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Question

Two condensers of capacitance $$4\mu F$$ and $$5\mu F$$ are joined in series. If the potential difference across $$5\mu F$$ is $$10V$$, then the potential difference across $$4\mu F$$ condenser is :



A
22.5V
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B
10V
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C
12.5V
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D
25V
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Solution

The correct option is C $$12.5V$$
When two capacitors are in series $$Q$$ present on then is same
$$\therefore\  Q\  on \ 5\mu F$$ is $$Q=5\times10^{-6}\times 10$$
$$Q=5\times 10^{-5}$$
$$\therefore Q$$ on $$4\mu F$$ is $$5\times 10^{-5}$$
$$\therefore V=\dfrac{Q}{C}$$
$$V=\dfrac{5\times 10^{-5}}{4\times 10^{-6}}$$
$$V=12.5\ V$$

Physics

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