Two identical 64.19 mF capacitorsare connected with superconducting wire through
ID: 1755031 • Letter: T
Question
Two identical 64.19 mF capacitorsare connected with superconducting wire through a switch, S, whichis initially open. A charge of 76.26 mCis initially stored on the capacitor on the left; the capacitor onthe right is initially uncharged. The two capacitors, conductingwires, and switch are all contained inside a rigid box composed ofadiabatic walls. How much electrical energy is initially storedinside the box?The switch now is closed and charge begins to flow from thecapacitor on the left to the capacitor on the right. Onceequilibrium is established again, how much electrical energy isstored inside the box?
The answer is incorrect. I used the equationU=1/2(Q^2/C) =1/2(76.26e-6C)^2/964.19e-6F I thought both answers would be the same because it hasadiabiatic walls? What happened to the difference in electrical energy storedinside the box before the switch was closed and after the switchwas closed?
Two identical 64.19 mF capacitorsare connected with superconducting wire through a switch, S, whichis initially open. A charge of 76.26 mCis initially stored on the capacitor on the left; the capacitor onthe right is initially uncharged. The two capacitors, conductingwires, and switch are all contained inside a rigid box composed ofadiabatic walls. How much electrical energy is initially storedinside the box?
The switch now is closed and charge begins to flow from thecapacitor on the left to the capacitor on the right. Onceequilibrium is established again, how much electrical energy isstored inside the box?
The answer is incorrect. I used the equationU=1/2(Q^2/C) =1/2(76.26e-6C)^2/964.19e-6F I thought both answers would be the same because it hasadiabiatic walls? What happened to the difference in electrical energy storedinside the box before the switch was closed and after the switchwas closed?
Explanation / Answer
1)let the two identical capacitors be C1 andC2.The capacitors C1 and C2 areconnected in series therefore the equivalent capacitance is (1/C) = (1/C1) + (1/C2) C1 = C2 = 64.19 F = 64.19 *10-6 F or (1/C) = (1/64.19) + (1/64.19) = (1 + 1/64.19) =(2/64.19) or C = (64.19/2) = 32.095 F when the two capacitors are connected in series the chargeflowing through the two capacitors is same therefore the charge onthe two capacitors is q = 76.26 C the electrical energy which is initially stored insidethe box is E = (1/2)CV2 = (1/2)C * (q/C)2 = (1/2) *(q2/C) 2)When equilibrium is established again the electrical energywhich is stored inside the box is same.This is because the chargeon the two capacitors is same when equilibrium is reached as thetwo capacitors are in series.Related Questions
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