Question

Two resistors, \( 4 \Omega \) and \( 6 \Omega \), are connected in parallel, and this combination is connected in series with a \( 2 \Omega \) 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 parallel resistors, calculate the equivalent resistance first, then proceed with the total resistance for the full circuit to find the power dissipated.

Answer 1

The Correct Option is A

First, calculate the equivalent resistance of the parallel resistors: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \] \[ R_{\text{parallel}} = \frac{12}{5} = 2.4 \, \Omega \] Now, the total resistance in the circuit is: \[ R_{\text{total}} = R_{\text{parallel}} + 2 \, \Omega = 2.4 + 2 = 4.4 \, \Omega \] Using Ohm’s law, the total current in the circuit is: \[ I = \frac{V}{R_{\text{total}}} = \frac{12}{4.4} \approx 2.73 \, A \] Finally, the total power dissipated is: \[ P = I^2 R_{\text{total}} = (2.73)^2 \times 4.4 \approx 32.7 \, W \] Thus, the correct power dissipated is \( 32.7 \, W \).