Question

A fully charged capacitor has a capacitance $C$. It is discharged through a small coil of resistance wire, embedded in a block of specific heat $s$ and mass $m$ under thermally isolated conditions. If the temperature of the block is raised by $\Delta T$, the potential difference $V$ across the capacitor initially is

• A. $\left(\frac{2 m s \Delta T}{C}\right)^{2}$
• B. $\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$
• C. $\left(\frac{2 m s \Delta T}{C}\right)$
• D. $2 m s \Delta T C$

Answer 1

Answer: (B)

Let $V$ be the potential across the capacitor when it is fully charged, then, energy stored in the capacitor will be
$=\frac{1}{2} C V^{2}$
(where $C=$ capacitance of capacitor)
When the capacitor is fully discharged, loss of energy is in the form of heat $=\Delta H$ As the system is thermally isolated so,
$\Delta H=\frac{1}{2} C V^{2}$
Here, $\Delta H=m s \Delta T$
$\therefore \frac{1}{2} C V^{2} = m s \Delta T $
or $V =\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$