# Overestimation Problem [duplicate]

I had asked this problem earlier but I was too vague in how I asked it so I'm making a new post to be more specific.

Problem: Say we have a series circuit with an ammeter, a cell with internal resistance and known emf, and a 10 ohm resistor. It is noticed that the resistor gets warmer. How would this affect the calculated value of the internal resistance?

Here is the official answer:

Here is my thinking: Since the resistor increases in temperature, it increases in resistance. Let's look at the equation ℰ = IR + Ir, where "ℰ" is emf, "I" is current, "R" is external resistance, and "r" is the internal resistance of the cell. if R increases and current is constant, in order to maintain constant emf, r must decrease. Where's the flaw in my thinking?

• Maybe it's just me, but you haven't actually listed your variables. In any question you shouuld write: it has a known emf $\epsilon$, and a 10 $\Omega$ resistor $R$. They are obvious to me here but I haven't a clue what $r$ is and that's the focus of your question. – Kraigolas May 20 '19 at 0:03
• The battery gets warmer too? That should make $r$ decrease. – Rodney Dunning May 20 '19 at 0:11
• I don't think it said anything about the battery getting warmer, just the 10 ohm resistor. – user225790 May 20 '19 at 0:12
• I'm confused about this overestimation of the r value, I don't get what the official answer is saying – user225790 May 20 '19 at 0:14
• Possible duplicate by OP: physics.stackexchange.com/q/480776/2451 – Qmechanic May 20 '19 at 2:20

We measure the current of the circuit to be $$I'$$ where $$I'$$ is the current resulting from the increase in resistance caused by the increase in temperature; in other words, $$I'>I$$ where $$I$$ is the current that would arise if the resistances did not change. Thus we have

$$R+r=\frac{\epsilon}{I'}>\frac{\epsilon}{I}$$

but we know that $$R=10 \Omega$$. We measure the current of the circuit, and in that way calculate the resistances (in this case). Because we know $$R=10\Omega$$ we attribute the higher resistance to $$r$$, so it appears as though $$r$$ is larger than it actually is, because we are including in our value of $$r$$ the change in the resistance $$R$$ from the temperature increase. In other words, if I write the new resistance that $$R$$ has after the temperature increase as $$R'=R+\Delta R$$, then through our calculations we find that

$$r=r_{\text{actual}}+\Delta R$$

which is larger than $$r_{\text{actual}}$$.

• wait isn't R>10Ω because the 10Ω resistor has been heated up? – user225790 May 20 '19 at 0:31
• I guess my main source of confusion is why would the person who is measuring the internal resistance even think that it is changing. Only the 10Ω resistor is changing resistance because only it is heating up in the circuit and nothing else. So why would the person think a lower current means a changing internal resistance when they noticed that the 10Ω resistor is heating up? Wouldn't that person connect the dots and realize the changing current is only due to the 10Ω resistor that's heating up? – user225790 May 20 '19 at 1:14
• In other words, why would that person think that the 10Ω resistor stays at 10Ω when that person saw it heat up. – user225790 May 20 '19 at 1:16
• If the person calculates a value for internal resistance that's overestimated, that person COULD NOT have known that the 10Ω resistor was ever heated up. But the problem states "it is noticed that the 10Ω resistor gets warmer" – user225790 May 20 '19 at 1:17
• I guess this problem is just poorly worded. – user225790 May 20 '19 at 1:19