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edit: I was very sleep deprived and my mind blanked out on the answer. Just find the voltage of the known resistor w/ the ohmemeter, and use ohm's law (i=v/r) to get the current, then use the current with the unknown r's voltage & ohm's law to get the resistance. Sorry to waste everyone's time.

I have an unknown resistor and a 1k resistor. I can only use a voltmeter, and I can place them in series. How would I go about finding the value of unknown resistor?

I have no idea where to even start with this- I have the voltage source (9v) but don't know what to do to find the rest. I suspect I have to get the current (as it is the same through all components) but I don't have the total resistance ( i=v/r) to get the current. If I calculate current with just the 1k resistor then it wouldn't be accurate.

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    $\begingroup$ @ Diggity Dungus - You should also have a voltage source, not only a voltmeter, that can only measure voltages. $\endgroup$
    – freecharly
    Commented Oct 11, 2016 at 2:48
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    $\begingroup$ Think of Ohms law and different connection with the unknown resistor so you get enough equation to solve for the unknown. $\endgroup$
    – freecharly
    Commented Oct 11, 2016 at 2:50
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    $\begingroup$ If the same current goes through both resistors, then the voltage drop over each resistor is proportional to the resistance. Now figure out the easy way to ensure they get the same identical current. $\endgroup$
    – hdhondt
    Commented Oct 11, 2016 at 3:01
  • $\begingroup$ Editors & Reviewers: Why was the edit (v4) approved? E.g. why delete the resistor tag? $\endgroup$
    – Qmechanic ♦
    Commented Oct 11, 2016 at 7:26
  • $\begingroup$ @Qmechanic This question was not about resistor. It did not ask anything about resistor particularly, like it's structure , use, etc. It was a simple Ohms aw application. $\endgroup$
    – Anubhav Goel
    Commented Oct 14, 2016 at 14:57

3 Answers 3

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Put the two resistors in series, then connect the voltage source across both of them. The voltage will drop across each in proportion to their resistance, because, being in series, they share the same current, and the voltage will be current times resistance.

So if resistance values are $R_1$ and $R_2$, and the voltages are $V_1$ and $V_2$, then you know that

$$\frac{V_1}{V_2}=\frac{R_1}{R_2}$$

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Your approach seems right - i.e. that a voltmeter measurement of the voltage across a know resistor can be used as an ammeter, which will then allow you to measure the resistance of an in-series resistor through a voltage measurement across the latter.

You also rightly point out that the voltmeter draws some current, and thus introduces a nonzero observer effect.

There are two approaches you can take here. Firstly, I think the approach you are likely meant to take in this conceptual problem is simply to neglect the current drawn by the voltmeter and do the calculations assuming ideal components. This is not an unrealistic assumption for many modern voltmeters, whose input impedance can approach $10^{10}\Omega$ (such as those that buffer their inputs through high input impedance operational amplifiers). At this level, the current drawn by the voltmeter is utterly dwarfed by that through the measured resistors.

You could also analyze the voltmeter to work out its impedance and thus its drawn current as a function of its reading. This knowledge will allow you to account precisely for the observer effect you are fearing. Although this assumes you may analyze the voltmeter in this way.

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  • $\begingroup$ You had applied over physics to this simple problem. $\endgroup$
    – Anubhav Goel
    Commented Oct 11, 2016 at 4:35
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    $\begingroup$ @AnubhavGoel What does that even mean? "If I calculate current with just the 1k resistor then it wouldn't be accurate." - the OP is clearly asking about inaccuracies introduced by the voltmeter itself. $\endgroup$
    – Selene Routley
    Commented Oct 11, 2016 at 4:59
  • $\begingroup$ Nah!! OP isn't asking that at all. OP is using secondary school physics involving only ohms law and series and parallel connections. He has no head for impedance and use of galvanometer as voltmeter or ammeter. He by above quote meant current he would get with 1k resistor and 9V battery is 9mA. He cannot use that current when he connect Unknown R to 9V, Since current would now change. He isn't looking for experimental limits. $\endgroup$
    – Anubhav Goel
    Commented Oct 11, 2016 at 5:09
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    $\begingroup$ Even my 40 year old analog voltmeter had a sensitivity of "50 kOhm/V", which would make it quite suitable for solving this problem. I agree with Anubhav - you are overthinking this... $\endgroup$
    – Floris
    Commented Oct 11, 2016 at 13:50
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The voltmeter might have internal currents and resistances that throw off the result slightly, but hopefully all that stuff is close to zero. We should assume that the voltmeter measures the change in voltage between two points without messing with the circuit.

Do you know the exact meaning for the three values in the formula "i=v/r"? v is the change in voltage between two points, i is the net current between those points, and r is the complete resistance of the circuit between them.

It is true for any two points on a static (non-changing) circuit.

You can algebraically manipulate i=v/r into different forms. v=ir and r=v/i are all equally true.

Ohm's law (the name for that formula) is extremely useful and versatile. You can use it to find the relative voltage of every point and the current through every wire in simple diagrams.

If you're confused, write it out instead of trying to do it in your head. Draw the diagram.

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