In the circuit shown in figure, calculate the charge on 2 μF capacitor in steady state (in μC).

In the circuit we distribute the potentials as shown in figure below with reference to a zero potential considered at negative terminal of 20 V battery.Writing nodal equations for x and y gives 2x+4(x−y)+4(x−10−20)+8(x−10−y+40)=09x−6y=−60⇒3x−2y=−20   ……..(i)and 4(y−x)+8(y−40−x+10)=03y−3x=60⇒y−x=20  …………(ii)Solving Eqs. (i) and (ii) givesx=20VThis gives charge on 2 μF capacitor asq2μF=2x=2×20=40μC