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
B
10V
C
12.5V
D
25V

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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