Water in an electric kettle connected to a 220 V supply takes 5 minutes to reach it's boiling point. How long will it take if the supply voltage falls to 200 V?

Given,

V = 220 V

time (t) = 5 min = 300 sec

From relation

$P = \dfrac{V^2}{R} \\[0.5em]$

and

Heat produced (H) —

$H = P \times t \\[0.5em] \Rightarrow H = (\dfrac{V^2}{R}) \times t \\[0.5em]$

Case 1 for V = 220 V

$H = (\dfrac{220^2}{R}) \times 300 \\[0.5em]$

    [Equation 1]

Case 2 for V = 200 V

$H = (\dfrac{200^2}{R}) \times t \\[0.5em]$

    [Equation 2]

Equating 1 and 2 we get,

$(\dfrac{220^2}{R}) \times 300 = (\dfrac{200^2}{R}) \times t \\[0.5em] \Rightarrow t = (\dfrac{220^2}{200^2}) \times 300 \\[0.5em] \Rightarrow t = 363 \text { sec} \\[0.5em] \Rightarrow t = 6.05 \text { min} \\[0.5em]$

Hence, time taken = 6.05 min