The total power dissipated in watts in the ci | vedclass

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(D) First,calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $3 \Omega$ resistors:$\frac{1}{R_p} = \frac{1}{6} + \frac{

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$54$

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(D) First,calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $3 \Omega$ resistors:$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2 \Omega$.Now,this $R_p$ is in series with the $4 \Omega$ resistor.The total equivalent resistance of the circuit is $R_{eq} = R_p + 4 \Omega = 2 \Omega + 4 \Omega = 6 \Omega$.The total power dissipated in the circuit is given by the formula $P = \frac{V^2}{R_{eq}}$,where $V = 18 \ V$.$P = \frac{18^2}{6} = \frac{324}{6} = 54 \ W$.Therefore,the correct option is $D$.

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

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