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  • A capacitor of capacitance  is charged to a potential while another capacitor of cap

A capacitor of capacitance C_{1} is charged to a potential V_{1} while another capacitor of capacitance C_{2} is charged to a potential difference V_{2}. The capacitors are now disconnected from their respective charging batteries and connected in parallel to each other.

(a) Find the total energy stored in the two capacitors before they are connected.

(b) Find the total energy stored in the parallel combination of the two capacitors.

(c) Explain the reason for the difference of energy in parallel combination in comparison to the total energy before they are connected.

 

 

 

 
 
 
 
 

Answers (1)

(a) Total energy before they are connected.

        E=\frac{1}{2}C_{1}V_{1}^{2}+\frac{1}{2}C_{2}V_{2}^{2}

(b) Let V be the potential across the parallel combination 

Applying conservation of charge, we can write(C_{1}+C_{2})V=C_{1}V_{1}+C_{2}V_{2}

 V=\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}

The total energy stored in parallel combination 

 E=\frac{1}{2}CV^{2}

E=\frac{1}{2}(C_{1}+C_{2})\left [ \frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}\right ]^{2}

 =\frac{1}{2}\frac{\left ( C_{1}V_{1}+C_{2}V_{2} \right )^{2}}{C_{1}+C_{2}}

(c) The difference of energy is due to the loss of energy which is due to the movement of charge during the sharing of the charge. 

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Safeer PP

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