Two capacitors with capacitances of 4.0 F and 6.0 F are connected in series across a 20 V battery. Determine the energy supplied by the battery.

Step-by-step solution

The equivalent capacitance C_ eq of the two capacitors connected in series is given by: C_ eq = C₁ C₂ C₁ + C₂ = 4.0 6.0 4.0 + 6.0 = 2.4 F The total charge Q drawn from the battery is: Q = C_ eq V = 2.4 F 20 V = 48 C The energy supplied by the battery is the work done in moving this charge across the potential difference V : E = Q V = 48 C 20 V = 960 J