This one is a variant on Challenge #1, but this one has some real practical utility.

--------------------------------------------------------
The following circuit is provided to you:
View attachment 372292

The voltage source consists of four new AA batteries in series.

R1 is a 1 MΩ, 1% resistor.

R2 is a black box consisting of network of several unknown resistors.

You are also provided with a voltmeter that has an unknown input resistance.

Using this meter, you are to determine the value of R2. You may only measure Vab and Vbc (i.e., you may not measure Vac, but can use it if you can calculate it from the other two).

What is your best estimate of the meter's nominal input resistance?
--------------------------------------------------------

Here are the two measurements so that everyone has the same data to work with:

Vab = 100.7 mV
Vbc = 3.29 V

Please limit yourselves to the same conditions that the teams would face, namely that they have a basic scientific calculator and pencil/paper. They can also bring a binder with whatever reference material they want, but it is unlikely that they would have anything in it that would directly help them with this problem.

EDIT: I didn't provide the circuit diagram, so it may not be obvious that it's the same one as in the other challenge. What's relevant is that Vab is the voltage across R1 and Vbc is the voltage across R2. Vac is the voltage across both, which is also the supply voltage.

 

My best estimate is that the meter's resistance is 1 MΩ.

 

My best estimate is that the meter's resistance is 1 MΩ.

And the value of R2?

 

My best estimate is that the meter's resistance is 1 MΩ.

Again, using first order assumptions,
assume Vs ≅ 6 V
assume R2 is very high, Rmeter is relatively low.

Since Vbc ≅ 0.5 Vs, Rmeter ≅ 1 MΩ

By using the voltage divider analysis, R1 || Rm = 0.5 MΩ
With R2 in MΩ units
Vs x (0.5) / (0.5 + R2) = 0.1
R2 = (5Vs - 0.5) MΩ

Hence R2 can be a range of values depending on Vs ≅ 2 x 3.29 V
If Vs = 6 V, R2 = 29.5 MΩ
if Vs = 6.6 V, R2 = 31.5 MΩ

My answer is R2 = 30 MΩ ±5%

 

Again, using first order assumptions,
assume Vs ≅ 6 V
assume R2 is very high, Rmeter is relatively low.

Since Vbc ≅ 0.5 Vs, Rmeter ≅ 1 MΩ

By using the voltage divider analysis, R1 || Rm = 0.5 MΩ
With R2 in MΩ units
Vs x (0.5) / (0.5 + R2) = 0.1
R2 = (5Vs - 0.5) MΩ

Hence R2 can be a range of values depending on Vs ≅ 2 x 3.29 V
If Vs = 6 V, R2 = 29.5 MΩ
if Vs = 6.6 V, R2 = 31.5 MΩ

My answer is R2 = 30 MΩ ±5%

Your answer seems to depend on what the supply voltage happens to be. What if you knew nothing about the supply voltage?

 

Your answer seems to depend on what the supply voltage happens to be. What if you knew nothing about the supply voltage?

You did state "The voltage source consists of four new AA batteries in series".

 

You did state "The voltage source consists of four new AA batteries in series".

Yes, I know, so I'm not saying you shouldn't have utilized that information. I'm just curious how you would have approached it if I hadn't provided that information.

For instance, your first assumption that R2 was very high is just that, an assumption. What if that assumption hadn't been correct?

Picking a number out of the hat, what if the meter's resistance had been 2 MΩ and R2 had also been 2 MΩ? That would also have put Vbc at half of the supply voltage. Would you have still proceed as you did and still conclude that R2 was ~30 MΩ?

 

If R2 and Rmeter were both 2 MΩ Vab would not be 100 mV.

 

If R2 and Rmeter were both 2 MΩ Vab would not be 100 mV.

I know that, it would have been ~1.65 V. But your reasoning didn't indicate any kind of a check that the 100 mV was or wasn't reasonable (which is not to say that you didn't do the check).

But that's still dodging the question of whether your approach relies on knowing the source voltage. What answer would your approach have gotten had Vab been 1.65 V? As near as I can tell, you would have gotten something between and 1.3 MΩ and 1.5 MΩ. And that's still using assumed values for Vs.

 

SOLUTION

NOTE: I can't enclose this in SPOILER tags because the equations won't render. So skip this post if you don't want to see the solution yet.

The easiest way to solve the problem is to leverage the results that were part of Challenge #1. Specifically that

\(
\frac{R_2}{R_1} \; = \; \frac{V_2}{V_1}
\)

This is independent of the meter input resistance and the applied voltage (provided they are the same for the two measurements).

Utilizing this, we have the result that

\(
R_2 \; = \; R_1 \cdot \frac{V_2}{V_1} \\
R_2 \; = \; 1\;M\Omega \cdot \frac{3.29\;V}{100.7\;mV} \; = \; 32.67\;M\Omega
\)

The value of R_1, per my 30+ year-old Fluke 27, is 1.004 MΩ, which would make R2 equal to 32.80 MΩ.

When I measured the resistance across the seven resistors that made up R_2 (in two segments so as to stay below the 30 MΩ limit), the measured values summed to 32.82 MΩ (agreement to 0.06%). Not bad, considering that the data was taken with a $5 DMM with an input resistance of only 1 MΩ.

The beauty is that, since the only real variable is the meter error, as long as you use the same meter, the errors largely cancel out because they are mostly systematic. Especially if you use the same range for both measurements (which I did not do).

This has some real practical utility. For instance, it let's you quickly and easily measure resistances well beyond the range of your multimeter (the max resistance my $5 meter can measure is 2 MΩ) and get a very good result provided you have a good reference resistor (and 1% tolerance is more than good enough for most situations). This can come in handy when you want build up large resistances that are known to good degree of accuracy in order to work with very small currents.