A heater is designed to operate with a power of `1000W` in a `100 V` line. It is connected in combination with a resistance of `10 Omega` and a resistance `R`, to a `100 V` mains as shown in figure. What will be the value of `R` so that the heater operates with a power of `62.5W`?

Explanation

(a)
Resistance of heater
`R_(h) = ((100)^(2))/(1000) = 10Omega`
Power consumed in water
`P_(h) = 62.5W`
Current through heater
`i_(h)^(2) = (P_h)/(R_h) = (62.5)/(10)`
`i_(h) = 2.5A`
`V_(h) = i_(h)R_(h) = 2.5 xx 10 = 25V`
`p.d` across resistance `10Omega = 100-25 = 75V`
Current in it = `75/10 = 7.5A`
Current in `R = 7.5-2.5 = 5A`
`25=5R rArr R = 5Omega`
(b) At `220 V` power consumed `= 100W`
Extra power = `100-40 = 60W`
Power consumed into light = `60/100xx60 = 36W = P_(1)`
Since `P prop V^(2)`
At `200V`
`(P')/(P) = ((V')/(V))^(2) implies (P')/(100) = ((200)/(220))^(2)`
`P' = 82.6W`
Extra power = `82.6 - 40 = 42.6W`
Power consumed into light = `60/100 xx 42.6 = 22.5W = P_(2)`
`%` change in light intensity = `((P_2)/(P_1)-1)xx100`
`=(22.5/36 - 1) xx 100`
`=-29%`
i.e, `29%` drop.