Question

A certain series R-C circuit is formed using resistance R, a capacitor without dielectric having a capacitance C = 2 F and a battery of emf E = 3 V. The circuit is completed and it is allowed to attain the steady state. After this, at t = 0 half the thickness of the capacitor is filled with a dielectric constant k = 2 as shown in the figure. The system is again allowed to attain a steady state. What will be the heat generated (in joule) in the capacitor between t = 0 and $\text{t}=\infty$ ?

• A. 3
• B. 5
• C. 2
• D. 6

Answer 1

Answer: (A)

Initial charge on capacitor = CE
Initial potential energy of capacitor = CE²/2
Now, $\text{C}={ϵ}_0\text{A/d}$
And new capacitance ${\text{C}}^{{}^{\prime }}=\frac{{ϵ}_0\text{A}}{\text{d/2}}$ , in series with $=\frac{{ϵ}_0\text{K}\text{A}}{\text{d/2}}$
$\Rightarrow {\text{ C}}^{\text{’}}=\frac{2{\in }_0\text{A}}{\text{d}}\text{in series with }\frac{4{\in }_0\text{A}}{\text{d}}$
$=\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}}\left(\frac{2\times 4}{2+4}\right)=\frac{4{\epsilon }_0\mathrm{A}}{3\mathrm{d}}=\frac{4}{3}\mathrm{C}$
$\Rightarrow$ New charge $=\frac{4}{3}\text{CE}$
New energy $=\frac{1}{2}\times \frac{4}{3}{\text{CE}}^2=\frac{2}{3}{\text{CE}}^2$
Now W battery $=\Delta \text{H}+\Delta \text{U}$
$\Rightarrow \text{E}\left(\frac{4}{3}\text{CE}-\text{CE}\right)=\Delta \text{H}+\frac{2}{3}{\left(\text{CE}\right)}^2-\frac{1}{2}{\left(\text{CE}\right)}^2$
$\Rightarrow \Delta \text{H}=\frac{1}{6}{\text{CE}}^2$ = 3