A 100 pF capacitor is connected to a 230V, 50 Hz A.C. source. The r.m.s. value of conduction current will be:

7.2 × 10–6 A 3.6 × 10–5 A 1.8 × 10–4 A 0.9 × 10–3 A

7.2 × 10–6 A

The r.m.s. value of conduction current I = $\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{\frac{1}{2 \pi \mathrm{nC}}}=2 \pi \mathrm{nCV}$ or  I = $2 \times 3.14 \times 50 \times 100 \times 10^{-12} \times 230$ = 7.2 × 10–6 A Hence the correct answer will be (A)