Q145 / 160 · TS EAMCET 2024 (10 May Shift 1) · Physics
A $10 \mu \mathrm{~F}$ capacitor is charged by a 100 V battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance $30 \mu \mathrm{~F}$. During this process, the electrostatic energy lost by the first capacitor is
- A$5 \times 10^{-2} \mathrm{~J}$
- B$1.25 \times 10^{-2} \mathrm{~J}$
- C$2.75 \times 10^{-2} \mathrm{~J}$
- Correct answerD$3.75 \times 10^{-2} \mathrm{~J}$
Solution
$\mathrm{C}_1=10 \mu \mathrm{~F}, \mathrm{~V}_1=100 \mathrm{~V}, \mathrm{C}_2=\mu \mathrm{F}, \mathrm{V}_2=0$
During sharing of capacitors, $\begin{aligned} & \text { Energy lost, } \Delta \mathrm{U}=\frac{1}{2}\left(\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\right)\left(\mathrm{V}_1-\mathrm{V}_2\right)^2 \\ & =\frac{1}{2}\left(\frac{10 \times 30}{10+30}\right)(100)^2 \times 10^{-6} \\ & =3.75 \times 10^{-2} \mathrm{~J} \end{aligned}$
During sharing of capacitors, $\begin{aligned} & \text { Energy lost, } \Delta \mathrm{U}=\frac{1}{2}\left(\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\right)\left(\mathrm{V}_1-\mathrm{V}_2\right)^2 \\ & =\frac{1}{2}\left(\frac{10 \times 30}{10+30}\right)(100)^2 \times 10^{-6} \\ & =3.75 \times 10^{-2} \mathrm{~J} \end{aligned}$
Hence, the correct answer is option: 'D'.