Final answer:
The power dissipated by the 16Ω resistor when connected to the 22.5V supply with 0.80Ω internal resistance is calculated using Ohm's Law and the power dissipation formula, resulting in 13.28W.
Step-by-step explanation:
To calculate the power dissipated by the 16Ω resistor, we need to find the current flowing through the circuit first. Using Ohm's Law V = IR, where V is the terminal voltage, I is the current, and R is the total resistance (the sum of internal and load resistances), we have:
22.5V = I × (16Ω + 0.80Ω)
Solving for I gives us:
I = 22.5V / (16Ω + 0.80Ω) = 22.5V / 16.8Ω = 1.3393A
Now, using the formula for power dissipation P = I²R, where P is power, I is current, and R is resistance, we get the power dissipated by the 16Ω resistor:
P = (1.3393A) ² × 16Ω = 13.28W
Therefore, the correct answer is a. 13.28W which represents the power dissipated by the 16Ω resistor.