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Let the voltage at junction of capacitor, b/w 2 resistances, b/w 3 resistances and at the negative of E volt be x, E, V1 and 0 respectively.Apply KCl at all the nodes respectively.The first eqn. is already given that is voltage across ccapicitor.which is:x – V2 = 10 volt.At x volt from kcl,the eqn. becomes,5x = 2V1 + E + 20....................(1)at V1 volt from kcl,3V1 = x + E...................................(2)at E volt,3E = x + 2V1....................(3)Solving the three eqns. we get,E = 24 volt.
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