Question

A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor is of same capacity is connected in parallel with it. Then energy stored in each capacitor is

• A. $\frac{U}{2}$
• B. $\frac{U}{4}$
• C. $4U$
• D. $2U$

Answer 1

The correct option is B

$\frac{U}{4}$

Step 1: Given

Energy of the battery: $U$

Let the capacity of the battery be $C$

The charge be $Q$

Step 2: Determine the charge

The charge will flow equally between both the capacitors as they both are connected in parallel.

Hence, charge in each capacitor: $\frac{Q}{2}$

Step 3: Determine the potential difference

The potential difference across two capacitors is given by,

$$\begin{gathered} V=\frac{Charge}{Capacity} \\ =\frac{Q}{2}\times \frac{1}{C} \\ =\frac{Q}{2C} \end{gathered}$$

Step 4: Determine the energy

The energy in each capacitor is given by,

$$\begin{gathered} U\text{'}=\frac{1}{2}C{V}^2 \\ =\frac{1}{2}C{\left(\frac{Q}{2C}\right)}^2=\frac{1}{2}C\frac{Q^2}{4C^2} \\ =\frac{1}{4}\times \frac{1}{2}\times \frac{Q^2}{C} \\ =\frac{1}{4}U \end{gathered}$$

(The energy in a capacitor can also be written as $U=\frac{1}{2}\frac{Q^2}{C}$)

Hence, option B is the correct option.