Final answer:
The power dissipated within the internal resistance of a 3.0 V battery with a 2.5 V terminal voltage connected to a resistor dissipating 0.5 W is 0.1 W.
Step-by-step explanation:
The student is asking about the power dissipated within the internal resistance of a battery when it is connected to a resistor. When a 3.0 V battery is connected to a resistor dissipating 0.5 W of power and the terminal voltage is 2.5 V, we can use the terminal voltage to determine the power dissipated by the internal resistance of the battery.
The power dissipated by the external resistor can be calculated using the formula P = V2 / R, where P is the power, V is the terminal voltage, and R is the resistance. In this case, we can find the current I by rearranging the power formula and solving for I (I = sqrt(P/R)). Since the power of the external resistor is 0.5 W and the terminal voltage is 2.5 V, the current through the resistor is I = sqrt(0.5W)/(2.5V) = 0.2 A.
Now, using the internal voltage drop (emf - terminal voltage), which is 3.0 V - 2.5 V = 0.5 V, we can find the power dissipated within the battery's internal resistance. The power dissipated by the internal resistance (r) is thus given by Pinternal = I2r. We have the current (I = 0.2 A) and the voltage drop, so the internal resistance can be calculated as r = (emf - terminal voltage)/I = 0.5V/0.2A = 2.5 Ω.
The power dissipated within the internal resistance can then be calculated as Pinternal = 0.2 A2 * 2.5 Ω = 0.1 W. Therefore, the correct answer to the power dissipated within the internal resistance of the battery is (a) 0.1 W.