A current of 3 , A flows through the 2 , Omeg | vedclass

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(B) The resistors $2 \, \Omega$,$4 \, \Omega$,and the series combination of $(1 \, \Omega + 5 \, \Omega)$ are connected in parallel. Therefore,the pot

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$5$

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(B) The resistors $2 \, \Omega$,$4 \, \Omega$,and the series combination of $(1 \, \Omega + 5 \, \Omega)$ are connected in parallel. Therefore,the potential difference across each branch is the same.The potential difference $V$ across the $2 \, \Omega$ resistor is given by $V = I 	imes R = 3 \, A 	imes 2 \, \Omega = 6 \, V$.Since the branches are in parallel,the same potential difference $V = 6 \, V$ is applied across the branch containing the $1 \, \Omega$ and $5 \, \Omega$ resistors in series.The total resistance of this branch is $R_{branch} = 1 \, \Omega + 5 \, \Omega = 6 \, \Omega$.The current $i_3$ flowing through this branch is $i_3 = \frac{V}{R_{branch}} = \frac{6 \, V}{6 \, \Omega} = 1 \, A$.The power dissipated in the $5 \, \Omega$ resistor is $P = i_3^2 	imes R = (1 \, A)^2 	imes 5 \, \Omega = 5 \, W$.

$A$ current of $3 \, A$ flows through the $2 \, \Omega$ resistor shown in the circuit. The power dissipated in the $5 \, \Omega$ resistor is ................. $W$.

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