Question

Given energy C1 and C2 of 2 microfarad and 3 microfarad respectively charged to a potential difference of 50volts and 100volts respectively. Calculate the loss in energy when two capacitors are connected in parallel

Answer 1

The initial energy of capacitors is

$W_1 = \dfrac{C_1U_1^2}{2} = \dfrac{2\cdot10^{-6}\cdot(50)^2}{2} =0.0025\,\mathrm{J} , \\ W_2 = \dfrac{C_2U_2^2}{2} = \dfrac{3\cdot10^{-6}\cdot(100)^2}{2} = 0.015\,\mathrm{J}.$

The total initial energy is $0.0175\,\mathrm{J}.$

When capacitors are connected in parallel, their voltage is equal and the total capacitance is C = 2+3=5 microfarad. The total initial charge is $Q = Q_1 + Q_2= C_1U_1 + C_2U_2 = 4\cdot10^{-4}\mathrm{C}.$

Therefore, the voltage will be $U = \dfrac{Q}{C_1+C_2} = \dfrac{4\cdot10^{-4}}{5\cdot10^{-6}} = 80\,\mathrm{V}.$

The total energy will be $W = \dfrac{C_1U^2}{2} + \dfrac{C_2U^2}{2} = \dfrac{(C_1+C_2)U^2}{2} = \dfrac{5\cdot10^{-6}\cdot80^2}{2} = 0.016\,\mathrm{J}.$ The loss of energy is 0.0015 J.