Solved: Power & Batteries Calc: Resistance, Factor Increase

[mer584]

Homework Statement

An Ordinary flashlight uses two D-cell 1.5V batteries connected in series. The bulb draws 450mA when turned on. A)Calculate the resistance of the bulb and the power dissipated. B) By what factor would the power increase if 4 D-cell batteries in series were used with the same bulb? (neglect heating effects)

Homework Equations

P=IV=I^2R=V^2/R V=IR

The Attempt at a Solution

I understand part A, and I got 1.4W for the power. For part B though, I keep getting that it's 2 times greater by doing P=IV where (.45A)(4 x 1.5) =2.7. but I know the answer is supposed to be 4 times. I'm pretty sure the current stays constant and that you add voltages, so I'm confused why it would be 4 and not 2.

 

[anathema]

Any help would be appreciated!

You are correct in thinking that the current stays constant and that the voltage increases when batteries are connected in series. However, remember that power is also dependent on voltage and resistance. In this case, the resistance of the bulb remains the same, but the voltage across it increases when more batteries are added in series.

Using the equation P=V^2/R, we can see that the power is directly proportional to the square of the voltage. So, when 4 D-cell batteries are used in series, the voltage across the bulb would be (4 x 1.5V) = 6V. Plugging this into the equation, we get P=(6V)^2/R=4P, where P is the power dissipated by the bulb with 2 D-cell batteries.

Therefore, the power would increase by a factor of 4 when 4 D-cell batteries are used in series with the same bulb. I hope this helps clarify your confusion. Keep up the good work!

 

[Moetasim]

I would like to clarify that the power dissipated in a circuit is directly proportional to the voltage and the current, and is not affected by the number of batteries in series. Therefore, the power dissipated in the circuit with 4 D-cell batteries in series would still be 1.4W, the same as in the circuit with 2 D-cell batteries in series. The factor by which the power would increase is actually 2, not 4. This can be seen by using the equation P=V^2/R, where the voltage would increase by a factor of 2 (from 3V to 6V) and the resistance would remain the same, resulting in a factor of 2 for the power increase. It is important to note that this calculation neglects any heating effects from the increased voltage, which may affect the actual power dissipated in the circuit.