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A circuit has section AB as shown in figure. The emf of the source equals to E = 10 V, the capacitor capacitances are equal to $C_{1} = 1.0 μ F a n d C_{2} = 2.0 μ F$ and the potential difference $V_{A} - V_{B} = 5.0 V$ . Find the voltage across each capacitor.

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5V,10V

15V, 10V

10V,5V

10V,15V

answer is C.

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Detailed Solution

Q shown in figure is in $μ C .$

Now, $V_{A} - \frac{q}{1} + 10 - \frac{q}{2} = V_{B}$

Or $V_{A} - V_{B} = \frac{3 q}{2} - 10 = 5$

$∴ q = 10 μ C$

Now, $V = \frac{q}{C}$ across each capacitor.

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