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, $\text{C} = \epsilon _{0} \text{A/d}$
And new capacitance $\text{C}^{'} = \frac{\epsilon _{0} \text{A}}{\text{d/2}}$ , in series with $= \frac{\epsilon _{0} \text{K} \text{A}}{\text{d/2}}$
$\Rightarrow \text{ C}^{\text{'}} = \, \frac{2 \in _{0 \, \, \, } \text{A}}{\text{d}} \text{in series with } \frac{4 \in _{0} \text{A}}{\text{d}}$
$=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\frac{2 \times 4}{2+4}\right)=\frac{4 \varepsilon_0 \mathrm{~A}}{3 \mathrm{~d}}=\frac{4}{3} \mathrm{C}$
$\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