Question: Q. 2. A capacitor of unknown capacitance is connected across a battery of $V$ volt. A charge of $360 \mu \mathrm{C}$ is stored in it. When the potential across the capacitor is reduced by $120 \mathrm{~V}$, the charge stored in the capacitor becomes $120 \mu \mathrm{C}$. Calculate $V$ and the unknown capacitance. What would have been the charge on the capacitor if the voltage were increased by 120 V? U [Delhi (Comptt.) I 2017]
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Solution:
Ans. Calculation of $V$ and unknown capacitance 2 Calculation of charge when voltage is increased by $120 \mathrm{~V}$ Capacitance
Also,
$$ \begin{aligned} C & =\frac{Q_{1}}{V_{1}} \ C & =\frac{Q_{2}}{V_{2}} \text { and } C=\frac{Q_{3}}{V_{3}} \ \frac{360 \mu C}{V} & =\frac{120 \mu C}{(V-120)} \end{aligned} $$
So,
$$ 3 V-360 V=V $$
$$ V=180 \mathrm{~V} $$
$$ C=\frac{360 \mu V}{180 V}=2 \mu F $$
$$ 2 \mu F=\frac{Q_{3}}{300} $$
$$ Q_{3}=600 \mu C $$
Detailed Answer :
(i) Let the initial yoltage $=V$ volts
Charge stored, $Q_{1}=360 \mu \mathrm{C}$
$$ \begin{equation*} Q_{1}=C V_{1}=C V \tag{i} \end{equation*} $$
Now the changed potential
$$ \begin{align*} & V_{2}=(V-120) \text { volts } \ & Q_{2}=120 \mu C \ & Q_{2}=C V_{2} \tag{ii} \end{align*} $$
Now divide equation (i) by (ii), we have:
$$ \begin{aligned} \frac{Q_{1}}{Q_{2}} & =\frac{C V_{1}}{C V_{2}} \ \frac{360 \mu C}{120 \mu C} & =\frac{V}{V-120} \end{aligned} $$
On solving, we get $V=180$ volts
Now value of unknown capacitance $C$ can be calculated as
$$ \begin{aligned} C & =\frac{Q_{1}}{V}=\frac{360 \times 10^{-6}}{180} \mathrm{~F} \ & =2 \times 10^{-6} \mathrm{~F}=2 \mu \mathrm{F} \end{aligned} $$
(ii) When the voltage applied increases to $120 \mathrm{~V}$, then
$$ V_{3}=180+120=300 \mathrm{~V} \quad 1 / 2 $$
Finally the charge stored in the capacitor will be :
$$ \begin{align*} Q_{3} & =C V_{3} \ & =2 \times 10^{-6} \times 300 C \ & =600 \mu C \end{align*} $$
