By evaluating i^2 R , dt , determine how the energy dissipated as heat relates to the energy stored in the capacitor when it is charged by connecting it to a battery via a resistor.

Step-by-step solution

Let the capacitance be C , resistance be R , and the voltage of the battery be V . The current in the circuit at any time t during charging is given by: i = V R e^ -t/RC The energy dissipated as heat across the resistor is: H = ₀^ i^2 R , dt Substituting the expression for i : H = ₀^ ( V R e^ -t/RC )^2 R , dt H = V^2 R ₀^ e^ -2t/RC , dt H = V^2 R [ e^ -2t/RC -2/RC ]₀^ H = V^2 R ( 0 - ( - RC 2 ) ) H = 1 2 C V^2 The energy stored in the capacitor when fully charged is: U = 1 2 C V^2 Thus, the energy dissipated as hea