InterviewSolution
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InterviewSolution
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A capacitor of capacitance `C_(1) = 1.0 muF` carrying initially a voltage `V = 300 V` is connected in parallel with an uncharged capacitor of capacitance `C_(2) = 2.0 muF`. Find the increment of the electric energy of this system by the moment equilibrium is reached. Explain the result obtained. |
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Answer» Charge contained in the capacitor fo capacitance n`C_(1)` is `q = C_(1) varphi` and the energy, stored in it : `U_(i) = (q^(2))/(2 C_(1)) = (1)/(2) C_(1) varphi^(2)` Now, when the capacitor are connected in parallel, equivalent capacitance of the system, `C = C_(1) + C_(2)` and hence, energy stored in the system : `U_(f) = (C_(1)^(2) varphi^(2))/(2 (C_(1) + C_(2)))`, as charge remains conserved during the process. So, increment in the energy, `Delta U = (C_(1)^(2) varphi^(2))/(2) ((1)/(C_(1) + C_(2)) - (1)/(C_(1))) = (-C_(2) C_(1) varphi^(2))/(2 (C_(1) + C_(2))) = 0.03 mJ` |
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