In the circuit shown, E=15V,r=1Ω,R1=2Ω,R2=4Ω,R3=15Ω and C=6μF. Find the charge (in μC) on the top right capacitor at steady state.

Current through battery , i=ER2+r=154+1=3APotential difference across battery, V=E-iR=15-31=12VThis potential difference is divided equally on both capacitor on top branch due to symmetry.So, Potential difference across the capacitor  VC=122V=6VCharge on capacitor  =CVC=6×6=36μC