Question

A capacitor of capacitance $10\mu \text{F}$ is charged by connecting it to a battery of emf $5\text{V}$. The capacitor is now disconnected and reconnected to the battery with the polarity reversed. Find heat developed in the connecting wires.

• A. $50\text{mJ}$
• B. $500\text{mJ}$
• C. $50\mu \text{J}$
• D. $500\mu \text{J}$

Answer 1

The correct option is D $500\mu \text{J}$

When the capacitor is initially charged using the battery:

Charge on plate $\text{A}$ will be

$Q_i=CV=(10\times {10}^{-6})\times 5=50\mu \text{C}$


When the polarity of the battery is reversed:

Charge on plate $\text{A}$ become

$Q_f=-CV=-50\mu \text{C}$


The charge flows through the battery is,

$\Delta \text{q}=50-(-50)=100\mu \text{C}$

So, work done by the battery,

$W_B=\Delta \text{q V}=100\times {10}^{-6}\times 5=500\times {10}^{-6}\text{J}$

$\therefore W_B=500\mu \text{J}$

Now, since in both cases charge on the capacitor is not changing in magnitude, thus energy stored in both cases will remain the same.

So, heat dissipated = Work done by the battery $=500\mu \text{J}$.

Hence, option $\left(d\right)$ is correct.