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Electrical energy is transmitted at the rate of 2.2 MW through the line wire . The resistance of the line wire is `25Omega` .Calculate the percentaage of heat dissipation of the electrical energy for each line voltage :
110 kV

Text Solution

Verified by Experts

`VI=2.2MW=2.2xx10^(6)W`
`V=110kV=110000V,I=(2.2xx10^(6))/(110000)=20A`
`therefore "percentage of heat dissipation"=(I^(2)R)/(VI)xx100%`
=`((20)^(2)xx25)/(2.2xx10^(6))xx100%`
= `0.45%`
[Obviously , it is seen that dissipation of energy becomes less if the voltage is high.]
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