calculate the potential difference and the energy stored in the capacitor C2 in the circuit.given potential at A is 90v,C1=20microF,C2=30 micro F,C3=15microF.

the circuit is in series with a at one end and the capacitors connecetd in series

Three capacitors c1=20×10-6f,  c2=30×10-6f,    c3=15×10-6f are respectively connected in series.Therefore according to series connection of capacitors,1c=120+130+11510-6=3+2+46010-6=960×10-6     or     c=609×10-6=203×10-6f.Here c is net effective capacitance of the circuit. Suppose total charge Qflows through the network of capacitors connected in series.So,    Q=CV    Here V=90V, therefore, Q=203×10-6×90=6×10-4C.Now potential difference across c2 is,V2=Qc2      or  V2=6×10-430×10-6=20V.The energy stored by the capacitor c2=12c2v22=12×30×10-6×202=15×400×10-6=6×10-3J.

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