In the circuit shown in the figure,the 5,Omeg | vedclass

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(B) Let $I_1$ be the current through the $5\,\Omega$ resistance and $I_2$ be the current through the $(6+9)\,\Omega$ branch.Given that the heat develo

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$14.2$

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(B) Let $I_1$ be the current through the $5\,\Omega$ resistance and $I_2$ be the current through the $(6+9)\,\Omega$ branch.Given that the heat developed in the $5\,\Omega$ resistance is $20.00\,cal/s$,we have:$P_1 = I_1^2 R_1 = 20.00\,cal/s$$I_1^2 	imes 5 = 20 \implies I_1^2 = 4 \implies I_1 = 2\,A$.The potential difference across points $C$ and $D$ is $V_{CD} = I_1 	imes 5 = 2 	imes 5 = 10\,V$.The current through the upper branch $(6\,\Omega + 9\,\Omega)$ is $I_2 = \frac{V_{CD}}{6+9} = \frac{10}{15} = \frac{2}{3}\,A$.The total current $I$ flowing through the $2\,\Omega$ resistance is $I = I_1 + I_2 = 2 + \frac{2}{3} = \frac{8}{3}\,A$.The heat developed per second in the $2\,\Omega$ resistance is $P = I^2 R = \left(\frac{8}{3}ight)^2 	imes 2 = \frac{64}{9} 	imes 2 = \frac{128}{9} \approx 14.22\,cal/s$.

In the circuit shown in the figure,the $5\,\Omega$ resistance develops $20.00\,cal/s$ due to the current flowing through it. The heat developed in the $2\,\Omega$ resistance (in $cal/s$) is

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