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Electricity

Cool goIITian

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30 Mar 2009 12:18:17 IST

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consider a circuit with only capacitor and a battery..... no resistance of wire...now the cap will g

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consider a circuit with only capacitor and a battery..... no resistance of wire...now the cap will get instantly charged to cv...so energy in cap is 1/2cv^2......but energy supplied by battery is cv*v=cv^2......where does extra 1/2cv^2 goes....?

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ADITYA VEERENDRA PUTTA

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30 Mar 2009 12:25:25 IST

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ENERGY SUPPLIED BY THE BATTERY = ENERGY STORED IN THE CAPACITOR + TOTAL WORK DONE TO CHARGE THE CAPACITOR

==>> THE EXTRA 1/2 CV^2 U WERE REFFERING TO IS THE TOTAL WORK DONE TO CHARGE THE CAPACITOR.

sajal kumar

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30 Mar 2009 12:28:08 IST

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if u say this then if we consider a circuit with a resistance then where is the heat loss in resistor....?

pranav agrawal

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30 Mar 2009 12:29:36 IST

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the rest of the energy is lost in the form of heat!

ADITYA VEERENDRA PUTTA

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30 Mar 2009 12:30:32 IST

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I COULDNT GET U SAJAL

pranav agrawal

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30 Mar 2009 12:39:26 IST

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now sajal is offline!

ADITYA VEERENDRA PUTTA

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30 Mar 2009 12:41:27 IST

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K BUT WHAT DOES HE MEAN BY [if u say this then if we consider a circuit with a resistance then where is the heat loss in resistor....?]

shane bhatia

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30 Mar 2009 13:03:13 IST

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the value of energy supplied by battery wud change frm cv^2 to smthing else or remain same???

Sujit

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31 Mar 2009 19:01:53 IST

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assume at an instance charge on capacitor is "q".

now electric field exists due to this charge on the plates of capacitor which consequently leads to repulsive force on bringing more charges.

hence work needs to be done to overcome this force and this leads to loss of Q²/2C.

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