A 100 pF capacitor is connected to a 230V, 50 Hz A.C. source. The r.m.s. value of conduction current will be:
Answer & explanation
Correct answer: option 1
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)