A capacitor having capacitance \(50 \mu \mathrm{F}\) and initial charge \(5 \mathrm{mC}\) is connected…

A capacitor having capacitance \(50 \mu \mathrm{F}\) and initial charge \(5 \mathrm{mC}\) is connected across a resistor at \(t=0 .\) After one second the potential difference between the plates of the capacitor becomes \(70 \mathrm{~V}\). Then at \(\mathrm{t}=2 \mathrm{sec}\).
  1. \(24 \%\) of initial electrostatic energy will remain in the capacitor.
  2. \(20 \%\) of initial electrostatic energy will remain in the capacitor.
  3. \(49 \%\) of initial electrostatic energy will remain in the capacitor.
  4. \(60 \%\) of initial electrostatic energy will remain in the capacitor.

Solution

Initial Voltage \(=\mathrm{Q} / \mathrm{C}=5 \mathrm{mc} / 50\) microfarad \(=100 \mathrm{~V}\)
\(70=100 \mathrm{e}^{-2 / t}\)
\(\mathrm{~V}=100 \mathrm{e}^{-1 / \mathrm{t}}\)
i.e. \(\mathrm{V}=70 \mathrm{e}^{-1 / 2}\)
Hence \(\mathrm{V}=\frac{70 \times 70}{100}=49 \%\)
\(\mathrm{U} \%=\frac{\frac{1}{2} \mathrm{CV}^{2}}{\frac{1}{2} \mathrm{CV}_{0}^{2}} \times 100=\frac{49^{2}}{100^{2}} \times 100=24 \%\)

Asked in: JEE Mains - Capacitance - Chapter Test

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