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WAEC Physics 2012 Paper

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Question : 33

Total: 47

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

Solution:

C = $C_1 + C_2 = 2 + 2 = 4\,\mu\mathrm{F}$

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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