Four capacitors of capacitances $2 \mu \mathrm{F}, 3 \mu \mathrm{F}, 4 \mu \mathrm{F}$ and $x \mu \mathrm{F}$ are connected to a battery of emf $6 \mathrm{~V}$ and of negligible internal resistance, as shown in the figure. If the ratio of the charges on $x \mu \mathrm{F}$ and $4 \mu \mathrm{F}$ capacitances is $\frac{3}{8}$, then the value of $x$ is

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Aicharya · Accepted Answer

As x and z are in parallel

$\Rightarrow$ Let charges on $2 \mu \mathrm{F}, 3 \mu \mathrm{F}, 4 \mu \mathrm{F}$ and $x \mu \mathrm{F}$ are $q_2, q_3$, $q_4$ and $q_x$. $q_4=C_4 V=4 \mu \mathrm{F} \times 6=24 \mu \mathrm{C}$ $\Rightarrow$ As, given $\frac{q_x}{q_4}=\frac{3}{8} \Rightarrow q_x=24 \times \frac{3}{8}=9 \mu \mathrm{C}$ Potential difference on $x \mu \mathrm{F}$ $ V_1=\frac{q_x}{x}=\frac{9}{x}\left\{V=\frac{q}{C}\right\} $ Same $\frac{9}{x}$ volt will be across $2 \mu \mathrm{F}$ too. Now, remaining $\left(6-\frac{9}{x}\right)$ volt potential difference will drop across $3 \mu \mathrm{F}$. As $3 \mu \mathrm{F}$ and $(x+2) \mu \mathrm{F}$ are in series, so charge on them will be same $ \begin{aligned} q_3= & q_{(x+2)} \\ 3\left(6-\frac{9}{x}\right)= & (x+2) \frac{9}{x} \\ & 6 x-9=3 x+6 \Rightarrow 3 x=15 \Rightarrow x=5 \end{aligned} $

Four capacitors of capacitances $2 \mu \mathrm{F}, 3 \mu \mathrm{F}, 4 \mu \mathrm{F}$ and $x \mu…

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