Question

Two resistors, 4Ω and 6Ω, are connected in parallel, and this combination is connected in series with a 2Ω resistor to a 12V battery. What is the total power dissipated?

• A. 32.7 W
• B. 28.8 W
• C. 24.0 W
• D. 36.0 W

Hint:

For circuits with resistors in series and parallel, first find the equivalent resistance of the parallel combination, then combine it with the series resistors to calculate the total resistance and the power dissipated.

Answer 1

The Correct Option is A

The first step is to calculate the equivalent resistance of the parallel combination of the 4\Ω\ and 6\Ω\ resistors: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \] \[ R_{\text{eq}} = \frac{12}{5} = 2.4 \, \Omega \] Now, this equivalent resistance is in series with the 2Ω resistor. So the total resistance in the circuit is: \[ R_{\text{total}} = R_{\text{eq}} + 2 = 2.4 + 2 = 4.4 \, \Omega \] Now, we can use Ohm's law to find the total current in the circuit: \[ I = \frac{V}{R_{\text{total}}} = \frac{12}{4.4} = 2.73 \, \text{A} \] Finally, we can calculate the total power dissipated using the formula \( P = I^2 R \): \[ P = (2.73)^2 \times 4.4 = 7.46 \times 4.4 = 32.7 \, \text{W} \] Thus, the total power dissipated is 32.7 W.