How to Choose a Multimeter Range Before Taking a Reading

A power supply is expected to deliver about \(6\text{ V}\). Your digital multimeter has three DC voltage ranges: \(2\text{ V}\), \(20\text{ V}\), and \(200\text{ V}\). Which should you select before touching the probes to the circuit?

The \(2\text{ V}\) range looks attractive because it displays more decimal places. But it cannot measure \(6\text{ V}\). The \(200\text{ V}\) range can accommodate the reading, but may display less detail than necessary.

Choose the correct measurement function and terminals first. Then select a range above the largest voltage or current you reasonably expect. Once you have a stable reading, move to the lowest range that comfortably accommodates the measurement.

For a known, stable \(6\text{ V}\) circuit, the \(20\text{ V}\) range is the sensible choice. If the magnitude is uncertain but the circuit is known to be safe for your instrument, begin on a higher range and work down.

This method gives you a useful balance between avoiding an out-of-range reading and obtaining enough resolution. But there’s a complication: on some real meters, changing the range can also change the circuit being measured.

01Why choosing the smallest range first can fail

Imagine investigating the output of a low-voltage DC power supply.

The apparatus contains a power supply, a digital multimeter, and two test leads.

The connections run as follows:

• The black lead goes into the multimeter’s COM terminal and connects to the negative supply terminal.
• The red lead goes into the V/Ω terminal and connects to the positive supply terminal.
• The meter measures voltage across the supply terminals, so the connection is parallel to the supply output.
• The selector is set to DC voltage, indicated by \(V\) with a straight-line symbol.

The supply is expected to produce approximately \(6\text{ V}\).

What would the first design show?

Suppose you select the \(2\text{ V}\) range because it offers the finest resolution.

Would the meter display the supply voltage with extra decimal places?

No. The input exceeds the selected measuring range. Many digital multimeters display OL, 1, or another overrange indication, depending on the model.

That doesn’t mean the supply produces no voltage. It means the selected range cannot provide a valid numerical reading.

And it doesn’t mean deliberately testing an overrange condition is a good measurement strategy, particularly for current.

Revising the investigation

A better procedure begins with a stable, safely bounded low-voltage source.

Leave the electrical connections unchanged. Begin on the \(200\text{ V}\) range, then move to \(20\text{ V}\) once the first reading confirms that the voltage is sufficiently low.

Suppose the supply’s actual output is \(6.04\text{ V}\).

The following table gives illustrative results for a hypothetical manual-ranging meter. These are modelled display values, not observations from a real experiment.

The \(2\text{ V}\) result is included to illustrate the problem, not as a test to perform.

The physical voltage hasn’t changed from \(6.0\text{ V}\) to \(6.04\text{ V}\) when the dial moves. The meter is displaying the same input with different resolution.

Here, the \(20\text{ V}\) range has ten times finer display resolution than the \(200\text{ V}\) range:

\[
\frac{0.1\text{ V}}{0.01\text{ V}}=10
\]

The smaller increment allows changes of \(0.01\text{ V}\) to be displayed rather than only changes of \(0.1\text{ V}\).

That doesn’t automatically mean the reading is ten times more accurate. Accuracy concerns how close a measurement is to the true value, and depends on the instrument’s specifications, calibration, and measurement conditions.

Why the stable supply matters

Suppose you repeat this investigation using a battery whose terminal voltage is changing because its load changes during the experiment.

You measure \(6.1\text{ V}\) on one range and \(6.04\text{ V}\) on another.

Did changing the range cause the difference, or did the battery voltage change between measurements?

You cannot tell from those two readings alone.

The revised investigation uses a stable source and keeps the probe positions fixed. That controls two possible explanations for a difference between readings.

Even then, a real meter can introduce measurement effects of its own. We’ll return to that after establishing the basic range-selection method.

02Range, resolution, and accuracy are different things

A meter’s range tells you the magnitude it is designed to measure on the selected setting.

Its resolution is the smallest change represented by its display or scale.

Its accuracy describes how close its indication is to the actual value under specified conditions.

These aren’t interchangeable.

Why an analogue meter makes the problem easier to see

Imagine an analogue voltmeter with a pointer and a scale marked from 0 to 10.

You want to measure \(6\text{ V}\).

On a \(10\text{ V}\) range, the pointer sits at 60% of full-scale deflection.

On a \(50\text{ V}\) range, that same voltage produces only 12% of full-scale deflection.

On the larger range, the pointer is crowded near zero. Small changes in voltage produce smaller movements, making them harder to distinguish.

The sensible analogue choice is a range that gives a substantial deflection without sending the pointer beyond full scale.

The same broad principle applies to a manual-ranging digital meter: don’t waste measurement detail by selecting a needlessly large range.

But digital meters do not all have identical range intervals, resolution, or input characteristics. Always use the actual instrument’s specifications.

03How to choose the range before connecting the meter

The range decision begins with the measurement you intend to make.

For AC measurements, use the appropriate AC function rather than DC. The selected function matters just as much as the numerical range.

For resistance measurements, disconnect power and safely discharge any capacitors before measuring.

For an ordinary school laboratory circuit, use this sequence:

1. Identify the quantity. Decide whether you need voltage, current, or resistance, and select the appropriate AC or DC function.
2. Estimate the maximum value. Use the source voltage, circuit components, Ohm’s law, or known apparatus limits. Consider any changes that will occur during the experiment.
3. Check the instrument and connections. Confirm the correct input sockets, lead condition, range limits, current-input protection, and suitability for the circuit.
4. Choose a starting range. If the quantity has a reliable upper bound, use a range comfortably above it. If the exact value is uncertain, start higher, provided the circuit and instrument are independently known to be safe.
5. Refine the range. Once the reading is stable, move down to a range that provides useful resolution without approaching an expected peak or exceeding the instrument’s limits.

For current measurements, power down the circuit before changing how the ammeter is connected. Never put a meter configured for current measurement directly across a power source.

A high numerical range doesn’t make an incorrect connection safe. Nor does it make an unsuitable meter safe for mains electricity or an unknown high-energy circuit. Those measurements require appropriate equipment, training, and risk controls.

04The important exception: a real meter can change the circuit

So far, we’ve treated the voltmeter as though it draws no current and the ammeter as though it adds no resistance.

Those are useful idealisations for HSC circuit calculations.

Real instruments are not perfect.

This section introduces nonideal-meter models as a practical extension, rather than treating them as a universal feature of every multimeter.

An ammeter has some effective resistance, \(R_{\mathrm A}\), in its current-measuring path. Because the ammeter is connected in series, that resistance adds to the circuit.

A voltmeter has a finite input resistance, \(R_{\mathrm V}\). Because the voltmeter connects in parallel, it provides an additional path for current.

On some instruments, these resistances also depend on the selected range.

That creates a surprising possibility: the range showing the most decimal places may produce a worse estimate of the circuit’s original value.

05A voltmeter can affect a reading too

Imagine measuring a very small electrical signal produced by a high-resistance sensor.

A voltmeter connects across the sensor output. Ideally, the voltmeter draws no current.

In practice, its input resistance is very large but finite.

The effect resembles connecting an additional resistor across the output.

For a simple model with an ideal source of emf \(E\), a series resistance \(R_{\mathrm s}\), and no other connected load, the measured voltage is

\[
V_{\mathrm m}
=E\frac{R_{\mathrm V}}{R_{\mathrm s}+R_{\mathrm V}}
\]

Here:

• \(V_{\mathrm m}\) is the voltage across the connected meter, in volts.
• \(E\) is the source’s open-circuit emf, in volts.
• \(R_{\mathrm s}\) is the source resistance, in ohms.
• \(R_{\mathrm V}\) is the voltmeter input resistance, in ohms.

Suppose \(E=6.0\text{ V}\), \(R_{\mathrm s}=1.0\text{ M}\Omega\), and \(R_{\mathrm V}=10\text{ M}\Omega\).

The predicted connected reading is

\[
\begin{aligned}
V_{\mathrm m}
&=6.0\left(\frac{10}{1+10}\right)\\
&=5.45\,\mathrm{V}
\end{aligned}
\]

We can use megaohms in the fraction because both resistances have the same units.

The voltmeter reads approximately \(5.45\text{ V}\), even though the source has a \(6.0\text{ V}\) open-circuit emf.

That difference is much larger than a \(0.01\text{ V}\) display increment.

The meter isn’t necessarily faulty. It is measuring the voltage that exists while its own input resistance loads the source.

Most ordinary digital voltmeters have sufficiently high input resistance that this effect is small in low-resistance school circuits. It becomes important when the resistance of the source or circuit is comparable to the meter’s input resistance.

The following practice questions use both the basic ideal-meter method and explicitly specified practical extensions.

06Questions and solutions

Question 1

A student wants to measure the voltage of an isolated DC cell.

The cell’s voltage is known to lie between \(1.4\text{ V}\) and \(1.7\text{ V}\).

The digital multimeter has these manual DC voltage ranges:

a) Select the most appropriate range and explain why.

b) State how the red and black test leads should be connected.

c) If the actual terminal voltage is \(1.563\text{ V}\), predict the display on the selected range and on the \(20\text{ V}\) range. Assume the meter rounds to its nearest display increment.

Solution 1

The best range is \(2\text{ V}\), with the black lead in COM and the red lead in V/Ω.

The largest expected voltage is \(1.7\text{ V}\), below the \(2\text{ V}\) range’s limit. The \(20\text{ V}\) and \(200\text{ V}\) ranges are also large enough, but their displays provide less detail.

For a positive DC reading, the red probe touches the cell’s positive terminal and the black probe touches its negative terminal.

With an actual voltage of \(1.563\text{ V}\):

• On the \(2\text{ V}\) range, the display is \(1.563\text{ V}\).
• On the \(20\text{ V}\) range, the display rounds to \(1.56\text{ V}\).

Both ranges give valid indications. The extra digits on the \(2\text{ V}\) range are useful because the entire expected voltage interval fits within it.

The tempting mistake is to assume that a larger range is always better because it can accommodate a larger input. Once the expected maximum is safely covered, additional range may simply sacrifice resolution.

Question 2

An ideal \(6.0\text{ V}\) DC supply is connected to a \(200\,\Omega\) resistor in series with two parallel resistors, \(600\,\Omega\) and \(1200\,\Omega\).

A student wants to measure the current through the \(600\,\Omega\) branch and the voltage across that branch.

The available ranges are:

• DC voltage: 2 V, 20 V, 200 V.
• DC current: 2 mA, 20 mA, 200 mA.

Assume ideal meters when connected correctly.

a) Calculate the current through the \(600\,\Omega\) resistor and its potential difference.

b) Choose suitable voltage and current ranges and describe the correct meter connections.

c) Another student connects an ammeter directly across the \(600\,\Omega\) resistor, using the \(200\text{ mA}\) range. Assuming the ammeter has negligible resistance, predict the current through that ammeter and explain why it does not measure the original branch current.

Solution 2

The \(600\,\Omega\) branch carries approximately \(6.67\text{ mA}\), with \(4.0\text{ V}\) across it. The suitable ranges are \(20\text{ mA}\) and \(20\text{ V}\). Incorrectly placing the ammeter across the branch changes the circuit and can produce \(30\text{ mA}\) through the meter instead.

First, calculate the equivalent resistance of the parallel combination.

\[
\begin{aligned}
\frac{1}{R_{\mathrm p}}
&=\frac{1}{600}+\frac{1}{1200}\\
R_{\mathrm p}&=400\,\Omega
\end{aligned}
\]

Here, \(R_{\mathrm p}\) is the equivalent resistance of the two parallel branches.

The total resistance is

\[
R_{\mathrm{total}}=200+400=600\,\Omega
\]

The total current is

\[
I_{\mathrm{total}}
=\frac{6.0\,\mathrm{V}}{600\,\Omega}
=0.010\,\mathrm{A}
\]

So the source supplies \(10\text{ mA}\).

The potential difference across the parallel combination is

\[
\begin{aligned}
V_{\mathrm p}&=I_{\mathrm{total}}R_{\mathrm p}\\
&=(0.010\,\mathrm{A})(400\,\Omega)\\
&=4.0\,\mathrm{V}
\end{aligned}
\]

Both parallel resistors have this potential difference.

Therefore, the current through the \(600\,\Omega\) branch is

\[
\begin{aligned}
I_{600}&=\frac{V_{\mathrm p}}{600\,\Omega}\\
&=\frac{4.0\,\mathrm{V}}{600\,\Omega}\\
&=6.67\,\mathrm{mA}
\end{aligned}
\]

The \(20\text{ mA}\) range accommodates the branch current, while the \(2\text{ mA}\) range does not.

Similarly, \(20\text{ V}\) is appropriate for the \(4.0\text{ V}\) potential difference.

The ammeter must be inserted in series within the \(600\,\Omega\) branch. The voltmeter connects across its two terminals.

Now consider the incorrect ammeter connection.

An ideal ammeter connected directly across the \(600\,\Omega\) resistor creates a zero-resistance path between the two parallel junctions.

That path also bypasses the \(1200\,\Omega\) branch.

The remaining effective resistance is just \(200\,\Omega\), so

\[
\begin{aligned}
I_{\mathrm{wrong}}
&=\frac{6.0\,\mathrm{V}}{200\,\Omega}\\
&=0.030\,\mathrm{A}\\
&=30\,\mathrm{mA}
\end{aligned}
\]

The \(200\text{ mA}\) range could accommodate that current in the idealised model, but it would be measuring a different circuit.

In a real circuit, this connection could overload the current input or damage the apparatus. It must not be attempted.

The governing constraint isn’t simply that the current fits the selected range. The ammeter must also be in the correct part of the circuit.

Question 3

An ideal \(6.0\text{ V}\) DC source supplies two series resistors:

\[
R_A=1.0\text{ k}\Omega,\qquad
R_B=9.0\text{ k}\Omega
\]

A switch is connected in parallel across resistor B.

When the switch is open, current flows through both resistors. When it is closed, resistor B is bypassed by an ideal wire.

A voltmeter measures the voltage across resistor A. An ammeter measures the total current.

The available DC ranges are \(2\text{ V}\), \(20\text{ V}\), and \(200\text{ V}\), and \(2\text{ mA}\), \(20\text{ mA}\), and \(200\text{ mA}\).

A student wants to record the experiment without changing either meter’s range while operating the switch.

a) Calculate the voltage across resistor A and the total current with the switch open.

b) Calculate both quantities when the switch is closed.

c) Choose the most suitable fixed ranges for the whole investigation.

d) Explain why selecting the ranges from the first readings alone could produce misleading results.

Assume the meters are ideal and that all components can safely withstand the calculated conditions.

Solution 3

Use the \(20\text{ V}\) range and the \(20\text{ mA}\) range throughout. Closing the switch increases both the voltage across resistor A and the total current by a factor of ten.

With the switch open, both resistors are in series.

Their total resistance is

\[
R_{\mathrm{open}}
=1000+9000
=10000\,\Omega
\]

The current is

\[
\begin{aligned}
I_{\mathrm{open}}
&=\frac{6.0\,\mathrm{V}}{10000\,\Omega}\\
&=0.00060\,\mathrm{A}\\
&=0.60\,\mathrm{mA}
\end{aligned}
\]

The voltage across resistor A is

\[
\begin{aligned}
V_{A,\mathrm{open}}&=I_{\mathrm{open}}R_A\\
&=(0.00060)(1000)\,\mathrm{V}\\
&=0.60\,\mathrm{V}
\end{aligned}
\]

Initially, both quantities would fit within the smallest available ranges.

But this is not the circuit’s only state.

When the switch closes, resistor B is bypassed. Its terminals are joined by an ideal conductor, so its potential difference becomes zero.

The total resistance falls to \(1000\,\Omega\).

\[
\begin{aligned}
I_{\mathrm{closed}}
&=\frac{6.0\,\mathrm{V}}{1000\,\Omega}\\
&=0.0060\,\mathrm{A}\\
&=6.0\,\mathrm{mA}
\end{aligned}
\]

Resistor A now has the entire supply voltage across it:

\[
V_{A,\mathrm{closed}}
=(0.0060)(1000)
=6.0\,\mathrm{V}
\]

The comparison is:

A \(2\text{ V}\) voltmeter range and a \(2\text{ mA}\) ammeter range would both be exceeded after closing the switch.

The correct fixed choices are therefore \(20\text{ V}\) and \(20\text{ mA}\).

The tempting prediction is that removing resistor B somehow removes part of the current path, reducing the current.

But closing the switch doesn’t disconnect the circuit. It creates a low-resistance path around B while leaving A connected to the supply.

The total resistance decreases, so the current increases.

This illustrates a general experimental principle: choose the range for the largest value that can occur during the investigation, not merely the first value observed.

Question 4

A high-resistance sensor can be modelled as an ideal DC source of unknown emf \(E\), with an unknown series resistance \(R_{\mathrm s}\).

The sensor is stable and has no other load connected.

A hypothetical manual-ranging voltmeter has the following specifications:

When the voltmeter is connected across the sensor output:

• On the \(20\text{ V}\) range, it displays \(4.00\text{ V}\).
• On the \(200\text{ V}\) range, it displays \(8.0\text{ V}\).

For this theoretical problem, treat the stated readings as exact and ignore rounding uncertainty.

a) Determine the sensor’s open-circuit emf \(E\) and series resistance \(R_{\mathrm s}\).

b) Which voltage range gives a reading closer to the open-circuit emf? Explain why the number of displayed decimal places is misleading.

c) Predict the reading of an additional idealised voltmeter having an input resistance of \(2.0\text{ M}\Omega\).

d) Explain how this third reading could help evaluate the proposed sensor model.

Solution 4

The sensor has an open-circuit emf of \(12.0\text{ V}\) and series resistance of \(2.0\text{ M}\Omega\). The \(200\text{ V}\) range gives the closer estimate of the open-circuit voltage, even though it displays fewer decimal places.

Use the voltage-divider model for the meter connected to the sensor:

\[
V_{\mathrm m}
=E\frac{R_{\mathrm V}}{R_{\mathrm s}+R_{\mathrm V}}
\]

To simplify the working, express all resistances in megaohms and let \(r\) be the numerical value of \(R_{\mathrm s}\) in megaohms.

On the \(20\text{ V}\) range:

\[
4.00=E\frac{1}{r+1}
\]

Rearranging:

\[
E=4(r+1)=4r+4
\]

On the \(200\text{ V}\) range:

\[
8.0=E\frac{4}{r+4}
\]

Rearranging:

\[
E=2(r+4)=2r+8
\]

Because the same stable source must satisfy both equations:

\[
\begin{aligned}
4r+4&=2r+8\\
2r&=4\\
r&=2
\end{aligned}
\]

Therefore,

\[
R_{\mathrm s}=2.0\text{ M}\Omega
\]

Substituting into either equation:

\[
E=4(2)+4=12.0\text{ V}
\]

Now compare the displayed values with the source’s open-circuit emf.

The \(20\text{ V}\) range reads \(4.00\text{ V}\), which is \(8.00\text{ V}\) below the open-circuit value.

The \(200\text{ V}\) range reads \(8.0\text{ V}\), which is \(4.0\text{ V}\) below it.

The larger range is closer because its specified input resistance is higher, reducing the loading effect.

For the additional \(2.0\text{ M}\Omega\) voltmeter:

\[
\begin{aligned}
V_{\mathrm m}
&=12.0\left(\frac{2}{2+2}\right)\\
&=6.0\,\mathrm{V}
\end{aligned}
\]

A third reading of approximately \(6.0\text{ V}\) would support the model because it agrees with a prediction using the values inferred from the first two measurements.

It wouldn’t conclusively prove that the model is correct. Other source behaviours might also produce matching readings.

Repeated measurements, a check that the source remains stable, and comparison with other known input resistances would strengthen the evidence.

The central mistake is treating extra display digits as proof that the instrument is measuring the original, unloaded voltage more accurately.

Question 5

An ideal \(12.0\text{ V}\) source is connected across two identical \(1.0\text{ M}\Omega\) resistors, A and B, in series.

A manual-ranging voltmeter on its \(20\text{ V}\) range has an input resistance of \(1.0\text{ M}\Omega\).

One student measures three voltages, moving the same meter between positions:

1. Across resistor A alone: \(4.0\text{ V}\).
2. Across resistor B alone: \(4.0\text{ V}\).
3. Across the complete series pair: \(12.0\text{ V}\).

The student concludes that Kirchhoff’s voltage law has failed because \(4.0+4.0\neq12.0\).

A second student proposes connecting two identical voltmeters simultaneously, one across each resistor, and predicts that each will still read \(4.0\text{ V}\).

a) Show why the three sequential readings are possible without violating Kirchhoff’s voltage law. Determine the voltage across the unmeasured resistor during each individual resistor measurement.

b) Determine what each meter reads when two identical meters are connected simultaneously.

c) Calculate the total current supplied by the source before any meters are connected and when both meters are connected.

d) Does obtaining the original resistor voltages with both meters connected prove that the measurement has not disturbed the circuit? Explain.

Assume the source and resistors are ideal, apart from the explicitly specified voltmeter input resistances. Ignore display rounding.

Solution 5

Neither student’s conclusion is correct. Kirchhoff’s voltage law still holds, and two identical voltmeters connected simultaneously will each read \(6.0\text{ V}\), not \(4.0\text{ V}\).

More surprisingly, those simultaneous readings match the original resistor voltages even though connecting the meters doubles the total current supplied by the source.

The key is recognising that moving or adding a voltmeter changes the circuit.

Part a: Explain the sequential measurements

First consider the circuit with no voltmeter.

The two \(1.0\text{ M}\Omega\) resistors are equal and connected in series across an ideal \(12.0\text{ V}\) source.

Each resistor therefore has \(6.0\text{ V}\) across it.

Now connect the voltmeter across resistor A.

Its \(1.0\text{ M}\Omega\) input resistance is in parallel with resistor A, which is also \(1.0\text{ M}\Omega\).

Using the parallel-resistance formula:

\[
R_{\mathrm{parallel}}
=\frac{R_1R_2}{R_1+R_2}
\]

The equivalent resistance across A is

\[
\begin{aligned}
R_{\mathrm{A,eq}}
&=\frac{(1.0)(1.0)}{1.0+1.0}\\
&=0.50\text{ M}\Omega
\end{aligned}
\]

The total series resistance is now

\[
R_{\mathrm{total}}
=0.50+1.0
=1.50\text{ M}\Omega
\]

The current supplied is

\[
\begin{aligned}
I&=\frac{12.0\,\mathrm{V}}{1.50\text{ M}\Omega}\\
&=8.0\,\mu\mathrm{A}
\end{aligned}
\]

Here, \(\mu\mathrm{A}\) means microamperes, with one microampere equal to \(10^{-6}\text{ A}\).

The voltage across resistor A and the parallel voltmeter is

\[
\begin{aligned}
V_A&=IR_{\mathrm{A,eq}}\\
&=(8.0\,\mu\mathrm{A})(0.50\text{ M}\Omega)\\
&=4.0\,\mathrm{V}
\end{aligned}
\]

Resistor B, which has no voltmeter connected across it, has

\[
V_B=12.0-4.0=8.0\,\mathrm{V}
\]

The voltage law is satisfied:

\[
4.0+8.0=12.0\,\mathrm{V}
\]

Now move the same voltmeter across resistor B.

Because the resistors are identical, the arrangement is reversed.

This time:

\[
V_A=8.0\,\mathrm{V},\qquad
V_B=4.0\,\mathrm{V}
\]

Again, the voltage law holds.

When the voltmeter is connected across the complete series pair, it is directly across the ideal \(12.0\text{ V}\) supply. It reads \(12.0\text{ V}\). The ideal source maintains that terminal voltage even though the meter draws additional current.

The mistake was adding \(4.0\text{ V}\) measured in one circuit configuration to \(4.0\text{ V}\) measured in a different configuration.

Kirchhoff’s voltage law applies to potential differences around the same circuit at the same time. It does not say that readings obtained after changing the connections can automatically be combined.

Part b: Connect two voltmeters simultaneously

Now place one identical \(1.0\text{ M}\Omega\) voltmeter across each resistor.

Each resistor is shunted by an equal resistance.

Both sections therefore have equivalent resistance

\[
R_{\mathrm{eq}}=0.50\text{ M}\Omega
\]

The two equal sections are in series across \(12.0\text{ V}\).

Each section receives half the voltage:

\[
V_A=V_B=6.0\,\mathrm{V}
\]

So both voltmeters read \(6.0\text{ V}\).

The second student’s prediction of \(4.0\text{ V}\) each is wrong because connecting the second voltmeter changes the first voltmeter’s electrical conditions.

The first meter’s reading actually rises from \(4.0\text{ V}\) to \(6.0\text{ V}\) when the second meter is connected.

Part c: Compare the source currents

Before any meters are connected, the total resistance is

\[
R_0=1.0+1.0=2.0\text{ M}\Omega
\]

The initial source current is

\[
\begin{aligned}
I_0&=\frac{12.0\,\mathrm{V}}{2.0\text{ M}\Omega}\\
&=6.0\,\mu\mathrm{A}
\end{aligned}
\]

With both voltmeters connected, the two equivalent resistances add:

\[
R_{\mathrm{loaded}}
=0.50+0.50
=1.0\text{ M}\Omega
\]

Therefore,

\[
\begin{aligned}
I_{\mathrm{loaded}}
&=\frac{12.0\,\mathrm{V}}{1.0\text{ M}\Omega}\\
&=12.0\,\mu\mathrm{A}
\end{aligned}
\]

The source current doubles.

Part d: Has the measurement disturbed the circuit?

Yes. The source current has changed from \(6.0\,\mu\mathrm{A}\) to \(12.0\,\mu\mathrm{A}\), so the circuit is not operating identically.

Yet each resistor still has \(6.0\text{ V}\) across it.

Why?

Before measurement, the circuit consists of two equal series resistances, so each receives half the supply voltage.

With two identical meters connected, both sections are reduced by exactly the same factor. The sections remain equal.

The voltage division is therefore unchanged.

But their combined resistance is smaller, so the source supplies more current.

In fact, each actual \(1.0\text{ M}\Omega\) resistor still carries \(6.0\,\mu\mathrm{A}\). Each added voltmeter provides another \(6.0\,\mu\mathrm{A}\) current path through its own section.

This is the conceptual hinge: a meter can give the correct original voltage while still changing another physical quantity in the circuit.

Correct numerical agreement alone doesn’t prove the measurement was non-invasive.

The proper test is whether the instrument’s effect is negligible for every quantity relevant to the investigation.

07What choosing a range lets you investigate next

Before taking a reading, you’re making three decisions: what quantity to measure, how to connect the instrument, and what range can accommodate the expected result.

For a conventional manual-ranging meter in an ordinary low-resistance school circuit, a dependable approach is to predict the largest plausible reading, select a suitable range, and then refine it to obtain useful resolution.

If the value changes with time or switching conditions, choose a range that accommodates the entire experiment.

If the measured result differs unexpectedly from your prediction, don’t immediately assume the circuit theory is wrong. Check the selected function, terminals, connections, range, and whether the meter could be changing the circuit.

That last question leads naturally into measurement uncertainty, internal resistance, and experimental design: how can you determine not only what the instrument displays, but how closely that display represents the physical quantity you actually intended to measure?