Question

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

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

Answer 1

Answer: (B)

$2cal{s}^{-1}$

Let current drawn from source is $I$ and source voltage is $V$.
Current through lower branch as well as $5\Omega$ resistor is $I_5=\frac{V}{5}$
Current through lower branch as well as $4\Omega$ resistor is $I_4=\frac{V}{(4+6)}=\frac{V}{10}$
$\frac{I_4}{I_5}=\left(V/10\right)\left(5/V\right)=1/2$

Heat generated, $H=I^{2}Rt$
For $5\Omega$, resistor $H/t=10=I_5^2{R}_5=I_5^2\left(5\right)\Rightarrow I_5^2=2$
Thus, heat in $4\Omega$ resistor $=H/t=I_4^2{R}_4=\left(1/4\right)I_5^2\left(4\right)=\left(1/4\right)\left(2\right)\left(4\right)=2cal{s}^{-1}$