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A 4 muF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and isconnecled toanother uncharged 2 muf capacitor. How much electrostatic energy of the first capacilor is lost in the form of heat and electromagnetic radiation ? |
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Answer» Solution :Energy stored in capacitor of 4 `mu`F initially . `U_(i)=(1)/(2) CV^(2)= (1)/(2) xx4xx10^(-6)xx(200)^(2)` `U_(i) = 8xx10^(-2) J` CHARGE on capacitor of `4 muF` `Q =CV = 4xx10^(-6) xx200` `:. Q = 8xx10^(-4)` C Potential of combination of CAPACITORS `4muF` and `2 muF` `V_(1) = ("Total charge")/("Total CAPACITANCE") = (0+8xx10^(-4))/((4+2)xx10^(-6))` `:.V_(1)=(8xx10^(-4))/(6xx10^(-6))=(800)/(6) =(400)/(3) V ` Final electrostatic energy of combination `U_(f) = (1)/(2) (C_(1)+C_(2))V_(1)^(2)` `=(1)/(2)(4+2)xx10^(-6)xx((400)/(3))^(2)` `=(1)/(2)xx(6xx10^(-6)xx16xx10^(4))/(9)` `:. U_(f) =(16)/(3) xx10^(-2)J` `= 5.33xx1^(-2) J` Hence amount of electrostatic energy lost by capacitor of `4 muF ` `DeltaU = U_(i) -U_(f)` `= (8-5.33) xx10^(-2) J` `= 2.67xx10^(-2) J` |
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