Answer to Question #159602 in Electric Circuits for Quanta

Question

A capacitor of unknown capacitance has been charged to a potential difference of 100V and then disconnected from the battery. When the charged capacitor is then connected in parallel to an uncharged 10.0 mF capacitor, the pottential difference across the combination is 30.0 V. Calculate the unknown capacitance.

Answer 1

Let the capacitance of unknown capacitor be $C$ and the charge of unknown capacitor be $Q.$ Then, the unknown capacitor connected in parallel to an uncharged 10.0 mF capacitor. As a result, the charge will distributed between two capacitors. Let the charge on $C$ be $Q_1$ and the charge on 10.0 mF capacitor (let denote it as $C_1$) be $Q_2$. Then, according to the law of conservation of charge, we get:

$$Q=Q_1+Q_2.$$

Since, $Q=CV$ and the voltage drop is the same across each element in the parallel circuit, we can write:

$$CV=CV_1+C_1V_1,$$ $$C=\dfrac{C_1V_1}{V-V_1}=\dfrac{10\cdot10^{-6}\ F\cdot30\ V}{100\ V-30\ V}=4.28\ \mu F.$$