Question

A capacitor charged to initial charge of $\mathrm{2C}\epsilon$ is connected to a resistance $R$ and source of emf $\epsilon$ by a switch S. Positive plate of capacitor is connected to positive terminal of source. At t = 0 switch S is closed, then

• A. power delivered by source is $-\frac{{\epsilon }^2}{R}{e}^{-\frac{t}{RC}}$.
• B. charge on capacitor is $\frac{3}{2}\epsilon C$ at $t=3RC\ln \left(2\right)$.
• C. charge on capacitor is $\frac{3}{2}\epsilon C$ at $t=RC\ln \left(2\right)$.
• D. power delivered by source is $-\frac{{\epsilon }^2}{R}{e}^{-\frac{2t}{RC}}$.

Answer 1

Answer: (A), (C)

The correct options are
A power delivered by source is $-\frac{{\epsilon }^2}{R}{e}^{-\frac{t}{RC}}$.
C charge on capacitor is $\frac{3}{2}\epsilon C$ at $t=RC\ln \left(2\right)$.

$\frac{q}{C}=iR+\epsilon \frac{q-C\epsilon }{RC}=i=-\frac{dq}{dt}{\int }_{2\epsilon C}^q\frac{dq}{q-\epsilon C}=-{\int }_0^t\frac{dt}{RC}\ln \left(\frac{q-\epsilon C}{\epsilon C}\right)=-\frac{t}{RC}\Rightarrow \frac{q-\epsilon C}{\epsilon C}=e^{\frac{-t}{RC}}q=\epsilon C(1+e^{\frac{-t}{RC}})\Rightarrow i=-\frac{dq}{dt}=\frac{\epsilon }{R}{e}^{\frac{-t}{RC}}\frac{3\epsilon C}{2}=\epsilon C(1+e^{\frac{-t}{RC}})\Rightarrow e^{\frac{t}{RC}}=2t=RC\ln \left(2\right)Similarly,Power=VI\Rightarrow P_{source}=-\epsilon I=\frac{-{\epsilon }^2}{R}{e}^{\frac{-t}{RC}}$