Question

The initial connection of the two capacitors are shown below. The heat energy dissipated until the system achieves its steady state is

• A. $\frac{C{V}^2}{7}$
• B. $\frac{3C{V}^2}{5}$
• C. $\frac{5C{V}^2}{11}$
• D. $\frac{9C{V}^2}{22}$

Answer 1

The correct option is B $\frac{3C{V}^2}{5}$

Initially:


Let, $E$ is the final potential difference across each capacitor. So, from the conservation of charge,

$2CV+6CV=2CE+3CE$

$\Rightarrow 8CV=5CE$

$\Rightarrow E=\frac{8V}{5}$



Now, the final charge on the capacitors,

$q_{2c}=2C\times \frac{8V}{5}=\frac{16CV}{5}$

$q_{3c}=3C\times \frac{8V}{5}=\frac{24CV}{5}$

Now,
$\text{Heat dissipated}\left(H\right)=\text{Initial stored energy}\left(U_i\right)-\text{Final stored energy}{U}_f$

$U_i=\frac{1}{2}\left(2C\right)(V{)}^2+\frac{1}{2}(3C\left)\right(2V{)}^2=7C{V}^2$

$U_f=\frac{1}{2}\left(2C\right){\left(\frac{8V}{5}\right)}^2+\frac{1}{2}\left(3C\right){\left(\frac{8V}{5}\right)}^2=\frac{32C{V}^2}{5}$

$\therefore H=U_i-U_f=7C{V}^2-\frac{32C{V}^2}{5}=\frac{3C{V}^2}{5}$

Hence, option (b) is correct answer.