I'm looking at an exercise where a battery is powering two parallel light bulbs. Given are the battery's EMF ($12.0 V$) and its terminal voltage in this particular circuit ($11.8 V$), as well as the light bulbs' power draw ($4.0 W$ each).

In order to calculate the resistance of a light bulb, the official solution manual does

$R = \frac{V^2}{P} = \frac{(12.0 V)^2}{4.0 W}$

using the full EMF of $12.0 V$. But I'm confused - shouldn't it use $11.8 V$ in the formula, because that's the voltage that's actually felt by the bulbs?

  • $\begingroup$ The question was have been asking for total resistance? $\endgroup$ – PhysicsDave Nov 25 '18 at 14:20

From your description I agree that the problem is ambiguous and confusing. It is not clear if the 4 W is a nominal or an actual power draw. The nominal draw would be e.g. a manufacturer’s specification obtained under standard conditions (presumably 12 V in this case).

Clearly, the solution interpreted it as a nominal power draw. Without the full context of the book, including discussions earlier in the text itself, it is impossible for me to know if that is the standard assumption used or if there was some other context clue that would have indicated to you a nominal power rating was intended.

  • 1
    $\begingroup$ Now that I look at it with this in mind, the question does appear to refer to nominal power draw, yes (although it was hardly very clear when I first read it). $\endgroup$ – Drubbels Nov 26 '18 at 17:56
  • $\begingroup$ It is a good thing you asked! Now you will know what to look for in the future with this textbook $\endgroup$ – Dale Nov 26 '18 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.