In the circuit shown in the figure the heat produced in the 5ohm resistor due to the current flowing through it is 10cal/sec. The heat produced in 4ohm resistor, in cal/sec, is ___________ 

Solution:

$\begin{array}{l}P=\frac{V^2}{R}\Rightarrow \frac{V^2}{5}=10 \to \left(1\right)\\ V_1+V_2=V \to \left(2\right) \\ \frac{V_1}{4}=\frac{V_2}{6}\\ \Rightarrow 3V_1=2V_2 \to \left(3\right) \\ \text{ From }\left(2\right)\mathrm{\&}\left(1\right)\\ \Rightarrow 5V_1=2V\Rightarrow V_1=\frac{2V}{5}\end{array}$
$$\begin{gathered} \text{ so the heat produced in the }4\Omega \text{ resistor, P}=\frac{V_1^2}{4}=\frac{4V^2}{25}\times \frac{1}{4} \\ \left(\text{ from }\left(1\right)V^2=10\times 5=50{Volt}^2\right) \\ \Rightarrow P=\frac{V^2}{25}=\frac{1}{25}\times 50=2cal/sec \end{gathered}$$