WAEC Physics 2021 Paper

Concept:A capacitor in an a.c. circuit offers opposition called capacitive reactance.It is calculated using the formula:$$X_C = \frac{1}{2\pi f C}$$Explanation:From the diagram, the frequency of the a.c. supply is:$$f = \frac{500}{\pi}\,\mathrm{Hz}$$The capacitance of the capacitor is:$$C = 5\,\mu\mathrm{F} = 5 \times 10^{-6}\,\mathrm{F}$$Substitute these values into the formula:$$X_C = \frac{1}{2\pi \times \frac{500}{\pi} \times 5 \times 10^{-6}}$$Simplify the denominator step by step:$$2\pi \times \frac{500}{\pi} = 1000$$Therefore,$$X_C = \frac{1}{1000 \times 5 \times 10^{-6}} = \frac{1}{0.005}$$$$X_C = 200\,\Omega$$Answer:The reactance of the capacitor is $200\,\Omega$.Correct option: C.