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A 600pF capacitor is charged a 200V supply. It is then disconnnected from the supply and is connected to another uncharged 600pF capacitor. How much electro static energy is lost in the process. |
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Answer» Solution :`C_(1)=600pF, V_(1)=200V,""C_(2)=600pF, triangleE=?` `E_(1)=1/2 C_(1)V_(1)^(2)=1/2 xx 600 xx 10^(-12) xx 200 xx 200=12 xx 10^(-6)J` `E_(2)=1/2 (Q^(2))/(C_(1)+C_(2))=1/2 xx ((C_(1)V_(1))^(2))/(C_(1)+C_(2))=1/2 xx ((600 xx 10^(-12) xx 200)^(2))/(1200 xx 10^(-12))` `=1/2 xx ((600)^(2) xx (10^(-2))^(2) xx 200 xx 200)/(1200 xx 10^(-12))=(600 xx 600 xx 10^(-12) xx 100)/(6)=6 xx 10^(-6)J` `triangleE=6 xx 10^(-6)J` |
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