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(B) In the given circuit,the resistors $6 \,\Omega$ and $3 \,\Omega$ are connected in parallel. Their equivalent resistance $R_p$ is given by:$\frac{1

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$54$

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(B) In the given circuit,the resistors $6 \,\Omega$ and $3 \,\Omega$ are connected in parallel. Their equivalent resistance $R_p$ is given by:$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2 \,\Omega$.The equivalent circuit consists of this $R_p = 2 \,\Omega$ resistor in series with the $4 \,\Omega$ resistor and the $18 \,V$ battery.The total resistance of the circuit is $R_{eq} = R_p + 4 \,\Omega = 2 \,\Omega + 4 \,\Omega = 6 \,\Omega$.The total current $I$ flowing through the circuit is given by Ohm's law:$I = \frac{V}{R_{eq}} = \frac{18 \,V}{6 \,\Omega} = 3 \,A$.The total power dissipated in the circuit is given by $P = I^2 R_{eq}$ or $P = VI_{total}$.Using $P = VI_{total} = 18 \,V 	imes 3 \,A = 54 \,W$.

The total power dissipated in the circuit shown here is ............. $W$.

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