A 10F capacitor is charged to a potential difference of500V. It is then disconne

Question

A 10F capacitor is charged to a potential difference of500V. It is then disconnect from the voltage supply and connectedin parallel to a 40F capacitor, initially uncharged.Calculate the total energy stored in the capacitors (a) before (b) after connection. Explain the conservation of energy during theprocess. A 10F capacitor is charged to a potential difference of500V. It is then disconnect from the voltage supply and connectedin parallel to a 40F capacitor, initially uncharged.Calculate the total energy stored in the capacitors (a) before (b) after connection. Explain the conservation of energy during theprocess.

Explanation / Answer

The capacitor is C = 10F = 10 * 10-6 F The capacitor is charged to a potential difference of V =500V (a)The total energy stored in the capacitor beforedisconnecting from the voltage supply is E = (1/2)CV2 or E = (1/2) * 10 * 10-6 * (500)2 or E = 1.25 J (b)The total energy stored in the capacitor afterdisconnecting from the voltage supply is E = (1/2)CeqV2 Here,Ceq= C1 + C2= (10 +40)F = 50F = 50 * 10-6 F or E = (1/2) * 50 * 10-6 * (500)2 or E = 6.25 J The Law of conservation of charge states that the net chargeof an isolated system remains constant. If a system starts out with anequal number of positive and negative charges, there's nothing wecan do to create an excess of one kind of charge in that systemunless we bring in charge from outside the system (or remove somecharge from the system). or E = 6.25 J The Law of conservation of charge states that the net chargeof an isolated system remains constant. If a system starts out with anequal number of positive and negative charges, there's nothing wecan do to create an excess of one kind of charge in that systemunless we bring in charge from outside the system (or remove somecharge from the system).

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