Question

Water in an electric kettle connected to 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?

• A. 6.5 minutes
• B. 5.05 minutes
• C. 6.05 minutes
• D. 6.15 minutes

Answer 1

Answer: (C)

6.05 minutes

The resistance of the kettle remains constant.
Energy dissipated $=Pt$ or $VI\times t$ as $P=VI$.
Substituting, $I=\frac{V}{R}$ (Ohm's law) in the above formula, we get, $P=\frac{V^2}{R}\times t$

The energy supplied when the kettle is heated for $5$ minutes, that is, $300$ seconds at $220V$ is:

$P=\frac{{220}^2}{R}\times 300$ ....(1)

To provide the same energy at $200V$, let the time be $T$.

The energy supplied at $200V$ is given as $T\times \frac{{200}^2}{R}$ ......(2)

Equating both the equations we get,
$T\times \frac{{200}^2}{R}=\frac{{220}^2}{R}\times 300$
$T\times {200}^2=300\times {220}^2$

$T=300\times \frac{{220}^2}{{200}^2}$

$⟹T=363seconds$

Or $T=\frac{363}{60}=6.05$ minutes