Calculating Internal Resistance: Methods A and B

[Adriano25]

Homework Statement

[/B]
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

Homework Equations

Ohm's Law: V = I*R

The Attempt at a Solution

For method A, the current of the ammeter (I_I) splits into I_v (current of voltmeter) and I_RX (current of unknown resistor). Also there will be an unknown resistance of voltmeter (R_V) and an unknown resistance of ammeter (R_I).
I found a relationship using Kirchhoff's current law:
I_I = I_V + I_RX
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

 

[ehild]

Homework Statement

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

Method A
View attachment 113855
Voltmeter = 1.46 V
Ammeter = 0.24 A
E (emf of battery) = 1.48 V

Method B
View attachment 113856
Voltmeter = 1.48 V
Ammeter = 0.24 A
E (emf of battery) = 1.48 V

Homework Equations

Ohm's Law: V = I*R

The Attempt at a Solution

For method A, the current of the ammeter (I_I) splits into I_v (current of voltmeter) and I_RX (current of unknown resistor). Also there will be an unknown resistance of voltmeter (R_V) and an unknown resistance of ammeter (R_I).
I found a relationship using Kirchhoff's current law:
I_I = I_V + I_RX
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

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.

 

[Adriano25]

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.

 

[ehild]

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.

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...

 

[Adriano25]

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...

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

 

[ehild]

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

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?

 

[Adriano25]

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?

View attachment 113908

The potential difference between P and O would be V = I_i * R_i

 

[ehild]

The potential difference between P and O would be V = I_i * R_i

Yes, and what is the numerical value?

 

[Adriano25]

Yes, and what is the numerical value?

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

 

[ehild]

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

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

 

[Adriano25]

Ups..
-R_x*I_i - R_i * I_i + E = 0
R_i = 0.04 Ω

 

[ehild]

Ups..
-R_x*I_i - R_i * I_i + E = 0
R_i = 0.04 Ω

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?

 

[Adriano25]

Then it would just be the difference between QO and QP, so the voltage between P and O would be 0.02 V?

 

[ehild]

Then it would just be the difference between QO and QP, so the voltage between P and O would be 0.02 V?

YES!