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Question

(B) For an alternating current $(ac)$ source,the relationship between the peak voltage $(V_0)$ and the root mean square $(rms)$ voltage $(V_{rms})$ is

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$\sqrt{2}$ times the $rms$ value of the $ac$ source

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(B) For an alternating current $(ac)$ source,the relationship between the peak voltage $(V_0)$ and the root mean square $(rms)$ voltage $(V_{rms})$ is given by the formula: $V_{rms} = \frac{V_0}{\sqrt{2}}$.Rearranging this formula to solve for the peak voltage,we get: $V_0 = \sqrt{2} 	imes V_{rms}$.Therefore,the peak voltage is $\sqrt{2}$ times the $rms$ value of the $ac$ source.

The peak voltage of the $ac$ source is equal to:

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