In the circuit shown,the heat produced in 5 O | vedclass

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(A) The power dissipated in the $5 \Omega$ resistor is given by $P_2 = I_2^2 R_2 = 50 	ext{ J/s}$.Given $R_2 = 5 \Omega$,we have $I_2^2 	imes 5 = 50

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$5$

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(A) The power dissipated in the $5 \Omega$ resistor is given by $P_2 = I_2^2 R_2 = 50 	ext{ J/s}$.Given $R_2 = 5 \Omega$,we have $I_2^2 	imes 5 = 50$,which implies $I_2^2 = 10 	ext{ A}^2$,so $I_2 = \sqrt{10} 	ext{ A}$.The potential difference $V$ across the parallel branches is $V = I_2 	imes R_2 = \sqrt{10} 	imes 5 = 5\sqrt{10} 	ext{ V}$.The upper branch consists of $2 \Omega$ and $8 \Omega$ resistors in series,so the total resistance of the upper branch is $R_1 = 2 \Omega + 8 \Omega = 10 \Omega$.The current $I_1$ in the upper branch is $I_1 = \frac{V}{R_1} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2} 	ext{ A}$.The heat generated per second in the $2 \Omega$ resistor is $P_1 = I_1^2 	imes R_{2\Omega} = \left(\frac{\sqrt{10}}{2}ight)^2 	imes 2 = \frac{10}{4} 	imes 2 = \frac{20}{4} = 5 	ext{ J/s}$.

In the circuit shown,the heat produced in $5 \Omega$ resistance due to current $I_2$ is $50 \text{ J/s}$. Then,the heat generated per second in $2 \Omega$ resistance is (in $\text{ J/s}$)

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