# Resistance Measurement

Here is a problem that was recently sent to me:

To solve, I will refer to the following four configurations. The first two are the ones presented in the problem (with the switches in different state), and the third and fourth are more useful configurations.

Configuration 1:

The voltage across the voltmeter can be expressed in terms of the current through the ammeter:

$V = I \frac{R R_v}{R + R_v}$

The measured resistance in configuration 1 is therefore:

$R_{m1} =\frac{V}{I} = \frac{R R_v }{R + R_v} \\ \Delta_1 = \frac{R_v }{R + R_v} - 1 = -\frac{R}{R + R_v}$

Configuration 2:

The current through the ammeter can be expressed in terms of the voltage across the voltmeter:

$I = \frac{V}{R + R_A}$

The measured resistance in configuration 2 is therefore:

$R_{m2} = \frac{V}{I} = R + R_A \\ \Delta_2 = 1 + \frac{R_A}{R} - 1 = \frac{R_A}{R}$

There are two ways to answer part (b) of the question: using either configuration 3 or configuration 4.

Configuration 3:

Configuration 3 is a more useful form of configuration 1. Starting with the switch in position 2, we know the current through $R_v$ as well as the voltage across it, which gives us $R_v$. Then with the switch in position 1, we get $R_m = \frac{R R_v }{R + R_v}$ (which is the same result from configuration 1). And since the first measurement gave us $R_v$, we can solve for $R$.

Configuration 4:

Configuration 4 is a more useful form of configuration 2. Starting with the switch in position 1, we know the current through $R_A$ as well as the voltage across it, which gives us $R_A$. Then with the switch in position 2, we get $R_m = R + R_A$ (which is the same result from configuration 2). And since the first measurement gave us $R_A$, we can solve for $R$.

Note by Steven Chase
6 months, 1 week ago

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@Steven Chase i have posted a new discussion on mechanis. Please solve that problem.

- 3 months, 1 week ago

- 3 months, 1 week ago

At last did you get any quadratic equation for $R$ which you have not written in your solution ?

- 6 months, 1 week ago

This is the first problem of InPho!

- 6 months, 1 week ago

Did I get it right?

- 6 months, 1 week ago

I think you did!, I had no clue about how to do the second part, and spent a lot of time getting an accurate error for the first part (some current goes into the voltmeter, that thought messed me up, I ended up obtaining a huge expression for error even after using binomial approximations multiple times)

- 6 months, 1 week ago

The problem doesn't do a great job explaining what assumptions you are allowed to make. That annoyed me a bit

- 6 months, 1 week ago

I assume minimal assumptions was best, also could I ask you the solution for another problem of the same paper

- 6 months, 1 week ago

I can take a look, but I can't promise that I'll be able to solve it

- 6 months, 1 week ago

Thank you for trying, even if you get it or not (I hope you do)

- 6 months, 1 week ago

OK, it's just about bed time on my side of the planet, so I'll try tomorrow.

- 6 months, 1 week ago

It’s fine even if you give the answer a week later

- 6 months, 1 week ago

I assumed this would be easy, but got stuck in middle when I had to perform an integration between time and distance, I am now unclear of how to approach this now

- 6 months, 1 week ago

@Jason Gomez I think you have probably stucked in part (c).
Here is the solution:

- 6 months, 1 week ago

Also why is the acceleration taken as only centripetal acceleration?

- 6 months, 1 week ago

Thanks a lot! But how is the factor $(γ (2πf)^2)=a$ (and what is γ?)

- 6 months, 1 week ago

Oh nevermind, just realised that’s an r

- 6 months, 1 week ago

@Talulah Riley Here is my answer to the problem

- 6 months, 1 week ago