Question

An electric kettle has two heating filaments. One brings it to boil in $10\text{min}$ and the other in $15\text{min}$ when they are connected across the same potential difference. If the two heating filaments are connected in parallel, then water in kettle will boil in

• A. $6\text{min}$
• B. $8\text{min}$
• C. $25\text{min}$
• D. $5\text{min}$

Answer 1

The correct option is A $6\text{min}$

Let the potential given to both the heating elements be $V$.

As we know, heat generation will be

$Q=P\times t=\frac{V^2}{R}t........\left(1\right)$

Given, both the heating filaments produces same amount of heat in different time when applied voltage is same.

$\therefore H_1=H_2=Q$

From eq. $\left(1\right)$

$Q=\frac{V^2}{R_1}{t}_1=\frac{V^2}{R_2}{t}_2$

$\Rightarrow R_1=\frac{V^2}{Q}{t}_1$ and $R_2=\frac{V^2}{Q}{t}_2$

When they are connected in parallel across the same potential difference, they together produces the same amount of heat $\left(Q\right)$ in time $t$.

$Q=\frac{V^2}{R_{eff}}t\Rightarrow R_{eff}=\frac{V^2}{Q}t$

Effective resistance of parallel combination will be

$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}$

putting the value of parameters

$\frac{Q}{V^{2}t}=\frac{Q}{V^2{t}_1}+\frac{Q}{V^2{t}_2}$


$\Rightarrow \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$

$\therefore t=6\text{min}$

Hence, option $\left(a\right)$ is correct.

Key concept: $t=\frac{Q}{P_1+P_2}=\frac{Q}{\frac{Q}{10}+\frac{Q}{15}}=6\text{min}$