A network of four capacitors of capacity equal to C₁ = C, C₂ = 2C, C₃ = 3C and C₄=4C are con

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1. Electric Charges and Fields

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A network of four capacitors of capacity equal to $C_1 = C, C_2 = 2C, C_3 = 3C$ and $C_4=4C$ are conducted to a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is

A

$\frac{7}{4}$

B

$\frac{22}{3}$

C

$\frac{3}{22}$

D

$\frac{4}{7}$

Solution

$\mathrm{Q}_{4}=4 \mathrm{CV}$

$\mathrm{Q}_{2}=\left(\frac{6}{11} \mathrm{C}\right) \mathrm{V}=\frac{6 \mathrm{CV}}{11} $

$\Rightarrow \frac{\mathrm{Q}_{2}}{\mathrm{Q}_{4}}=\frac{6 \mathrm{CV}}{11} \times \frac{1}{4 \mathrm{CV}}=\frac{3}{22}$

Standard 12

Physics

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