WAEC Physics 2015 Paper

Concept:For the same brightness, the direct current voltage equals the root-mean-square (rms) value of the alternating current supply.Explanation:The lamp has the same brightness as with a direct current source of $6\,\text{V}$. So, the rms potential difference is $V_{\text{rms}} = 6\,\text{V}$.The peak potential difference of an AC supply is given by:$$V_0 = V_{\text{rms}} \times \sqrt{2}$$Substitute $V_{\text{rms}} = 6\,\text{V}$ and $\sqrt{2} \approx 1.414$:$$V_0 = 6 \times 1.414 = 8.484\,\text{V}$$Rounding gives approximately $8.5\,\text{V}$.Answer:$V_0 = 8.5\,\text{V}$, so the correct option is C.