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=IRRvR+RvV = I \frac{R R_v}{R + R_v}

The measured resistance in configuration 1 is therefore:

Rm1=VI=RRvR+RvΔ1=RvR+Rv1=RR+RvR_{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=VR+RA I = \frac{V}{R + R_A}

The measured resistance in configuration 2 is therefore:

Rm2=VI=R+RAΔ2=1+RAR1=RAR 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 Rv R_v as well as the voltage across it, which gives us Rv R_v . Then with the switch in position 1, we get Rm=RRvR+RvR_m = \frac{R R_v }{R + R_v} (which is the same result from configuration 1). And since the first measurement gave us Rv R_v , we can solve for R 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 RA R_A as well as the voltage across it, which gives us RA R_A . Then with the switch in position 2, we get Rm=R+RAR_m = R + R_A (which is the same result from configuration 2). And since the first measurement gave us RA R_A , we can solve for R 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.

Talulah Riley - 3 months, 1 week ago

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@Steven Chase

Talulah Riley - 3 months, 1 week ago

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@Steven Chase Thanks ,you rocked , your answers are correct.
At last did you get any quadratic equation for RR which you have not written in your solution ?

Talulah Riley - 6 months, 1 week ago

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This is the first problem of InPho!

Jason Gomez - 6 months, 1 week ago

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Did I get it right?

Steven Chase - 6 months, 1 week ago

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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)

Jason Gomez - 6 months, 1 week ago

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

Steven Chase - 6 months, 1 week ago

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

Jason Gomez - 6 months, 1 week ago

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

Steven Chase - 6 months, 1 week ago

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

Jason Gomez - 6 months, 1 week ago

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

Steven Chase - 6 months, 1 week ago

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

Jason Gomez - 6 months, 1 week ago

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@Steven Chase

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

Jason Gomez - 6 months, 1 week ago

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

Talulah Riley - 6 months, 1 week ago

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@Talulah Riley Also why is the acceleration taken as only centripetal acceleration?

Jason Gomez - 6 months, 1 week ago

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

Jason Gomez - 6 months, 1 week ago

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@Jason Gomez Oh nevermind, just realised that’s an r

Jason Gomez - 6 months, 1 week ago

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@Talulah Riley Here is my answer to the problem

Steven Chase - 6 months, 1 week ago

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