Question

Calculate the quantity of heat produced in a $20\Omega$ resistor carrying $2.5A$current in $5$ minutes.

Answer 1

Step 1 Given data:
$$\begin{gathered} Current\left(i\right)=2.5A \\ Resis\tan ce\left(R\right)=20\Omega \\ time\left(t\right)=5minutes=5\times 60sec=300seconds \end{gathered}$$

Step 2 Formula used:
Quantity of heat produced is given
$Heatproduced=\left(Potentialdifference\right)\times \left(current\right)\times \left(time\right)$
Numerically $H=V\times i\times t$
$H=i^2\times R\times t(fromOhm\text{'}slaw,V=i\times R)$
Where $V-$potential difference, $R-$resistance, $i-$current, $H-$heat produced, $t-$time

Step 3 Finding the quantity of heat produced:
Quantity of heat produced $\left(H\right)$$=i^2\times R\times t$
$\left(H\right)$$={\left(2.5\right)}^{2}A\times 20\Omega \times 300sec$
$\left(H\right)$$=37500J$
Since $1kJ=1000J$
therefore, $\left(H\right)$$=37.5kJ$
Hence, $37.5kJ$ heat is produced.