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Two capacitors, each of capacitance 2\(\mu\)F are co...

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 is C

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