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Question
Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point. How long would it have taken if the supply voltage had fallen to 200 V?
Solution
P = `V^2/R`
Heated gained =`(V^2/R)xx t`
`(V_1^2/R)xx t_1` = `(V_2^2/R)xx t_2`
`t_2 = (V_1/V_2)^2 xx t_1`
= `(220/200)^2 xx300 = 363s=6.05 min`
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