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(B) First, calculate the equivalent resistance of the parallel combination of $6\, \Omega$ and $3\, \Omega$ resistors:$R_{P} = \frac{6 	imes 3}{6 + 3

Vedclass · Accepted Answer

$54$

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(B) First, calculate the equivalent resistance of the parallel combination of $6\, \Omega$ and $3\, \Omega$ resistors:$R_{P} = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2\, \Omega$Now, the circuit consists of this equivalent resistance $R_{P} = 2\, \Omega$ 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 power dissipated in the circuit is given by the formula $P = \frac{V^{2}}{R_{eq}}$.Substituting the values, we get $P = \frac{18^{2}}{6} = \frac{324}{6} = 54\, W$.

The total power dissipated (in $watt$) in the circuit shown is

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