dt}$$V_{rms} = V_0 \sqrt{\frac{1}{2T} [t - \frac{\sin(2\omega t)}{2\omega}]_{0}^{T/2}} = V_0 \sqrt{\frac{1}{2T} [\frac{T}{2} - 0]} = \frac{V_0}{2}$Given $V_0 = 200 \ V$,the rms voltage is:$V_{rms} = \frac{200}{2} = 100 \ V$.

$A$ sinusoidal voltage of peak value $200 \ V$ is connected to a diode and resistor $R$ in the circuit shown so that half-wave rectification occurs. If the forward resistance of the diode is negligible compared to $R$,the rms voltage (in volt) across $R$ is approximately:

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