Question

Two resistors, 4 ohm and 6 ohm, are connected in parallel, and this combination is connected in series with a 2 ohm 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 resistors in parallel, the equivalent resistance is given by \( \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \). Then, use the total resistance to calculate the power dissipated.

Answer 1

The Correct Option is A

First, calculate the equivalent resistance of the two resistors in parallel: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \] Thus, \[ R_{\text{parallel}} = \frac{12}{5} = 2.4 \, \Omega \] Now, the total resistance in the circuit is: \[ R_{\text{total}} = R_{\text{parallel}} + 2 = 2.4 + 2 = 4.4 \, \Omega \] Now, use the formula for power: \[ P = \frac{V^2}{R_{\text{total}}} \] Substitute the values: \[ P = \frac{12^2}{4.4} = \frac{144}{4.4} = 32.7 \, \text{W} \] Thus, the total power dissipated is 32.7 W.