A sinusoidal voltage of peak value 200, V is | 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.The i

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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.The instantaneous voltage is $V(t) = V_0 \sin(\omega t)$ for $0 \le t \le T/2$ and $V(t) = 0$ for $T/2 The $rms$ voltage is defined as $V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V(t)^2 dt}$.Substituting the values: $V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T/2} (V_0 \sin(\omega t))^2 dt}$.Since $\int_{0}^{T/2} \sin^2(\omega t) dt = \frac{T}{4}$,we get $V_{rms} = \sqrt{\frac{V_0^2}{T} \cdot \frac{T}{4}} = \frac{V_0}{2}$.Given $V_0 = 200\, V$,the $rms$ voltage is $V_{rms} = \frac{200}{2} = 100\, V$.

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

$A$ 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 across $R$ is approximately.....$V$.

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