A heater boils \( x \) amount of water in time \( t_1 \) and another heater boils the same water in time \( t_2 \) having resistances \( R_1 \) and \( R_2 \) respectively. The ratio \( \displaystyle\frac{t_1}{t_2} \) will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{R_1}{R_2} \)
Boiling the same amount of water needs the same heat \( H \) in both cases. A heater is run from the mains supply, so both heaters work at the same voltage \( V \), and by Joule's law of heating
\( \displaystyle H = \frac{V^{2}}{R}\,t \)
\( \displaystyle \frac{V^{2}}{R_1}\,t_1 = \frac{V^{2}}{R_2}\,t_2 \)
\( \displaystyle \frac{t_1}{t_2} = \frac{R_1}{R_2} \)
A heater with a larger resistance draws less power at the same voltage, so it takes longer.