The total power dissipated in Watts in the ci | vedclass

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(C) First, identify the circuit configuration. The $6 \, \Omega$ and $3 \, \Omega$ resistors are connected in parallel.Their equivalent resistance $R_

Vedclass · Accepted Answer

$54$

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(C) First, identify the circuit configuration. The $6 \, \Omega$ and $3 \, \Omega$ resistors are connected in parallel.Their equivalent resistance $R_p$ is given by:$R_p = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega$This parallel combination is in series with the $4 \, \Omega$ resistor.Therefore, 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 calculated using the formula $P = \frac{V^2}{R_{eq}}$, where $V = 18 \, V$ is the voltage of the battery.$P = \frac{18^2}{6} = \frac{324}{6} = 54 \, W$Thus, the total power dissipated is $54 \, W$.

The total power dissipated in Watts in the circuit shown here is

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