# When is the voltmeter resistance significant when calculating the resistance of a resistor?

I have begun university level physics and am trying to understand errors, which I wasn't really exposed to before.

In one of my questions, I know that the actual resistance of a resistor with unknown resistance, $R$, in series with a standard resistor with resistance $R_s$, is

$R=R_s\frac{V_1}{V_2}$

where $V_1$ is the potenetial difference across the unknown resistor and $V_2$ across the standard resistor.

If, to measure the unknown resistor I connect a voltmeter of resistance $r$ across the standard resistor, then the unknown resistor resistance is given by

$R=\frac{R_sr}{R_s+r}\frac{V_1}{V_2}$

Now I am asked to consider whether the voltmeter resistance has a significant effect on the estimate of $R$ if: \begin{align} R_s&=1.20\rm\,k\Omega \\ r&=10\rm\,k\Omega \end{align}

and $V_1,V_2$ values are accurate to about 1 part in 1000.

So far, I have calculated $R=\frac{R_sr}{R_s+r}$ to be 1199.8560... and I thought I could use this and the error in $V_1/V_2$ somehow to find and error In$R$ and consoder if it is significant? However I was didn't think this could be right because the value of $R_s$ is given to 3 significant figures and the value of $r$ to 1 or 2 significant figures- and I don't know which. So I m not sure how this impacts the problem and how to go about estimating the impact of $r$.

Also, as an aside I was also wondering if there was a specific reason why the voltmeter was connected against the known resistor and not the unknown resistor? My guess is that this question is not valid because we do not know whether the series combination is powered by a known power supply (in which case only 1 voltmeter is needed and it doesn't matter where it goes) or and unknown power supply in which case I wold also need to connect a voltmeter across the unknown resistor, however this problem is just not included in the question?

• When multiplying or dividing, the rule of thumb is to keep as many significant figures in the results as are contained in the number with the least number of significant figures in the calculation, in this case 1 or 2 significant figures as per the resistance of the voltmeter.
– NaOH
Nov 13, 2016 at 13:59
• The simplest way to approach this is to see that your "standard resistor" $R_s$ has actually been replaced by a parallel combination of $R_s$ and $r$, so its actual value $R'_s$ in the circuit is given by $1/R'_s = 1/R_s + 1/r$. Note, you seem to have made a mistake calculating $R = 1199.8560$ - the values of $R_s$ and $r$ are both given in $k\Omega$. Nov 13, 2016 at 21:55

If your standard/reference resistance $R_s$ is in parallel with your voltmeter, the effective resistance is given by the parallel combination of $R_s$ and $r$. In parallel, the easy rule is "small resistance wins." Since you have $R_s = 1.2\rm\,k\Omega$ and $r = 10\mathrm{\,k\Omega} \approx 10 R_s$, the effective resistance including the meter is about 10% less than $R_s$:
$$R_s' = \left( \frac{1}{R_s} + \frac1r \right)^{-1} = R_s \left( 1+\frac{R_s}{r} \right)^{-1} \approx \frac{R_s}{1.1} \approx 1.1\rm\,k\Omega$$
This must be a galvanometer, with a needle that is deflected by current stolen from the circuit under test; digital voltmeters typically have $r=10\rm\,M\Omega$. When I did this experiment as a freshman, its title was "why your galvanometer sucks." The error in neglecting such a high-impedance meter in parallel with a current-carrying circuit element would be $\sim \frac{R_s}{r} \approx 10^{-3}$ -- much better.
If you put the galvanometer in parallel with your unknown resistance $R$, the current drawn will be dominated by the current into the meter unless $R\ll r$, so large impedances become invisible.