Question

In the circuit shown in figure, the heat produced in the $5\Omega$ resistor due to the current flowing through it is $10{\text{cal s}}^{-1}$. The heat generated in the $4\Omega$ resistor is

• A. $1{\text{cal s}}^{-1}$
• B. $2{\text{cal s}}^{-1}$
• C. $3{\text{cal s}}^{-1}$
• D. $4{\text{cal s}}^{-1}$

Answer 1

The correct option is B $2{\text{cal s}}^{-1}$

Let current in $5\Omega$ resistor be $I_1$ and in $4\Omega$ resistors be $I_2$, then $I_2/I_1=1/2$.

$\therefore I_1=2I_2$

Given that heat generated in the $5\Omega$ resistor is $10{\text{cal s}}^{-1}$.

$\therefore I_1^2\times 5=10{\text{cal s}}^{-1}$

$\Rightarrow 4I_2^2\times 5=10{\text{cal s}}^{-1}(\because I_{!}=2I_2)$

$\Rightarrow I_2^2\times 4=2{\text{cal s}}^{-1}$

Therefore, heat generated in the $4\Omega$ resistor will be

$I_2^2\times R=I_2^2\times 4=2{\text{cal s}}^{-1}$

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

Key Concept: The heat generated in $\text{Joules}$ when a current of $I\text{amperes}$ flows through a resistance of $R\text{Ohm}$ for $t\text{seconds}$ is given by $H=I^{2}Rt\text{Joule}=\frac{I^{2}Rt}{4.2}\text{cal}$ This relation is known as Joule's law of electrical heating.