Heater- $1$ , takes $3$ minutes to boil a given amount of water. Heater- $2$ takes $6$ minutes. If $t_{1}$ and $t_{2}$ are the time taken to boil the same amount of water, when the two heaters are connected in series and parallel, respectively, then $\frac{t_{1}}{t_{2}} = ?$

$[$ Assume same cell is connected in all the cases $]$

\frac{9}{2}

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\frac{2}{9}

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\frac{4}{9}

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\frac{9}{4}

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Solution

The correct option is A $\frac{9}{2}$
Here, the heat required $($ say $H)$ to boil water is the same.

Let $P_{1}$ and $P_{2}$ are the powers of the heaters.

Then, $H = P_{1} t_{1} = P_{2} t_{2}$

So,
$P_{1} = \frac{H}{t_{1}} = \frac{H}{3}$

$P_{2} = \frac{H}{t_{2}} = \frac{H}{6}$

In series,

$P = \frac{P_{1} P_{2}}{P_{1} + P_{2}} - \frac{(\frac{H}{3} \times \frac{H}{6})}{(\frac{H}{3} + \frac{H}{6})} = \frac{H}{9}$

Now,
$t_{1} = \frac{H}{P} = \frac{H}{(\frac{H}{9})} = 9 minutes$

In parallel,

$P = P_{1} + P_{2} = \frac{H}{3} + \frac{H}{6} = \frac{H}{2}$

$∴ t_{2} = \frac{H}{P} = \frac{H}{\frac{H}{2}} = 2 minutes$

Further,

$\frac{t_{1}}{t_{2}} = \frac{9}{2}$

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Heater-1, takes 3 minutes to boil a given amount of water. Heater-2 takes 6 minutes. If t1 and t2 are the time taken to boil the same amount of water, when the two heaters are connected in series and parallel, respectively, then t1t2=? [Assume same cell is connected in all the cases]