Two capacitors, each of capacitance 2\(\mu\)F are connected in parallel. If the p.d across them is 120V, calculate the charge on each capacitor
- 6.0 x 10-5C
- 1.2 x 10-4C
-
2.4 x 10-4C
- 4.8 x 10-4C
The correct answer is: C
Explanation
C = \(C_1 + C_2 = 2 + 2 = 4 \mu F\)
but Q = CV
= 4 x 10\(^{-6}\) x 120
= 4.8 x 10\(^{-4}\)
\(\therefore\) Each of the capacitor has a charge of \(\frac{4.8 \times 10^{-4}}{2}\)
= 2.4 x 10\(^{-4}\) C