Physics - Capacitor Question with Solution | TestHub

In steady state, there will be no current in the capacitor branch. Net resistance of the circuit $R=1+1+0.5=2.5\Omega$ Current drawn from the cell $i=\frac{V}{R}=\frac{2.5}{2.5}=1\mathrm{A}$ Potential drop across two parallel branches $\begin{array}{rl}V& =E-\text{ ir }=2.5-1\times 0.5\\ & =2.5-0.5=2.0\mathrm{V}\end{array}$ So, charge on the capacitor plates $q=CV=5\times 2=10\mu \mathrm{C}$