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(B) The circuit shown is a half-wave rectifier circuit with a capacitor filter.In a half-wave rectifier,the capacitor charges to the peak value of the

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$200$

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(B) The circuit shown is a half-wave rectifier circuit with a capacitor filter.In a half-wave rectifier,the capacitor charges to the peak value of the input alternating voltage.The relationship between the peak voltage $(V_{peak})$ and the root-mean-square voltage $(V_{rms})$ is given by:$V_{peak} = \sqrt{2} 	imes V_{rms}$Given that $V_{rms} = 141.4\,V$,we calculate the peak voltage as:$V_{peak} = \sqrt{2} 	imes 141.4$Since $\sqrt{2} \approx 1.414$,we have:$V_{peak} = 1.414 	imes 141.4 = 200\,V$Therefore,the maximum potential difference across the condenser is $200\,V$.

An alternating voltage of $141.4\,V$ (rms) is applied to a vacuum diode as shown in the figure. The maximum potential difference across the condenser will be......$V$

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