WAEC - Physics (2012 - No. 33)
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
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
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