The question in textbook:-
A 4uF capacitor is charged by connecting it to 200V supply.It is then connected to an uncharged 2uF capacitor.How much electrostatic energy of the first capacitor is lost in form of heat and electromagnetic radiation?
My question is that why the energy must be lost if the wires in the circuit are assumed to be of 0 resistance.
The energy must simply be divided according to the capacitance of both the capacitor.
Explain in detail.
When a capacitor charges another capacitor why it dissipates energy in form of heat?
2016-01-05 5:11 am
回答 (2)
2016-01-05 7:24 am
before :
Q = C1*V1 = 4*200 = 0.8 mCoulomb
E1 = 1/2C1V1^2 = 2*10^-6*4*10^4 = 8*10^-2 joule
after : just Q is maintained
0.8 = (C1+C2)*V2
V2 = 0.8*10^-3/(6*10^-6) = 800/6 = 400/3 volt
E2 = 1/2(C1+C2)*V2^2 = 3*10^-6*16/9*10^4 = 5.333*10^-2 joule
lost energy ΔE = E1-E2 = 8-5.333 = 2.667 *10^-2 joule
Q = C1*V1 = 4*200 = 0.8 mCoulomb
E1 = 1/2C1V1^2 = 2*10^-6*4*10^4 = 8*10^-2 joule
after : just Q is maintained
0.8 = (C1+C2)*V2
V2 = 0.8*10^-3/(6*10^-6) = 800/6 = 400/3 volt
E2 = 1/2(C1+C2)*V2^2 = 3*10^-6*16/9*10^4 = 5.333*10^-2 joule
lost energy ΔE = E1-E2 = 8-5.333 = 2.667 *10^-2 joule
2016-01-05 6:57 am
Capacitors have dielectric material between their plates which is polarized. This is work and energy is dissipated in the process.
收錄日期: 2021-04-21 16:10:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160104211122AAAYmhM