Question

an electric heater of resistance 8 ohm draws 15 amphere, from the service main in 2 hour. caulculate the rate at which heat is developed in the heater

Answer 1

USE FORMULA OF HEATING EFFECT OF CURRENT :



$$\begin{gathered} H=i^{2}rthereHisheatdeveloped,iiscurrent,risresis\tan ce,andtistime. \\ {\left(15\right)}^2\times 8\times 2\left(hour\right)asalldatagiveninquestion. \\ 225\times 8\times 2\times 3600(second) \\ 1800\times 7200 \\ 12960000joule. \end{gathered}$$

hence, 12960000 joule of energy is consumed

but remember here according to question it has been asking rate of heat developed means power consumed not the energy consumed

hence power = energy/time

= $$\begin{gathered} \frac{12960000joule}{7200second} \\ =1800watt. \end{gathered}$$