If the supply voltage drop from 220 V to 200 V , what would be percentage reduction in heat produced by a 220V - 1000W heater ? Neglect the change of resistance. If the change of resistance is taken into consideration would the reduction of heat produced be smaller or larger than the previously calculated value ? Explain.

Explanation

We , know , `P=(V^(2))/(R)`
So , `R=(V^(2))/(P)`
If the variation of resistance is ignored ,
in the first case , `R=(V_(1)^(2))/(P_(1)),`
in the second case , `R=(V_(2)^(2))/(P_(2)),`
i.e ,
`(V_(1)^(2))/(P_(1)^(2))=(V_(2)^(2))/(P_(2)^(2))"or, "(P_(1))/(P_(2))=(V_(2)^(2))/(V_(1)^(2))`
`"or, "(P_(2)-P_(1))/(P_(1))=(V_(2)^(2)-V_(1)^(2))/(V_(1)^(2))=((V_(2)+V_(1))(V_(2)-V_(1)))/(V_(1)^(2))` ltbrge `((200+220)(200-220))/(220xx220)`
`(P_(2)-P_(1))/(P_(1))=(420xx20)/(220xx220)=-0.1736` (approx.)
So , the power of the heater will reduce by 17.36 % .
i.e., the perecentage reduction of heat produced is 17.36% .
In the above calcuation ,decrease of resistance with voltage drop was ignored ,But actually with decrease in temperature of the coil ,its resistance also decrease .Hence according to the relation `P=(V^(2))/(R)` ,power increases , i.e., heat supplied by the heater also increases So , the reduction of heat produced will be lower than 17.36 % .