A 4 muF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 mu F capacitor. How much electrostatic energy of the first capacitor is dissipated in the form of heat and electromagnetic radiation ?

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Answer Text

Solution :
Capacitance of a charged capacitor , `C_(1)=4muF=4xx10^(-6) F` <br> supply voltage, `V_(1)=200V` <br> electrostatic energy stored in `C_(1)` in given by <br> `E_(1)=1/2 C_(1)V_(1)^(2)` <br> `=1/2xx4xx10^(-6)xx(200)^(2)` <br> `=8xx10^(-2)J` <br> Capacitance of an uncharged capacitor, `C_(2)=2muF=2xx10^(-6)F`<br> when `C_(2)` is connected to the circuit, the potential acquired by it is `V_(2)` <br> According to the conservation of charge, intial charge capacitor `C_(1)` is equal to the final charge on capacitors `C_(1)` and `C_(2)` <br> `:. V_(2)(C_(1)+C_(2))=C_(1)V_(1)` <br> `V_(2)xx(4+2)xx10^(-6)=4xx10^(-6)xx200` <br> `V_(2)=400/3 V` <br> Electrostatic energy for the combination of two capacitors is given by, <br> `E_(2)=1/2(C_(1)+C_(2))V_(2)^(2)` <br> `=1/2(2+4)xx10^(-6)xx(400/3)^(2)` <br> `=5.33xx10^(2)J` <br> Hence , amount of electrostatic energy lost by capacitor `C_(1)` <br> `=E_(1)-E_(2)` <br> `=0.0800.0533 =0.0267` <br> `=2.67xx10^(-2)J`

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