In the given circuit,the 5, Omega resistor di | vedclass

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(B) The circuit consists of a $12\, \Omega$ resistor in series with a parallel combination of two branches. The first branch has a resistance of $9\,

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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. The first branch has a resistance of $9\, \Omega + 6\, \Omega = 15\, \Omega$ and the second branch has a resistance of $5\, \Omega$.Let $i_1$ be the current through the $5\, \Omega$ resistor and $i_2$ be the current through the $15\, \Omega$ branch. Since they are in parallel,the potential difference across them is the same: $i_1 	imes 5 = i_2 	imes 15 \implies \frac{i_1}{i_2} = \frac{15}{5} = 3 \implies i_1 = 3i_2$.The power dissipated in the $5\, \Omega$ resistor is $P_5 = i_1^2 	imes 5 = 45\, W$.$i_1^2 = \frac{45}{5} = 9 \implies i_1 = 3\, A$.Since $i_1 = 3i_2$,we have $3 = 3i_2 \implies i_2 = 1\, A$.The total current $i$ flowing through the $12\, \Omega$ resistor is $i = i_1 + i_2 = 3\, A + 1\, A = 4\, A$.The power dissipated in the $12\, \Omega$ resistor is $P_{12} = i^2 	imes 12 = (4)^2 	imes 12 = 16 	imes 12 = 192\, W$.

In the given circuit,the $5\, \Omega$ resistor dissipates $45\, J/s$ of heat due to the current flowing through it. The power dissipated per second in the $12\, \Omega$ resistor will be .............. $W$.

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