Two capacitors of capacitances 4.0μF and 6.0μF are connected in series with a battery of 20V.

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Nisa · Answer 1 · 2018-05-31T01:16:57+0000

∴The E_q Capacitance C_eq = 2.5μF

∴ The energy supplied by the battery to each plate

E = (1/2) CV² = (1/2) × 2.4 × 20² = 480μJ

∴ The energy supplies by the battery to capacitor = 2 × 480 = 960μJ