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A 4 \mathrm{\mu F} capacitor is connected to a battery of potential difference 20 \mathrm{~V}. The capacitor get charged and after that battery is removed and charged capacitor is connected to another uncharged capacitor  C_2 of capacitance 10 \mathrm{\mu F}. The charge on \mathrm{C}_2 in equilibrium condition is 14 \mathrm{\mu F}.

Option: 1

10.3 \mathrm{ \mu F} .


Option: 2

12.3 \mathrm{ \mu F} .


Option: 3

14.3 \mathrm{ \mu F} .


Option: 4

14.6 \mathrm{ \mu F} .


Answers (1)

best_answer

When switch S_2 is closed and S_1 remain open, charging of G is take place.

\begin{aligned} Q=C V & \\ Q=C V & =(4 \times 20) \mu \mathrm{F} \\ Q & =80 \mathrm \mu F \end{aligned}

When battery is removed means S_2 is open and S_1 is closed. Then at equilibrium let voltage across both C_1 and C_2  is V.

Using equilibrium of charge - 

\begin{aligned} & Q_1=c_1 V, \quad Q_2=c_2 V \\ & =4 \mathrm{~V} \mathrm{ \mu F}, \quad=10 \mu \mathrm{F} \times \mathrm{V}=10 \mathrm{~V} \mu \mathrm{F} \\ & \end{aligned}

Q=Q_1+Q_2

\begin{aligned} & 20=4 \mathrm{~V}+10 \mathrm{~V} \\ & \Rightarrow 14 \mathrm{~V}=20 \Rightarrow \mathrm{V}=\frac{20}{14}=\frac{10}{7} \\ & Q_2=C_2 \mathrm{~V}=10 \times \frac{10}{7} \mu \mathrm{F}=\frac{100}{7} \mu \mathrm{F} \simeq 14.3 \mathrm{ \mu F} . \end{aligned}

Posted by

Ritika Jonwal

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