A capacitor of unknown capacity is connected across a battery of V volt. The charge stored in it is Q coulomb. When potential across the capacitor is reduced by V 1 volt, the charge stored in it bec

Question

A capacitor of unknown capacity is connected across a battery of V volt. The charge stored in it is Q coulomb. When potential across the capacitor is reduced by V 1 ​ volt, the charge stored in it becomes Q 1 ​ coulomb. The potential V is

TestHub · Accepted Answer

Let the capacitance of the capacitor be $C$ .

Initially, the charge stored is $Q$ at potential $V$ .

$Q = C V \Rightarrow C = \frac{Q}{V}$

When the potential is reduced by $V_{1}$ , the new potential is $V - V_{1}$ and the new charge is $Q_{1}$ .

$Q_{1} = C (V - V_{1})$

Substituting the value of $C$ :

$Q_{1} = \frac{Q}{V} (V - V_{1})$

$Q_{1} V = Q V - Q V_{1}$

$Q V - Q_{1} V = Q V_{1}$

$V (Q - Q_{1}) = Q V_{1}$

$V = \frac{Q V _{1}}{Q - Q _{1}}$

Answer: $\frac{Q V _{1}}{Q - Q _{1}}$

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