A 4.0- mF capacitor is discharged through a 4.0- kΩ resistor. How long will it take for the capacitor to lose half its initial stored energy?
Here energy storedin acapacitor \(E=0.5 \mathrm{CV}_{0}^{2}\)
Voltage across capacitor \(\mathrm{V}=\mathrm{V}_{0} e^{-\frac{1}{R}}\)
Energystoredin capacitor during discharging \(\mathrm{E}=0.5 \mathrm{CV}^{2}\)
We have
\(E=\frac{E_{0}}{2}\)
\(\frac{E_{0}}{2}=0.5 \mathrm{CV}^{2}\)
\(\frac{E_{0}}{2}=0.5 C\left(V_{0} e^{\frac{-t}{R C}}\right)^{2}\)
\(\frac{E_{0}}{2}=E_{0} e^{\frac{-2 t}{R C}}\)
\(0.5=e^{\frac{-2 t}{R C}}\)
\(\ln (0.5)=\frac{-2 t}{R C}\)
\(t=\frac{-R C}{2} \ln (0.5)\)
\(=-\frac{4 \times 10^{3} \times\left(4 \times 10^{-3}\right)}{2} \ln (0.5)\)
\(=5.54 \mathrm{~s}\)
A 4.0- mF capacitor is discharged through a 4.0- kΩ resistor. How long will it take...
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