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  • A   capacitor is charged by a 200 V supply. It is then disconnected from t

A  4 \mu \mathrm{F} capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharge 2 \mu \mathrm{F}capacitor. How much electrostatic energy of the first capacitor is dissipated in the form of heat and electromagnetic radiation?

Option: 1

2.67 \times 10^{-2} \mathrm{~J}


Option: 2

2.67 \times 10^{-4} \mathrm{~J}


Option: 3

3.67 \times 10^{-2} \mathrm{~J}


Option: 4

3.67 \times 10^{-4} \mathrm{~J}


Answers (1)

best_answer

Here,\mathrm{ C_1=4 \mu F=4 \times 10^{-6} \mathrm{~F}, \mathrm{~V}_1=200 \mathrm{~V} }
Initial electrostatic energy stored in \mathrm{\mathrm{C}_1 } is
\mathrm{\begin{aligned} & \mathrm{u}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}_1^2=\frac{1}{2} \times 4 \times 10^{-6} \times 200 \times 200 \\ & \mathrm{u}_1=8 \times 10^{-2} \mathrm{~J} \end{aligned} }
When \mathrm{ 4 \mu \mathrm{F} } capacitor is connected to uncharged capacitor of \mathrm{2 \mu \mathrm{F}, } charge flows and both acquire a common potential which is given by
\mathrm{\begin{aligned} & \mathrm{V}=\frac{\text { Total ch arge }}{\text { Total capacity }} ; V=\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_1+\mathrm{C}_2} \\ & \mathrm{~V}=\frac{4 \times 10^{-6} \times 200}{(4+2) \times 10^{-6}}=\frac{800}{6} \mathrm{~V} \end{aligned} }
\mathrm{\therefore \quad } Final electrostatic energy of both capacitors
\mathrm{\mathrm{u}_2=\frac{1}{2}\left(\mathrm{C}_1+\mathrm{C}_2\right) \mathrm{V}^2=\frac{1}{2} \times 6 \times 10^{-6} \times \frac{800}{6} \times \frac{800}{6}=5.33 \times 10^{-2} \mathrm{~J} }
\mathrm{\therefore \quad } Energy dissipated in the form of heat and electromagnetic radiation is
\mathrm{\mathrm{u}_1-\mathrm{u}_2=\left(8 \times 10^{-2}-5.33 \times 10^{-2}\right) \mathrm{J}=2.67 \times 10^{-2} \mathrm{~J} }

Posted by

Pankaj Sanodiya

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