Question

For the circuit shown in the figure-3.305, find the charge stored on capacitor in steady state :

• A. $\frac{R C}{R+R_{0}} E$
• B. $\frac{R C}{R_{0}}\left(E-E_{0}\right)$
• C. Zero
• D. $\frac{R C}{R+R_{0}}\left(E-E_{0}\right)$

Answer 1

Answer: (D)

In steady state current only flows through the outer loop which is given as
$i=\frac{\left[E-E_{0}\right]}{R+R_{0}}$
Writing the equation of potential drop from left side of capacitor to right side of it from the lower branch containing $R_{0}$ gives
$V_{a}-E+E_{0}+i R_{0}=V_{b}$
In above equation we considered left plate of capacitor is $a$ and right one is $b$
$\Rightarrow V_{a}-V_{b}=\left(E-E_{0}\right)-i R_{0}$
$\Rightarrow V_{a}-V_{b}=\left(E-E_{0}\right)\left[1-\frac{R_{0}}{R+R_{0}}\right]$
$\Rightarrow V_{a}-V_{b}=\frac{R(E-E)}{R+R_{0}}$
Thus steady state charge on capacitor is given as
$q=C\left(V_{a}-V_{b}\right)=\frac{C R\left(E-E_{0}\right)}{R+R_{0}}$