1. The problem statement, all variables and given/known data The Problem is as follows :- https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-snc6/224892_2277870043526_218311126_n.jpg [Broken] Three capacitors of 2 μF, 3μF, and 5μF are indulgently charged with batteries of emf’s 5V,20V and 10V respectively. After disconnected from the batteries these capacitors are joined as shown in the figure with their positive polarity plates are connected to A and negative polarity plates are earthed. Now a battery of 20 V and an uncharged capacitor of 4μF are joined to A as shown with the switch S. When the switch is closed find (i) The potential of junction A, (ii) Final charges on all four capacitors. 2. Relevant equations & The attempt at a solution as the question says Initially the charges on 2 μF, 3μF, and 5μF are 10 μc, 60 μc, 50 μc respectively. After the connection to ‘A’ charge will flow in all the capacitor in such a manner to make the potential of connected plates same. For to do this I used the formula of common potential V_C=(C_1 V_1+〖 C〗_2 V_2+〖 C〗_3 V_3)/(〖 C〗_1+〖 C〗_2+〖 C〗_3 ) After using this formula I got the potential of Point ‘A’ as 12 V. Now a 20 V battery (+)ve terminal is connected to point A hence the potential should be 20 + 12 = 32V. But the answers are as follows: (i) 100/7 V (ii) Charges - 28.56 μc, 42.84 μc, 71.4 μc, 22.88 μc Please friends I am totally stuck in this problem please help me in finding right lane.