A sinusoidal voltage of peak value 200,V is c | vedclass

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

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$100$

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(B) For a half-wave rectifier,the output voltage across the resistor $R$ exists only for the positive half-cycle of the input sinusoidal voltage.Let the input voltage be $V(t) = V_0 \sin(\omega t)$,where $V_0 = 200\,V$.The rms voltage $V_{rms}$ is defined as $\sqrt{\frac{1}{T} \int_0^T V(t)^2 dt}$.For half-wave rectification,the voltage across $R$ is $V_0 \sin(\omega t)$ for $0 \le t Thus,$V_{rms} = \sqrt{\frac{1}{T} \int_0^{T/2} (V_0 \sin(\omega t))^2 dt} = \sqrt{\frac{V_0^2}{T} \int_0^{T/2} \sin^2(\omega t) dt}$.Using $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$,we get $V_{rms} = \sqrt{\frac{V_0^2}{T} \cdot \frac{T}{4}} = \frac{V_0}{2}$.Substituting $V_0 = 200\,V$,we get $V_{rms} = \frac{200}{2} = 100\,V$.

$A$ 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 value of voltage across $R$ is approximately (in volt):

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