JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Capacitance

A capacitor C is fully charged with voltage

V_{0}

. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance

\frac{C}{2}

. The energy loss in the process after the charge is distributed between the two capacitor is:

Answer: (A)

Solution

The charged capacitor holds

Q_{0} = C V_{0}

and stores

U_{i} = \frac{1}{2} C V_{0}^{2}

. When it is joined in parallel to an uncharged capacitor of value

\frac{C}{2}

the charge is conserved and both settle to a common voltage

V

, so

C V_{0} = (C + \frac{C}{2}) V

gives

V = \frac{2 V _{0}}{3}

. The final stored energy is

\frac{1}{2} \cdot \frac{3 C}{2} \cdot (\frac{2 V _{0}}{3})^{2} = \frac{1}{3} C V_{0}^{2}

, so the energy dissipated is

U_{i} - U_{f} = \frac{1}{2} C V_{0}^{2} - \frac{1}{3} C V_{0}^{2} = \frac{1}{6} C V_{0}^{2}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option