Question

For a E-R-C series circuit $E=10volt;C=1mF;R=100\Omega$ then pick correct statements. Switch is closed at t = 0. (Initially capacitor was uncharged)

• A. Rate of energy storage in capacitor is maximum at $t=0.0693sec$
• B. Rate of energy storage in capacitor is maximum at $t=0.053sec$
• C. Potential across capacitor and resistor become same at $t=0.0693sec$
• D. Potential across capacitor and resistor become same at $t=0.053sec$

Answer 1

Answer: (A), (C)

The correct options are
A Rate of energy storage in capacitor is maximum at $t=0.0693sec$
C Potential across capacitor and resistor become same at $t=0.0693sec$

The equation of potential and charge during charging are
$V\left(t\right)=E(1-e^{-t/CR})$ and $q=q_0(1-e^{-t/CR})$
Now the stored energy in the capacitor is $U\left(t\right)=\frac{1}{2}C{V}^2\left(t\right)=\frac{1}{2}C{E}^2(1-e^{-t/CR})$
Rate of stored energy is $\frac{dU}{dt}=C{E}^2(1-e^{-t/CR})\left(1/CR\right)e^{-t/CR}=\frac{E^2}{R}(e^{-t/CR}-e^{-2t/CR})$
For maximum rate of stored energy,
$\frac{d^{2}U}{d{t}^2}=\frac{E^2}{R}[-(1/CR)e^{-t/CR}+(2/CR\left)e^{-2t/CR}\right]=0$
or $2e^{-2t/CR}=e^{-t/CR}\Rightarrow 2=e^{t/CR},t=CRln2$
Thus the time for maximum rate of stored energy is $t=CRln2={10}^{-3}\times 100\times ln2=0.0693s$
At $t=CRln2$ potential across R is $V_R=E(1-e^{-ln2})=E(1-1/2)=E/2$
At $t=CRln2$ potential across R is $V_C=\frac{q_0}{C}(1-e^{-ln2})=E(1-1/2)=E/2$