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
-
A.
6.0 x 10-5C -
B.
1.2 x 10-4C -
C.
2.4 x 10-4C -
D.
4.8 x 10-4C
Correct Answer: Option 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