JAMB Physics 2017 Paper

Concept:The peak voltage is the maximum value of the alternating supply, and it is related to the r.m.s value by a factor of $\sqrt{2}$.Explanation:The r.m.s value is the effective value of an alternating voltage.The formula connecting r.m.s and peak voltage $V_o$ is:$$V_{\mathrm{r.m.s}} = \frac{V_o}{\sqrt{2}}$$To find the peak value, rearrange the formula:$$V_o = V_{\mathrm{r.m.s}} \times \sqrt{2}$$Substitute the given r.m.s value:$$V_o = 240 \times \sqrt{2}$$Since $\sqrt{2} \approx 1.414$:$$V_o = 240 \times 1.414 = 339.5\ \text{V}$$This is approximately $340\ \text{V}$.Answer:B. $340\ \text{V}$