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A capacitor of an unknown capacitance is connected to a battery of $\mathrm{V}$ volts. Charge stored inside the capacitor is $720 \mu \mathrm{C}$ . When the potential across the capacitor is reduced by $240 \mathrm{~V}$ , the charge stored becomes $240 \mu \mathrm{C}$ . Determine the charge stored inside the capacitor, if voltage applied gains an increment of $240 \mathrm{~V}$ .
Option: 1
$1200 \mu \mathrm{C}$

Option: 2
$2400 \mu \mathrm{C}$

Option: 3
$3600 \mu \mathrm{C}$

Option: 4
$7200 \mu C$

Answers (1)

best_answer

Initial charge $=720 \mu \mathrm{C}$

$Q_1=C V_1$ $.........(1)$

Changed potential,

$V_2=V-240$

$Q_2=240 \mu C$

$Q_2=C V_2$ $............(2)$

Dividing (1) by (2) -

$\frac{Q_1}{Q_2}=\frac{V_1}{V_2}$

$\frac{720}{240}=\frac{V}{V-240}$

$V=360 \mathrm{~V}$

$C=\frac{Q_1}{V_1}=2 \times 10^{-6} F$

New voltage $=360+240=600 \mathrm{~V}$

Charge $=C V(\text { new })$

$\begin{aligned} & =2 \times 10^{-6} \times 600 C \\ = & 1200 \mu C \end{aligned}$

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Sanket Gandhi

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