In the following circuit,the 18,Omega resisto | vedclass

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(B) First,simplify the circuit. The two $12\,\Omega$ resistors in parallel have an equivalent resistance of $R_p1 = (12 	imes 12) / (12 + 12) = 6\,\O

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$10$

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(B) First,simplify the circuit. The two $12\,\Omega$ resistors in parallel have an equivalent resistance of $R_p1 = (12 	imes 12) / (12 + 12) = 6\,\Omega$. The three $9\,\Omega$ resistors in parallel have an equivalent resistance of $R_p2 = 9 / 3 = 3\,\Omega$. The upper branch has a total resistance of $R_{upper} = 6\,\Omega + 3\,\Omega = 9\,\Omega$. The lower branch has a resistance of $18\,\Omega$. Let $i_1$ be the current through the $18\,\Omega$ resistor and $i_2$ be the current through the upper branch $(9\,\Omega)$. Since they are in parallel,the potential difference is the same: $i_1 	imes 18 = i_2 	imes 9$,which gives $i_2 = 2i_1$. The total current $i = i_1 + i_2 = i_1 + 2i_1 = 3i_1$. The power developed in the $18\,\Omega$ resistor is $P_{18} = i_1^2 	imes 18 = 2\,W$. Thus,$i_1^2 = 2 / 18 = 1/9$,so $i_1 = 1/3\,A$. The total current $i = 3 	imes (1/3) = 1\,A$. The power developed across the $10\,\Omega$ resistor is $P_{10} = i^2 	imes 10 = (1)^2 	imes 10 = 10\,W$.

In the following circuit,the $18\,\Omega$ resistor develops $2\,J/s$ due to the current flowing through it. The power developed across the $10\,\Omega$ resistor is .............. $W$.

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