In a parallel circuit with resistances of 8Omega and 20Omega connected to a 12V power supply, what is the total power dissipated? A 25.2 W B 32.4 W c 2.10 W D 4.40 W

To find the total power dissipated in a parallel circuit, we first need to calculate the total resistance in the circuit. In a parallel circuit, the total resistance is given by the formula:<br /><br />\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \]<br /><br />where \( R_1 \) and \( R_2 \) are the resistances of the two branches.<br /><br />Given:<br />\[ R_1 = 8\Omega \]<br />\[ R_2 = 20\Omega \]<br /><br />Substitute these values into the formula:<br /><br />\[ \frac{1}{R_{\text{total}}} = \frac{1}{8} + \frac{1}{20} \]<br /><br />To add these fractions, find a common denominator:<br /><br />\[ \frac{1}{8} = \frac{5}{40} \]<br />\[ \frac{1}{20} = \frac{2}{40} \]<br /><br />So,<br /><br />\[ \frac{1}{R_{\text{total}}} = \frac{5}{40} + \frac{2}{40} = \frac{7}{40} \]<br /><br />Now, take the reciprocal to find \( R_{\text{total}} \):<br /><br />\[ R_{\text{total}} = \frac{40}{7} \approx 5.71\Omega \]<br /><br />Next, use Ohm's Law to find the total current \( I_{\text{total}} \) in the circuit:<br /><br />\[ I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12V}{5.71\Omega} \approx 2.10A \]<br /><br />Finally, calculate the total power dissipated using the formula:<br /><br />\[ P_{\text{total}} = V \times I_{\text{total}} = 12V \times 2.10A = 25.2W \]<br /><br />Therefore, the total power dissipated in the circuit is:<br /><br />\[ \boxed{25.2W} \]<br /><br />So, the correct answer is A) 25.2 W.