0
$\begingroup$

I have what I assume is a very basic question on Ohm's law. Let's say that we have $n$ equal light bulbs in a series with a battery. We know that $$ U - IR - IR - \cdots -IR= U - nIR = 0.$$ Solving for the current gives $$I = \frac{U}{nR}.$$ From Ohm's law we know that $$P = RI^{2}=\frac{U^2}{nR},$$ but we also know Ohm's law can be written as $$P=UI=\frac{U^2}{nR},$$ but that does not give the same answer as the other expression. Why don't they give the same answer?

What am I missing?

The original question is from this exam that I found on the Internet. It is question 1a). http://aesop.phys.utk.edu/ph231/fin.pdf and solution can be found here http://aesop.phys.utk.edu/ph231/fins.pdf

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ $P=UI$ is not an expression of Ohm's Law. It's a general expression for power consumed by a circuit branch. Where Ohm's Law only applies to ideal linear resistors, $P=UI$ applies to any branch of a lumped circuit (including inductors, capacitors, diodes, transformers, transistors, etc) $\endgroup$ – The Photon Aug 13 '19 at 16:10
  • 1
    $\begingroup$ Also, can you give an example of how using $P=UI$ gives a different answer from using the series combination formula? It's not even clear what question you're trying to get an answer to, and if power was not mentioned as part of the problem statement, how you could use $P=UI$ to get a solution. $\endgroup$ – The Photon Aug 13 '19 at 16:14
  • $\begingroup$ I think you are confusing power of a single bulb with power of whole system. $\endgroup$ – Droid Celestial Aug 13 '19 at 16:15
  • $\begingroup$ Please edit to show your working that gives a different answer for the two methods. $\endgroup$ – The Photon Aug 13 '19 at 16:17
  • $\begingroup$ @ThePhoton. I have edited it now $\endgroup$ – user5744148 Aug 13 '19 at 16:22
4
$\begingroup$

What I'm I missing?

Aren't you failing to distinguish between the power delivered to a single (series connected) bulb

$$P_{bulb} = RI^2 = R\frac{U^2}{N^2R^2}= \frac{1}{N}\frac{U^2}{NR}$$

and the power delivered by the source to the entire load of $N$ series connected bulbs?

$$P_{source} = UI = U\frac{U}{NR} = \frac{U^2}{NR} = N\cdot\frac{1}{N}\frac{U^2}{NR} = N\cdot P_{bulb}$$

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Ah beat me to it :) $\endgroup$ – BioPhysicist Aug 13 '19 at 16:32
  • $\begingroup$ Thanks! That was what I was looking for $\endgroup$ – user5744148 Aug 13 '19 at 16:38
2
$\begingroup$

Your two power equations are calculating different things.

With the equation $P = R I^2$, you have to be careful what you're using for resistance.

$R$ in this case is the resistance across one lightbulb. In the case of $P = IU$, your voltage drop is across the entire circuit.

If we make the first equation across the entire circuit as well, we find the equivalent resistance $R_{eq} = N \times R$, then you find $P_{total} = UI = NRI^2 = \frac {U^2}{NR}$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for your help! $\endgroup$ – user5744148 Aug 13 '19 at 16:39
1
$\begingroup$

Your answer is different from the original solution in the manner that you are calculating $P$ for the whole circuit instead of each light bulb's.

The power supplied in the circuit is $P = NRI^2 = \frac{U^2}{NR}$. If you want to calculate the power supplied in each light bulb using the voltage, you need to calculate each light bulb's voltage:

$$ U = N V_R$$ $$ V_R = \frac{U}{N} $$

Thus, each light bulb's power is

$$ P = V_R I = V_R \frac{V_R}{R} = \frac{U^2}{N^2 R} $$

Improve this answer
$\endgroup$

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.