Four capacitors of capacitances 2 μ F, 3 μ F, 4 μ F and x μ F are connected to a battery of emf 6 ~V and of negligible internal resistance, as shown in the figure. If the ratio of the charges on x μ F and 4 μ F

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Quantrex Academy · Accepted Answer

Correct answer: B. 5

As x and z are in parallel Four capacitors of capacitances 2 μ F, 3 μ F, 4 μ F and x μ F are connected to a battery of emf 6 ~V and of negligible

⇒ 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 \lleft(6-\frac{9}{x})

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\lleft(6-\frac{9}{x})= \text{ and } (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 μ F, 3 μ F, 4 μ F and x μ F are...

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