Question

Find final energy stored in the capacitor in steady state :

• A. $\frac{1}{4}C{E}^2$
• B. $\frac{1}{2}C{E}^2$
• C. $C{E}^2$
• D. $\frac{1}{8}C{E}^2$

Answer 1

Answer: (B)

$\frac{1}{2}C{E}^2$

In steady state, no current will flow through C so no potential drop across $R_2$ and the potential across $R_1$ is $E$ and it will be also potential across C.

Thus, final energy , $U=\frac{1}{2}C{V}^2=\frac{1}{2}C{E}^2$