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(B) The circuit consists of a $12 \, \Omega$ resistor in series with a parallel combination of two branches. One branch contains a $5 \, \Omega$ resis

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$192$

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(B) The circuit consists of a $12 \, \Omega$ resistor in series with a parallel combination of two branches. One branch contains a $5 \, \Omega$ resistor,and the other branch contains resistors of $9 \, \Omega$ and $6 \, \Omega$ in series (total $15 \, \Omega$).Let $i_1$ be the current through the $5 \, \Omega$ resistor and $i_2$ be the current through the $9 \, \Omega$ and $6 \, \Omega$ branch.Since the branches are in parallel,the potential difference across them is equal: $i_1 	imes 5 = i_2 	imes (9 + 6) = i_2 	imes 15$.Therefore,$\frac{i_1}{i_2} = \frac{15}{5} = 3$,which means $i_1 = 3i_2$.The power developed in the $5 \, \Omega$ resistor is $P_5 = i_1^2 	imes 5 = 45 \, W$.$i_1^2 = \frac{45}{5} = 9 \Rightarrow i_1 = 3 \, A$.Since $i_1 = 3i_2$,we have $3 = 3i_2$,so $i_2 = 1 \, A$.The total current $i$ flowing through the $12 \, \Omega$ resistor is $i = i_1 + i_2 = 3 + 1 = 4 \, A$.The power developed in the $12 \, \Omega$ resistor is $P_{12} = i^2 	imes 12 = (4)^2 	imes 12 = 16 	imes 12 = 192 \, W$.

In the following circuit,the $5 \, \Omega$ resistor develops $45 \, J/s$ due to the current flowing through it. The power developed per second across the $12 \, \Omega$ resistor is ............. $W$.

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