# Homework Help: Finding Internal Resistance of a Battery

1. Apr 6, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
A real battery can be modeled as an ideal voltage source in series with a resistor. A voltmeter with input resistance of $1000\Omega$ measures the voltage of a 1.5V worn out battery as 0.9V. What is the internal resistance of the battery.

2. Relevant equations
$V=IR$

3. The attempt at a solution

After I had already drawn the circuit I realized it's wrong to have a wire with nothing on it along the right side but it's painful to draw the circuits so I'll just mention to ignore that section rather than redraw it.

I'm quite lost as to how to solve this question, the battery has a voltage of 1.5V so $V_0=1.5V$ but the voltmeter is reading 0.9V. Now without the right hand portion of the circuit $R_v$ and $R_i$ are in series so the current is given by $i=\frac{v_0}{R_v+R_i}$, now from KVL $v_0=i(R_v+R_i)$ but my $i$ will just cancel out everything and I'll end up with essentially nothing. Not really sure what to do with the voltmeter reading 0.9V.

2. Apr 6, 2016

### haruspex

Having found the expression for the current, as you have, figure out all the voltage drops.

3. Apr 6, 2016

### Potatochip911

Ok then the voltage drop across $R_i$ is $V_{Ri}=\frac{v_0}{R_v+R_i}R_i$ and across $R_v$ is $V_{Rv}=\frac{v_0}{R_v+R_i}R_v$, I'm not really sure what to do with these though, KVL gives $v_0=V_{Rv}+V_{Ri}$ but this doesn't seem to be leading to the solution.

4. Apr 6, 2016

### haruspex

What voltage drop does the voltmeter measure?

5. Apr 6, 2016

### Potatochip911

Ok I think I understand now, I believe the voltmeter measures the drop across $v_0$ and $R_i$ so $0.9V=V_0-iR_i$?

6. Apr 6, 2016

### haruspex

Refer back to the voltage drops you listed in post #3. The equations you had for those were more useful. But which one does the voltmeter measure?
It may be that your diagram is confusing you. What is the purpose of the fat resistor in the voltmeter?

7. Apr 6, 2016

### haruspex

On second thoughts, sorry, you can use what you posted in post #5. It's just not quite the route I had in mind, but it comes to the same.

8. Apr 6, 2016

### ehild

Not only that wire on the right is wrong: The real voltmeter can be considered as an ideal one (with infinite internal resistance) connected parallel with its resistance. And the voltmeter measures the voltage across its internal resistance.

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9. Apr 6, 2016

### Potatochip911

@haruspex @ehild Ok I think I finally got the answer, since $R_v$ is in parallel with the $R_i$ branch voltage across $R_i$ is the same as the voltage measured by the voltmeter. Now the current in the circuit will be given by $i=\frac{v_0}{R_{eq}}=\frac{v_0(R_i+R_v)}{R_iR_v}$ and voltage across $R_v$ will be 0.9V since this is what's measured by the voltmeter and we obtain $$0.9V=iR_v=\frac{V_0(R_i+R_v)}{R_i} \\ \frac{0.9R_i}{v_0}=R_i+R_v \\ R_i=\frac{R_v}{1-0.9/v_0}=2500\Omega$$

10. Apr 6, 2016

### ehild

NO, it is wrong. Ri and Rv are in series, and the voltmeter measures the voltage across Rv. The ideal voltmeter is not part of the circuit, it just shows the voltage between its terminals.

11. Apr 6, 2016

### Potatochip911

Ah yes I guess I forgot the voltmeter is just along a wire. Now current is given by $i=\frac{v_0}{R_i+R_v}$ and then voltage across $R_v$ will be $V_{Rv}=iR_v=\frac{v_0R_v}{R_i+R_v}$ which will be equal to the voltmeter reading of 0.9V, after solving gives $R_i=\frac{R_v(v_0-0.9V)}{0.9V}=666.67\Omega$, I can't comprehend why it measures voltage across $R_v$ though since then our expression for $R_i$ makes no sense if we let $R_v\to\infty$

12. Apr 6, 2016

### haruspex

Let me take you back to what you had in post #3
Relate the measured voltage to that.

13. Apr 6, 2016

### haruspex

As Rv increases, i decreases, increasing the measured voltage. In the limit the measured voltage is v0.

14. Apr 6, 2016

### Potatochip911

The final equation I obtained was (not sure if the correct word) independent of $i$ and only the resistors mattered. Is it due to the measured voltage 0.9V?
@ehild
Edit: Seems to just be a property of the fact that the resistance is in series with the battery's internal resistance? Does it even make sense that the internal resistance of the battery depends on the internal resistance of the DMM?

Last edited: Apr 6, 2016
15. Apr 6, 2016

### ehild

The voltage across Rv is the same as the voltage across the terminals of the battery. If the voltmeter resistance tends to infinity, the current tends to zero, the voltage read by the voltmeter is $\lim_{R_v \rightarrow∞} \frac {V_0 R_v}{R_v+R_i}=V_0$
Or you can think the way that infinite Rv means zero current, then there is no voltage drop across Ri and the voltmeter reads the emf of the battery , 1.5 V.

16. Apr 6, 2016

### haruspex

Yes. $V_{Rv}=\frac{v_0R_v}{R_i+R_v}$, so as $R_v\rightarrow \infty$, $V_{R_V}\rightarrow v_0$

17. Apr 6, 2016

### ehild

The internal resistance of the battery does not depend on the resistance of the voltmeter, but the voltage measured depends on it. You would get a higher terminal voltage with a better voltmeter and you would read the emf with an ideal voltmeter.

18. Apr 6, 2016

### Potatochip911

It seems to me from KVL that the voltage across the terminals of the battery is equal to the voltage across both resistors, $v_0=i(R_i+R_v)$?

Ok I understand now, I was taking the limit as $R_v\to\infty$ for a function with a value of $v_R$ that would only be possible if $R_v$ was finite.

19. Apr 6, 2016