Question

The circuit shown in figure- $3.343$ is closed at $t=0$. Calculate the total amount of heat generated in $R_{2}$ during the time capacitor gets fully charged:

• A. $\frac{200}{3} \, \mu J$
• B. $\frac{400}{3}\, \mu J$
• C. $\frac{800}{3} \, \mu J$
• D. $400\, \mu J$

Answer 1

Answer: (A)

Current in $R_{2}$ will be half of the current in transient duration which charges the capacitor and it is given as
$i=\frac{1}{2} \times \frac{20}{3} e^{-\frac{t}{2 \times 3}}=\frac{10}{3} e^{-\frac{t}{6}}$
Heat produced in resistor $R_{2}$ is given as
$H=\int\limits_{0}^{\infty} i^{2} R d t$
$\Rightarrow H=\frac{100}{9} \int\limits_{0}^{\infty} e^{-\frac{t}{3}} \times 2 d t$
$\Rightarrow H=\frac{200}{9}\left|\frac{e^{-\frac{t}{3}}}{\left(-\frac{1}{3}\right)}\right|_{0}^{\infty}$
$\Rightarrow H=\frac{200}{3}\, \mu J$
Above heat can also be directly calculated as total heat dissipated is half of the work done by the battery in above case and heat is distributed in same ratio of resistances in the two resistances $R_{1}$ and $R_{2} / 2$ in series. Students are advised to verify the result using this method also.