Corrected exercises: lab measurements, uncertainties and graphical analysis (203-SN2-RE)

The $20.00$ V range displays two decimals: the last figure is the hundredth of a volt, so ONE DIGIT IS WORTH $0.01$ V. The word “digit” never means a significant figure, it means one unit of the last displayed place, and it changes with the range. It is the first thing to look for in a datasheet.

The datasheet gives two terms to ADD. Proportional term: $0.5\,\%\times 12.47=0.0624$ V. Digit term: $2\times 0.01=0.02$ V. Total: $\Delta U=0.0624+0.02=0.0824$ V, which is rounded to ONE significant figure, $0.08$ V. The value is aligned on the same decimal: $U=(12.47\pm 0.08)$ V. Writing $(12.4700\pm 0.0824)$ V would be wrong by excess of precision: the four digits of the display are not four correct digits.

$\dfrac{\Delta U}{U}=\dfrac{0.0824}{12.47}=0.0066$, that is $0.66\,\%$. A common multimeter sits between half a percent and two percent: this is the order of magnitude to keep in mind to decide whether a measurement is worth refining.

On the $200.0$ V range, one digit is worth $0.1$ V. Proportional term: $0.5\,\%\times 12.5=0.0625$ V, almost unchanged. Digit term: $2\times 0.1=0.2$ V, that is TEN TIMES more. Total $0.2625$ V, rounded to $0.3$ V, so $U=(12.5\pm 0.3)$ V and a relative uncertainty of $2.1\,\%$. The same voltage, the same instrument, and a measurement three times worse.

The rule: ALWAYS STEP DOWN TO THE SMALLEST RANGE THAT DOES NOT SHOW OVERLOAD. The digit term is set by the range, not by the voltage measured; measuring $12$ V on a $200$ V range means paying the uncertainty of a $200$ V measurement. This is the most common error of the first lab, and it leaves no visible trace in the notebook: the display looks just as sharp.

First convert the current to amperes, $I=28.4$ mA $=0.0284$ A, otherwise the resistance comes out in kilo-ohms without you noticing. $R=\dfrac{6.42}{0.0284}=226.06$ Ω. We provisionally keep all the decimals: the rounding is decided at the end, once the uncertainty is known.

Each factor contributes its relative uncertainty. For the voltage: $\dfrac{0.05}{6.42}=0.00779$, that is $0.78\,\%$. For the current: $\dfrac{0.3}{28.4}=0.01056$, that is $1.06\,\%$. The two add, whether the quantity is in the numerator or the denominator: $\dfrac{\Delta R}{R}=0.00779+0.01056=0.01835$, that is $1.84\,\%$. The fact that $I$ is in the denominator changes NOTHING about the sign: an uncertainty is never subtracted.

$\Delta R=226.06\times 0.01835=4.15$ Ω, rounded to $4$ Ω. The value is aligned on the unit: $R=(226\pm 4)$ Ω. The decimals of $226.06$ disappear, because an uncertainty of $4$ Ω makes the tenths digit meaningless.

The current, without hesitation: it contributes $1.06\,\%$ against $0.78\,\%$ for the voltage, that is $58\,\%$ of the total. This is the reasoning you do BEFORE going back to the lab: stepping the ammeter down one range, or increasing the current, will almost halve the uncertainty, whereas a finer voltmeter would gain almost nothing.

The student treats the uncertainties as if they were themselves a voltage and a current, and applies Ohm's law to them. An uncertainty is not a physical quantity of the circuit: it is the width of an interval. It propagates by a rule on RELATIVE uncertainties, never by substituting it into the formula. The result obtained, $167$ Ω, is moreover of the same order as the resistance itself, which should have raised an alarm: an uncertainty of $74\,\%$ on a measurement made with a multimeter is physically absurd.

$R_{s}=100+220=320$ Ω. It is a SUM, so the absolute uncertainties add: $\Delta R_{s}=5+11=16$ Ω. Result: $R_{s}=(320\pm 16)\ \Omega$, which can be written $(3.20\pm 0.16)\times 10^{2}\ \Omega$ if you want to keep only the useful figures.

$\dfrac{16}{320}=0.050$, that is $5.0\,\%$. Since both resistors have the same relative tolerance of $5\,\%$, the sum keeps it: this is a general property of sums of quantities of the SAME sign.

$R_{p}=\dfrac{100\times 220}{100+220}=\dfrac{22000}{320}=68.75$ Ω. Order-of-magnitude check, to be made every time: an equivalent resistance in parallel is always SMALLER than the smaller of the two, so below $100$ Ω, and here it is a little more than two thirds of $100$ Ω. It is consistent.

The two partial derivatives: $\left(\dfrac{R_{2}}{R_{1}+R_{2}}\right)^{2}=\left(\dfrac{220}{320}\right)^{2}=0.4727$ and $\left(\dfrac{R_{1}}{R_{1}+R_{2}}\right)^{2}=\left(\dfrac{100}{320}\right)^{2}=0.0977$. Hence $\Delta R_{p}=0.4727\times 5+0.0977\times 11=2.363+1.074=3.44$ Ω, and $\dfrac{\Delta R_{p}}{R_{p}}=\dfrac{3.44}{68.75}=0.050$, again $5.0\,\%$. We write $R_{p}=(69\pm 3)\ \Omega$.

With $\Delta R_{1}=tR_{1}$ and $\Delta R_{2}=tR_{2}$, we get $\Delta R_{p}=\dfrac{R_{2}^{2}}{(R_{1}+R_{2})^{2}}\,tR_{1}+\dfrac{R_{1}^{2}}{(R_{1}+R_{2})^{2}}\,tR_{2}=\dfrac{t\,R_{1}R_{2}(R_{2}+R_{1})}{(R_{1}+R_{2})^{2}}=t\,\dfrac{R_{1}R_{2}}{R_{1}+R_{2}}=t\,R_{p}$. In other words: when all the components of a circuit carry the SAME relative tolerance, any series or parallel combination keeps it. This is what makes the $5\,\%$ tolerance usable directly on a whole network, without recalculating. The symmetric trap: a student who added the relative uncertainties for the parallel case would find $10\,\%$, double the truth.

$P=\dfrac{12.0^{2}}{47}=\dfrac{144}{47}=3.064$ W.

The power grows with $U$ and DECREASES with $R$, which is in the denominator. To maximize $P$, you therefore take $U$ at its maximum AND $R$ at its MINIMUM: $U_{max}=12.2$ V with $R_{min}=45\ \Omega$. This is the one delicate point of the method, and the one people miss: taking both high bounds would give a value that is neither the maximum nor the minimum.

$P_{max}=\dfrac{12.2^{2}}{45}=\dfrac{148.84}{45}=3.3076$ W and $P_{min}=\dfrac{11.8^{2}}{49}=\dfrac{139.24}{49}=2.8416$ W. We check in passing that the central value $3.064$ W lies between the two, and roughly in the middle: if it did not, that would be a sign that a bound was badly chosen.

$\Delta P=\dfrac{3.3076-2.8416}{2}=\dfrac{0.4660}{2}=0.2330$ W, rounded to $0.2$ W. Result: $P=(3.1\pm 0.2)$ W. The central value is rounded to the tenth, since the uncertainty is on the tenth.

For $P=\dfrac{U^{2}}{R}$, the voltage appears SQUARED, so its relative uncertainty counts TWICE: $\dfrac{\Delta P}{P}=2\dfrac{\Delta U}{U}+\dfrac{\Delta R}{R}=2\times\dfrac{0.2}{12.0}+\dfrac{2}{47}=0.0333+0.0426=0.0759$, that is $7.59\,\%$. Hence $\Delta P=3.064\times 0.0759=0.233$ W: the two methods give the same number to three decimals. They are not equivalent in general, the extreme-values method being exact and the relative-uncertainty method a first-order approximation, but as long as the relative uncertainties stay below a few percent, the gap is invisible. Forgetting the factor $2$ of the square would drop the result to $5.9\,\%$, an underestimate of one quarter: this is the number-one trap of the chapter.

$B=\dfrac{4\pi\times 10^{-7}\times 500\times 2.40}{0.320}=4\pi\times 10^{-7}\times 3750=4.712\times 10^{-3}$ T, that is $4.71$ mT. The intermediate calculation $\dfrac{NI}{L}=\dfrac{500\times 2.40}{0.320}=3750$ A/m is the linear density of ampere-turns: it is this that sets the field, not the three quantities taken separately.

$\dfrac{\partial B}{\partial I}=\dfrac{\mu_{0}N}{L}=\dfrac{4\pi\times 10^{-7}\times 500}{0.320}=1.9635\times 10^{-3}$ T/A. Contribution: $1.9635\times 10^{-3}\times 0.05=9.82\times 10^{-5}$ T, that is $0.098$ mT. Since $B$ is proportional to $I$, you could also write directly $\dfrac{\Delta I}{I}\times B=\dfrac{0.05}{2.40}\times 4.712=0.098$ mT: the two readings are the same.

$\dfrac{\partial B}{\partial L}=-\dfrac{\mu_{0}NI}{L^{2}}=-\dfrac{B}{L}$, whose absolute value is $\dfrac{4.712\times 10^{-3}}{0.320}=1.4726\times 10^{-2}$ T/m. Contribution: $1.4726\times 10^{-2}\times 0.002=2.95\times 10^{-5}$ T, that is $0.029$ mT. The MINUS SIGN disappears: an uncertainty is never subtracted, you add the contributions in absolute value, because you are looking for the worst case.

$\Delta B=0.098+0.029=0.127$ mT, rounded to $0.13$ mT since the first significant figure is a $1$ and you then keep two. Result: $B=(4.71\pm 0.13)$ mT, and $\dfrac{\Delta B}{B}=\dfrac{0.127}{4.712}=2.7\,\%$. The current accounts for $2.1$ percentage points and the length for $0.6$: it is the ammeter that should be improved, not the measuring tape.

Because $\Delta N=0$. Counting $500$ turns gives exactly $500$, not $500\pm\text{something}$: it is an INTEGER, not a measurement. The same reasoning holds for $\mu_{0}$, which is a defined constant, and for any exact factor, such as the $2$ of a diameter converted to a radius. A quantity with no uncertainty does not contribute, even when it multiplies everything else. Conversely, if the turns were wound by hand and counted to within $\pm 5$, you would have to add $\dfrac{5}{500}=1\,\%$.

The sum is $50.23$ V, so $\bar U=\dfrac{50.23}{10}=5.023$ V. We provisionally keep the thousandths, the mean of a sample legitimately having one more figure than the individual measurements.

The deviations from the mean are $+0.007$, $+0.047$, $-0.043$, $-0.013$, $+0.067$, $-0.063$, $+0.017$, $-0.023$, $+0.037$, $-0.033$ V. Their sum is zero, which is the first check to make. The sum of squares is $1.601\times 10^{-2}\ \mathrm{V}^{2}$, so $s=\sqrt{\dfrac{1.601\times 10^{-2}}{9}}=\sqrt{1.779\times 10^{-3}}=0.0422$ V, that is $42$ mV. The denominator is $N-1=9$, not $10$: the mean has already used up one degree of freedom.

$\dfrac{s}{\sqrt{N}}=\dfrac{0.0422}{\sqrt{10}}=\dfrac{0.0422}{3.162}=0.0133$ V, that is $13$ mV. This is the TYPE A uncertainty, the one that comes from the observed scatter, and it is three times smaller than the scatter of a single measurement.

The display resolution is $0.01$ V, so a single reading is already known to better than $\pm 0.005$ V from that alone. But the real scatter is $42$ mV, EIGHT TIMES more. The ten readings are therefore not spread over one or two neighbouring digits: they sprawl over thirteen digits. It is the SOURCE that fluctuates, not the instrument that hesitates. A student who wrote $U=(5.03\pm 0.005)$ V after a single reading would have underestimated the uncertainty by a factor of eight.

Yes, they must be combined: $13$ mV is a type A uncertainty, $45$ mV a TYPE B uncertainty, coming from the datasheet, and both are present at the same time. Adding them, $\Delta U=0.045+0.013=0.058$ V, that is $0.06$ V; combining them in quadrature, $\sqrt{0.045^{2}+0.013^{2}}=0.047$ V, that is $0.05$ V. In both conventions, the instrument term LARGELY dominates. Practical conclusion: going from ten to forty measurements would divide $13$ mV by two, so it would gain less than $7$ mV out of a total of $58$; it is not worth the time. Repeating a measurement only reduces the random part, and stops being useful as soon as the instrumental part exceeds it. To really gain, you need a better instrument, not more rows in the table.

You plot the CURRENT on the horizontal axis and the VOLTAGE on the vertical axis, because the current is what you impose and the voltage is what you read. Ohm's law is then written $U=RI$, of the form $Y=aX$, and the SLOPE IS THE RESISTANCE, in volts per ampere, that is in ohms. Be careful to convert milliamperes to amperes before calculating the slope: keeping the mA, you would get $0.22$ and a resistance a thousand times too small.

With $\bar I=0.015$ A and $\bar U=3.304$ V, we form $\sum(I_{i}-\bar I)(U_{i}-\bar U)=5.495\times 10^{-2}$ and $\sum(I_{i}-\bar I)^{2}=2.5\times 10^{-4}$, hence $a=\dfrac{5.495\times 10^{-2}}{2.5\times 10^{-4}}=219.8\ \Omega$. Quick check without a calculator: the last point gives $\dfrac{5.52}{0.025}=220.8\ \Omega$, of the right order.

$b=\bar U-a\bar I=3.304-219.8\times 0.015=3.304-3.297=0.007$ V. It SHOULD be zero, since with no current there is no voltage across a pure resistor. The $7$ mV found is at the noise level of the measurements, hence consistent with zero. A clearly nonzero intercept, say $0.2$ V, would signal something else in the circuit: a zero offset of the voltmeter, a stray electromotive force, or a diode in series.

$\dfrac{|219.8-220|}{220}=\dfrac{0.2}{220}=0.0009$, that is $0.09\,\%$. This is well within the $5\,\%$ tolerance of the component: the measurement confirms the marking, and confirms it a hundred times better than the tolerance required.

The line predicts $U=219.8\times 0.040+0.007=8.80$ V. We measure $9.05$ V, that is $0.25$ V more, a gap of $2.9\,\%$, well beyond the noise of the first five points. The point DOES NOT LINE UP. The cause is physical: at $40$ mA, the resistor dissipates $P=UI=0.36$ W, it heats up, and the resistance of a metallic conductor RISES with temperature; the local ratio is $\dfrac{9.05}{0.040}=226\ \Omega$. The lesson is the one of the chapter: a single measurement would have given $226\ \Omega$ without the slightest warning signal, and it would have been written down as the value of the resistance. It is the GRAPH, and only the graph, that shows the drop-off and lets you say from which current Ohm's law stops being linear. An isolated point can never contradict a model: it has nothing to be compared with.

FALSE. The number of digits displayed is a feature of the DISPLAY ELECTRONICS, not of the accuracy. An instrument specified $\pm(0.5\,\%+2)$ displaying $12.47$ V is known to $\pm 0.08$ V: the last displayed digit is already uncertain. Correct version: “the number of significant figures of a result is decided from the uncertainty, calculated from the datasheet, and it is almost always smaller than the number of digits displayed”.

FALSE, and doubly so. First, the formula $R_{p}=\dfrac{R_{1}R_{2}}{R_{1}+R_{2}}$ is not a product: it contains a sum in the denominator, so the relative-uncertainty rule does not apply as it stands. Second, the result is numerically wrong: for $(100\pm 5)\ \Omega$ and $(220\pm 11)\ \Omega$, adding the relative uncertainties would give $10\,\%$ whereas the exact calculation gives $5\,\%$, the DOUBLE of the truth. Correct version: “you propagate with partial derivatives; when both resistors have the same relative tolerance, the parallel combination keeps it”.

FALSE. A real ammeter has an internal resistance, small but not zero, and inserting it in series INCREASES the total resistance, so it LOWERS the current. The error is larger the lower the resistance of the circuit. It is not a random error: it always has the same sign and repeating does not wipe it out. Correct version: “an ammeter disturbs the circuit; you choose one whose resistance is much smaller than that of the circuit, and you correct for it when that is not possible”.

FALSE. The number of decimals a calculator outputs has no connection with the uncertainty: it is a precision of CALCULATION, not an accuracy of MEASUREMENT. The slope inherits the uncertainty of the points, and you estimate it by the half-range of the slopes of the EXTREME LINES, or by the formula for the uncertainty on the slope. Correct version: “the regression gives the best slope, not its uncertainty; that is read from the scatter of the points around the line”.

FALSE. Two different numbers can perfectly well designate the same quantity: that is precisely what the uncertainty allows. Here, the measurement covers $[222, 230]\ \Omega$ and it indeed does not include $220$. But the marked value is not an EXACT value: it carries the manufacturer's tolerance of $5\,\%$, hence $[209, 231]\ \Omega$. The two intervals overlap widely, so the values are CONSISTENT. Correct version: “two values are consistent if their intervals overlap; comparing the central numbers decides nothing”. This is the costliest reasoning error of the chapter, because it leads to declaring an experiment failed when it succeeded.

The current also flows through the INSIDE of the cell, which has a resistance $r$; it loses the voltage $rI$ there. What remains available at the terminals is therefore $\varepsilon-rI$, and it decreases linearly with the current. The relation has the form $Y=b+aX$ with $X=I$ and $Y=U$: the SLOPE is $-r$, so the internal resistance is the OPPOSITE of the slope, and the INTERCEPT is $\varepsilon$, the voltage you would read at zero current, that is with the cell disconnected.

$\bar I=0.30$ A and $\bar U=1.428$ V. We form $\sum(I_{i}-\bar I)(U_{i}-\bar U)=-0.045$ and $\sum(I_{i}-\bar I)^{2}=0.10$, hence $a=\dfrac{-0.045}{0.10}=-0.45$ V/A. The internal resistance is $r=-a=0.45\ \Omega$. The negative sign of the slope is NOT an error: it carries the physics, and taking it positive would give a cell that gets stronger when you draw on it.

$\varepsilon=\bar U-a\bar I=1.428-(-0.45)\times 0.30=1.428+0.135=1.563$ V. Notice that you do NOT get it by measuring the open-circuit voltage with a voltmeter: the voltmeter itself draws a small current. Extrapolating to zero current is the right method, and it is one of the rare cases where the intercept carries the sought result rather than a check.

$\Delta r=\dfrac{|{-0.48}|-|{-0.42}|}{2}=\dfrac{0.48-0.42}{2}=0.03\ \Omega$. So $r=(0.45\pm 0.03)\ \Omega$, a relative uncertainty of $6.7\,\%$. Extreme lines are the standard graphical tool: you draw the steepest and the shallowest lines that still pass through all the uncertainty bars, and the half-range of their slopes gives the uncertainty.

The interval of the measurement is $[1.54, 1.58]$ V, and $1.5$ V does not belong to it: strictly speaking, the measurement and the label are INCONSISTENT. That does not mean the experiment failed. A nominal cell value is not an expected value in the metrological sense: it is a category name, like the $220\ \Omega$ of a resistor, and a new alkaline cell commonly delivers $1.55$ to $1.60$ V, the value falling toward $1.5$ V and then below as it discharges. The correct conclusion is therefore: “the measured electromotive force, $(1.563\pm 0.020)$ V, is that of a new cell, above its nominal value, which is expected”. Comparing a measurement with a commercial label without knowing what the label guarantees is the most common error of judgment in lab reports.

Taking the natural logarithm of both sides of $u=U_{0}e^{-t/\tau}$: $\ln u=\ln U_{0}+\ln\left(e^{-t/\tau}\right)=\ln U_{0}-\dfrac{t}{\tau}$. It indeed has the form $Y=b+aX$ with $Y=\ln u$ and $X=t$: the SLOPE is $-\dfrac{1}{\tau}$ and the INTERCEPT is $\ln U_{0}$. So you do not plot $u$ against $t$, which would give a curve from which nothing precise can be read, but $\ln u$ against $t$.

The logarithms are $2.1972$, $1.7951$, $1.3962$, $0.9932$, $0.5933$. The successive differences, $-0.402$, $-0.399$, $-0.403$, $-0.400$, are almost identical: this is the signature of a true exponential, and it is a check to make before any calculation. The regression gives $\bar t=10$ s, $\overline{\ln u}=1.3950$, $\sum(t_{i}-\bar t)(\ln u_{i}-\overline{\ln u})=-20.048$, $\sum(t_{i}-\bar t)^{2}=250$, hence $a=\dfrac{-20.048}{250}=-0.0802$ s$^{-1}$.