# Homework Help: Find internal resistance

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1. Feb 28, 2017

1. The problem statement, all variables and given/known data

Calculate the internal resistance of V & A based on methods A & B below:

Method A

Voltmeter = 1.46 V
Ammeter = 0.24 A
E (emf of battery) = 1.48 V

Method B

Voltmeter = 1.48 V
Ammeter = 0.24 A
E (emf of battery) = 1.48 V

2. Relevant equations

Ohm's Law: V = I*R

3. The attempt at a solution

For method A, the current of the ammeter (II) splits into Iv (current of voltmeter) and IRX (current of unknown resistor). Also there will be an unknown resistance of voltmeter (RV) and an unknown resistance of ammeter (RI).
I found a relationship using Kirchhoff's current law:
II = IV + IRX
Then, I tried to use Kirchhoff's voltage law through both loops, but I'm ending up with too many unknowns. Am I going on the right path?

Also, I'm not sure if I could find the unknown Rx by just using Ohm's Law:
Method A:
Rx = V / I
Rx = 1.46 V / 0.24 A = 6.08 Ω

Method B:
Rx = V / I
Rx = 1.48 V / 0.24 A = 6.17 Ω

2. Feb 28, 2017

### ehild

No, your simple method is not correct. You have to calculate with the internal resistances, Rv and Ri. Include them into the drawings, Rv parallel with the voltmeter and Ri in series with the ammeter.

3. Feb 28, 2017

From method 1, would that mean that I can find an equivalent resistance from Rv and Rx using the parallel circuit formula?
Also, from method 2, I could find an equivalent resistance from Ri and Rx using the series circuit formula?
I'm still not sure how that would help me in solving the problem.

4. Feb 28, 2017

### ehild

Yes.
In case A, the voltage difference E-Uv falls on the internal resistance of the ammeter. You know also the current flowing through Ri, so you can calculate it. In case B, you know the voltage across Rx+Rv, and you also know the current. Apply Ohm's Law to get Rx+Ri. But you know Ri already...

5. Feb 28, 2017

I'm sorry, I don't follow you. Could you please reexplain it? Thank you.

6. Feb 28, 2017

### ehild

What is that you do not understand?
Look at the figure A. The meters are replaced by ideal ones, together with the internal resistances. The EMF of the (ideal) battery is 1.48 V, and the voltmeter reads 1.46 V. What is the potential difference between P and O?

7. Feb 28, 2017

The potential difference between P and O would be V = Ii * Ri

8. Feb 28, 2017

### ehild

Yes, and what is the numerical value?

9. Feb 28, 2017

But we don't know Ri.
I was thinking that since Rx and Rv are in parallel, they share the same voltage, so V1 = 1.46.
Then, using Kirchhoff's voltage law:
-V1 * Ii - Ri * Ii + E = 0
so, Ri = 4.7 Ω ?

10. Feb 28, 2017

### ehild

V1*Ii is not voltage, you can not add it to voltages.

11. Feb 28, 2017

Ups..
-Rx*Ii - Ri * Ii + E = 0
Ri = 0.04 Ω

12. Mar 1, 2017

### ehild

No. Rx and Ri are not in series, they have different currents.
You know the KVL. The voltages add in a loop. The voltage between Q and P is shown by the voltmeter, it is 1.46 V. The voltage between Q and O is the same as the emf of the battery. What is the numerical value of the voltage between P and O?

13. Mar 1, 2017