In the given circuit the potential difference across the 4-5- F capacitor is

C=9× 4.513.5μ F or C=3 μ F Q=CV=3× 12 μ C=36 μ C Potential difference across 4.5 μ F =364.5 V=8 V Time saving solution V_1V_2=C_2C_1=94.5=21 V_1+V_2=12 V_1=8 V ...

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Byju's Answer

Standard XI

Physics

Parallel and Series Combination of Capacitors

In the given ...

Question

In the given circuit the potential difference across the $4.5 μ F$ capacitor is

\frac{8}{3} V

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4 V

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6 V

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8 V

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Solution

The correct option is D $8 V$
$C = \frac{9 \times 4.5}{13.5} μ F$ or $C = 3 μ F$
$Q = C V = 3 \times 12 μ C = 36 μ C$
Potential difference across $4.5 μ F$
$= \frac{36}{4.5} V = 8 V$

Time saving solution
$\frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}} = \frac{9}{4.5} = \frac{2}{1}$
$V_{1} + V_{2} = 12$
$V_{1} = 8 V$

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In the given circuit the potential difference across the 4.5μF capacitor is

The correct option is D 8 V C=9×4.513.5μF or C=3 μF Q=CV=3×12 μC=36 μC Potential difference across 4.5 μF =364.5 V=8 V Time saving solution V1V2=C2C1=94.5=21 V1+V2=12 V1=8 V

In the given circuit the potential difference across the $4.5 μ F$ capacitor is