Why a Voltmeter Changes the Circuit It Measures

Connect a voltmeter across the 2.0 kΩ resistor in a voltage divider containing a 4.0 kΩ resistor and a 9.0 V supply. You expect a reading of 3.00 V. But a voltmeter with an input resistance of 10.0 kΩ would actually read approximately 2.65 V.

The meter is connected correctly. The calculation is correct. So why is the reading different?

A voltmeter can change the circuit because it has a finite resistance and draws current. That extra current changes the resistance of the circuit, which can change the potential difference you are trying to measure.

This effect is called voltmeter loading. Understanding it means knowing when a meter can be trusted to measure a voltage without significantly disturbing the circuit.

01Why does a voltmeter change the circuit?

Start with the way a voltmeter is connected.

A voltmeter measures the potential difference between two points. To do this, it is connected in parallel with the component being measured.

For example, if you want to measure the potential difference across a resistor, you connect one voltmeter terminal to each end of that resistor.

The important detail is that the voltmeter becomes another branch of the circuit.

The ideal voltmeter model

In HSC Physics, you generally begin by treating a voltmeter as an ideal instrument.

An ideal voltmeter has infinite resistance:

\[
R_v \rightarrow \infty
\]

Using Ohm’s law:

\[
I_v=\frac{V}{R_v}
\]

where:

• \(I_v\) is the current through the voltmeter, measured in amperes (A).
• \(V\) is the potential difference across the voltmeter, measured in volts (V).
• \(R_v\) is the voltmeter’s resistance, measured in ohms (\(\Omega\)).

If the resistance is infinite, the current is zero.

This means an ideal voltmeter draws no current from the circuit. Connecting it in parallel does not change the circuit’s current distribution or the potential difference being measured.

The practical voltmeter model

Real voltmeters do not have infinite resistance.

They have a large but finite input resistance. This means a small current flows through the meter whenever a potential difference exists across its terminals.

For a practical voltmeter, we can model the instrument as an ordinary resistor, \(R_v\), connected in parallel with the component being measured.

The equivalent resistance of two parallel resistors is:

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

For a voltmeter connected across a resistor \(R\):

\[
R_{\text{new}}=
\frac{RR_v}{R+R_v}
\]

Notice something important: the equivalent resistance is smaller than the resistance of either individual resistor.

The voltmeter has created an additional path for current. The total current through the parallel combination is now shared between the component and the meter.

The meter is therefore no longer just observing the circuit. It is participating in it.

02When does the voltmeter actually change the measured voltage?

There is an important distinction here.

A voltmeter draws current whenever it has a potential difference across its terminals. However, this does not mean that its reading must always change.

The effect depends on the rest of the circuit.

Consider two situations.

Situation 1: A resistor connected directly across an ideal voltage source

Suppose a 6.0 kΩ resistor is connected directly across an ideal 12.0 V supply.

A voltmeter is then connected in parallel with the resistor.

Both the resistor and the voltmeter are connected across the same two supply terminals. An ideal voltage source maintains a constant potential difference of 12.0 V across those terminals.

The meter draws extra current, but the voltage across the resistor remains 12.0 V.

Situation 2: A resistor connected through another resistor

Now place a 4.0 kΩ resistor between the supply and the measured resistor.

The extra current drawn by the voltmeter must also pass through the 4.0 kΩ resistor. This increases the potential difference across that resistor, leaving less voltage across the measured component.

The voltmeter reading decreases.

This is why the statement that a voltmeter always reduces the voltage is incorrect.

The more useful question is whether the rest of the circuit can maintain its original potential difference when the meter draws additional current.

03Understanding voltmeter loading in a voltage divider

Voltage dividers are particularly useful for understanding why a voltmeter can give an incorrect reading.

Consider a circuit containing two resistors connected in series across a 9.0 V supply:

• \(R_1=4.0\text{ k}\Omega\)
• \(R_2=2.0\text{ k}\Omega\)

We want to measure the potential difference across \(R_2\).

Before connecting the voltmeter

First, calculate the potential difference across \(R_2\) without a voltmeter.

The total resistance is:

\[
R_{\text{total}}=R_1+R_2
\]

\[
R_{\text{total}}=4.0+2.0
=6.0\text{ k}\Omega
\]

The circuit current is:

\[
I=\frac{V}{R_{\text{total}}}
\]

\[
I=\frac{9.0}{6000}
=1.50\text{ mA}
\]

The potential difference across \(R_2\) is:

\[
V_2=IR_2
\]

\[
V_2=0.00150\times2000
\]

\[
\boxed{V_2=3.00\text{ V}}
\]

This is the voltage we want to measure.

After connecting a practical voltmeter

Now connect a voltmeter with an input resistance of 10.0 kΩ across \(R_2\).

The voltmeter and \(R_2\) are in parallel, so their equivalent resistance is:

\[
R_{\parallel}=
\frac{R_2R_v}{R_2+R_v}
\]

Substituting:

\[
R_{\parallel}=
\frac{2.0\times10.0}{2.0+10.0}
\]

\[
R_{\parallel}=1.67\text{ k}\Omega
\]

The lower branch of the voltage divider has changed from 2.0 kΩ to approximately 1.67 kΩ.

The total circuit resistance is now:

\[
R_{\text{total}}=4.0+1.67
\]

\[
R_{\text{total}}=5.67\text{ k}\Omega
\]

The new circuit current is:

\[
I=\frac{9.0}{5667}
\]

\[
I\approx1.59\text{ mA}
\]

The potential difference across the parallel branch is:

\[
V_{\parallel}=IR_{\parallel}
\]

\[
V_{\parallel}=0.00159\times1667
\]

\[
\boxed{V_{\text{meter}}\approx2.65\text{ V}}
\]

The voltmeter now reads 2.65 V instead of the original 3.00 V.

Why did the reading decrease?

There are two connected effects.

First, the voltmeter provides an additional path for current through the lower branch. This reduces the equivalent resistance of that branch.

Second, the reduced branch resistance increases the total current flowing through the circuit. More current passes through the upper 4.0 kΩ resistor, increasing the potential difference across it.

Consequently, less of the supply voltage remains across the lower branch.

The meter has changed the voltage division, which is why it measures a different potential difference from the original unloaded circuit.

Notice that the voltmeter does not need to draw a huge current to cause a noticeable error. In this example, its current is approximately 0.265 mA, but that is enough to change the reading by about 11.8%.

04Investigating the loading effect

You can investigate this effect by keeping a voltage divider unchanged and comparing voltmeters with different input resistances.

Consider the same 9.0 V divider:

• Upper resistor: 4.0 kΩ
• Lower resistor: 2.0 kΩ
• Ideal supply: 9.0 V

The voltmeter is always connected across the lower resistor.

What should the original design show?

If the voltmeter were ideal, its infinite resistance would mean that no current flows through it.

The potential difference across the lower resistor would remain 3.00 V.

Now consider what happens when the meter has different finite input resistances.

The following table contains illustrative calculated predictions, not recorded experimental measurements. The supply and resistor values are assumed to be exact, and the supply is treated as ideal.

The pattern is clear: as the voltmeter’s resistance increases, its reading approaches the original 3.00 V.

What could make the real investigation less reliable?

There are a few things to control.

1. Resistor tolerances: Actual resistors may not have exactly their labelled resistances.
2. Supply resistance: A real power supply may have some output resistance, which can also affect the reading.
3. Meter accuracy: Different meters may have different calibration errors, separate from their loading effects.
4. Input resistance: The actual input resistance should be obtained from the meter specifications rather than assumed.

A better investigation keeps the same divider and supply, uses meters with known input resistances, and compares repeated readings.

If a high-input-resistance meter is used as a reference, its reading should be closer to the unloaded voltage. However, even this meter has finite resistance, so its reading is not necessarily the exact unloaded value.

The experiment supports the idea that the voltmeter changes the circuit, but it is important to control other variables before attributing every difference to loading.

05Practical extension: How large does the voltmeter resistance need to be?

You now know that increasing the voltmeter’s input resistance reduces its effect on the circuit.

But how large is large enough?

A useful way to answer this is to consider the resistance of the circuit as seen from the two points where the voltmeter is connected.

Call this resistance \(R_s\), the effective source resistance.

For a simple resistive DC circuit, the rest of the circuit can be represented by an equivalent voltage source \(V_0\) in series with \(R_s\).

Connecting a voltmeter of resistance \(R_v\) across the output gives:

\[
V_{\text{measured}}=
V_0\frac{R_v}{R_s+R_v}
\]

The percentage loading error is:

\[
\text{Error}=
\frac{V_0-V_{\text{measured}}}{V_0}
\times100\%
\]

Substituting the voltage equation gives:

\[
\text{Error}=
\frac{R_s}{R_s+R_v}
\times100\%
\]

This equation gives us a useful decision rule.

For example, if the voltmeter resistance is 100 times the effective source resistance:

\[
R_v=100R_s
\]

Then:

\[
\text{Error}=
\frac{R_s}{101R_s}\times100\%
\]

\[
\boxed{\text{Error}\approx0.99\%}
\]

So, for a simple resistive DC circuit, making the voltmeter resistance approximately 100 times the effective source resistance keeps the loading error below about 1%.

This is a practical rule of thumb, not a universal requirement for every measurement.

Which resistance should you compare?

This is where students sometimes make a mistake.

You might think that you only need to compare the voltmeter resistance with the resistance of the component being measured.

However, you should compare it with the effective resistance of the rest of the circuit, viewed from the voltmeter’s connection points.

For a simple voltage divider with an ideal supply, this effective resistance is:

\[
R_s=R_1\parallel R_2
\]

For the earlier 4.0 kΩ and 2.0 kΩ divider:

\[
R_s=\frac{4.0\times2.0}{4.0+2.0}
\]

\[
R_s=1.33\text{ k}\Omega
\]

A 10.0 kΩ voltmeter is only about 7.5 times this resistance.

That is why its loading effect is noticeable, even though its resistance is five times larger than the 2.0 kΩ resistor being measured.

This equivalent-circuit model is a practical extension of the ideal voltmeter model. It is particularly useful when working with batteries, voltage dividers, and other circuits where the source cannot maintain a perfectly constant terminal voltage.

It also connects to the distinction between a battery’s EMF and its terminal potential difference. When internal resistance is present, the terminal voltage can change as the current changes. You can explore that relationship in How to Distinguish EMF from Potential Difference in Circuits.

07Questions and solutions

The following questions progress from straightforward applications of voltmeter resistance to situations where several parts of a circuit interact.

Question 1

A 6.0 V ideal supply is connected directly across a 3.0 kΩ resistor. A voltmeter with an input resistance of 30.0 kΩ is connected in parallel with the resistor.

Calculate:

a) The voltmeter reading.

b) The current through the voltmeter.

c) The total current supplied by the source.

Solution 1

The voltmeter reads 6.0 V, draws 0.200 mA, and the total current supplied is 2.20 mA.

Part a: Voltmeter reading

The voltmeter is connected directly across an ideal voltage source.

Therefore, the potential difference across its terminals is fixed at 6.0 V.

\[
\boxed{V_v=6.0\text{ V}}
\]

Part b: Current through the voltmeter

Using Ohm’s law:

\[
I_v=\frac{V_v}{R_v}
\]

\[
I_v=\frac{6.0}{30000}
\]

\[
\boxed{I_v=0.200\text{ mA}}
\]

Part c: Total current

First, calculate the current through the resistor:

\[
I_R=\frac{6.0}{3000}
\]

\[
I_R=2.00\text{ mA}
\]

The total current is the sum of the parallel branch currents:

\[
I_{\text{total}}=I_R+I_v
\]

\[
I_{\text{total}}=2.00+0.200
\]

\[
\boxed{I_{\text{total}}=2.20\text{ mA}}
\]

The key point is that the voltmeter changes the total current but does not change the voltage across the resistor.

The ideal voltage source maintains its potential difference regardless of the additional current drawn by the meter.

Question 2

A 15.0 V ideal supply is connected to two resistors in series:

• \(R_1=5.0\text{ k}\Omega\)
• \(R_2=10.0\text{ k}\Omega\)

A voltmeter with an input resistance of 20.0 kΩ is connected across \(R_2\).

Calculate:

a) The potential difference across \(R_2\) without the voltmeter.

b) The reading obtained when the voltmeter is connected.

c) The percentage loading error.

Solution 2

The original voltage is 10.0 V, but the practical voltmeter reads approximately 8.57 V, giving a loading error of 14.3%.

Part a: Voltage without the voltmeter

The total resistance is:

\[
R_{\text{total}}=5.0+10.0
\]

\[
R_{\text{total}}=15.0\text{ k}\Omega
\]

The circuit current is:

\[
I=\frac{15.0}{15000}
=1.00\text{ mA}
\]

The potential difference across \(R_2\) is:

\[
V_2=IR_2
\]

\[
V_2=0.00100\times10000
\]

\[
\boxed{V_2=10.0\text{ V}}
\]

Part b: Voltage with the voltmeter

First, calculate the equivalent resistance of the lower parallel branch.

\[
R_{\parallel}=
\frac{10.0\times20.0}{10.0+20.0}
\]

\[
R_{\parallel}=6.67\text{ k}\Omega
\]

The new total resistance is:

\[
R_{\text{total}}=5.0+6.67
\]

\[
R_{\text{total}}=11.67\text{ k}\Omega
\]

The new circuit current is:

\[
I=\frac{15.0}{11667}
\]

\[
I\approx1.286\text{ mA}
\]

The voltage across the parallel branch is:

\[
V_{\parallel}=IR_{\parallel}
\]

\[
V_{\parallel}=0.001286\times6667
\]

\[
\boxed{V_{\text{meter}}=8.57\text{ V}}
\]

Part c: Percentage loading error

The loading error is the difference between the original voltage and the measured voltage, expressed as a percentage of the original voltage.

\[
\text{Error}=
\frac{10.0-8.57}{10.0}\times100
\]

\[
\boxed{\text{Error}=14.3\%}
\]

The error is significant because the voltmeter’s resistance is only twice that of the measured resistor.

The tempting assumption is that a 20.0 kΩ meter must have negligible influence on a 10.0 kΩ resistor. However, the meter is not connected in series with the resistor. It is connected in parallel, reducing the lower branch’s equivalent resistance to 6.67 kΩ.

Question 3

Two circuits use the same 5.00 V ideal supply, 100 kΩ load resistor, and 1.00 MΩ voltmeter.

In Circuit A, the load resistor is connected directly across the supply.

In Circuit B, an additional 900 kΩ resistor is placed in series between the positive supply terminal and the load resistor. The load returns directly to the negative supply terminal.

In both circuits, the voltmeter is connected across the load resistor.

Calculate the voltmeter reading in each circuit. Explain why the same voltmeter produces different readings.

Assume all resistances are exact and the supply is ideal.

Solution 3

Circuit A gives a reading of 5.00 V, while Circuit B gives approximately 0.459 V.

The difference comes from the series resistance in Circuit B.

Circuit A: Direct connection to the ideal supply

The load and voltmeter are connected directly across the ideal 5.00 V supply.

Therefore:

\[
\boxed{V_A=5.00\text{ V}}
\]

The voltmeter draws current, but the ideal source maintains the voltage.

Circuit B: Additional series resistance

The voltmeter is now connected in parallel with the 100 kΩ load.

Calculate the equivalent resistance:

\[
R_{\parallel}=
\frac{100\times1000}{100+1000}
\]

\[
R_{\parallel}=90.9\text{ k}\Omega
\]

The total resistance is:

\[
R_{\text{total}}=900+90.9
\]

\[
R_{\text{total}}=990.9\text{ k}\Omega
\]

The current supplied is:

\[
I=\frac{5.00}{990900}
\]

\[
I\approx5.05\ \mu\text{A}
\]

The voltage across the load and voltmeter is:

\[
V_B=IR_{\parallel}
\]

\[
V_B=5.05\times10^{-6}\times90900
\]

\[
\boxed{V_B\approx0.459\text{ V}}
\]

Why are the readings different?

In Circuit A, the ideal source maintains the voltage across the load, so the voltmeter cannot change that voltage.

In Circuit B, the additional 900 kΩ resistor creates a substantial resistance between the supply and the measurement points.

The voltmeter draws current through this series resistor. This produces a voltage drop across it, reducing the voltage available across the load.

Without the voltmeter, Circuit B would have a voltage of:

\[
V_{\text{unloaded}}=
5.00\times\frac{100}{900+100}
\]

\[
V_{\text{unloaded}}=0.500\text{ V}
\]

The voltmeter reduces this to approximately 0.459 V, an error of about 8.26%.

The important lesson is that the effect of a voltmeter depends on the resistance of the entire circuit supplying the measured component, not simply the resistance of that component.

Question 4

A battery has an EMF of 12.0 V and an unknown internal resistance \(r\).

Its external circuit contains two resistors in series:

• \(R_1=2.00\text{ k}\Omega\)
• \(R_2=4.00\text{ k}\Omega\)

A voltmeter with an input resistance of 8.00 kΩ is connected across \(R_2\) and reads 6.00 V.

Assuming steady DC conditions and exact component values:

a) Determine the battery’s internal resistance.

b) Calculate the voltage that an ideal voltmeter would read across \(R_2\) if the practical voltmeter were replaced by an ideal one.

Solution 4

The battery’s internal resistance is approximately 0.667 kΩ, and an ideal voltmeter would read 7.20 V.

Part a: Determine the internal resistance

First, calculate the equivalent resistance of the lower parallel branch.

\[
R_{\parallel}=
\frac{4.00\times8.00}{4.00+8.00}
\]

\[
R_{\parallel}=2.667\text{ k}\Omega
\]

The measured voltage across this branch is 6.00 V.

Using Ohm’s law, the total current entering the parallel branch is:

\[
I=\frac{6.00}{2667}
\]

\[
I=2.25\text{ mA}
\]

This is also the current through the battery’s internal resistance and the upper series resistor.

The total circuit resistance can therefore be calculated from the battery EMF:

\[
R_{\text{total}}=\frac{E}{I}
\]

\[
R_{\text{total}}=
\frac{12.0}{0.00225}
\]

\[
R_{\text{total}}=5.333\text{ k}\Omega
\]

The total resistance consists of the internal resistance, the upper resistor, and the parallel combination.

\[
5.333=r+2.00+2.667
\]

Rearranging:

\[
r=5.333-4.667
\]

\[
\boxed{r=0.667\text{ k}\Omega}
\]

Part b: Determine the ideal voltmeter reading

An ideal voltmeter has infinite resistance, so no current flows through the meter.

The lower branch now consists only of the 4.00 kΩ resistor.

The total circuit resistance becomes:

\[
R_{\text{total}}=0.667+2.00+4.00
\]

\[
R_{\text{total}}=6.667\text{ k}\Omega
\]

The new circuit current is:

\[
I=\frac{12.0}{6667}
\]

\[
I=1.80\text{ mA}
\]

The voltage across \(R_2\) is:

\[
V_2=IR_2
\]

\[
V_2=0.00180\times4000
\]

\[
\boxed{V_2=7.20\text{ V}}
\]

The ideal voltmeter reads 7.20 V, compared with the practical meter’s reading of 6.00 V.

The trap here is to assume that the measured 6.00 V represents the voltage across the 4.00 kΩ resistor before the meter was connected.

It does not.

The practical meter draws current of its own:

\[
I_v=\frac{6.00}{8000}
=0.750\text{ mA}
\]

The current through the actual 4.00 kΩ resistor is:

\[
I_2=\frac{6.00}{4000}
=1.50\text{ mA}
\]

Together, these currents give the total of 2.25 mA.

When the meter is removed, the circuit current decreases, reducing the voltage drops across the internal resistance and upper resistor. More voltage is consequently available across \(R_2\).

Question 5

A stable linear DC circuit is connected to two output terminals, A and B.

A student investigates the potential difference between these terminals using two different voltmeters.

• Voltmeter 1 has an input resistance of 10.0 MΩ and, when connected alone, reads 4.80 V.
• Voltmeter 2 has an input resistance of 1.00 MΩ and, when connected alone, reads 4.00 V.

Assume that the rest of the circuit can be represented by an ideal voltage source of EMF \(E\) in series with an effective resistance \(R_s\). The circuit does not change between measurements.

The student then connects both voltmeters simultaneously across A and B.

The student claims:

“The 10.0 MΩ voltmeter is much closer to an ideal voltmeter, so it should continue reading approximately 4.80 V. The 1.00 MΩ meter will probably give a different reading because it has a lower resistance.”

Determine:

a) The equivalent source EMF, \(E\).

b) The effective source resistance, \(R_s\).

c) The equivalent resistance of the two voltmeters when connected simultaneously.

d) The reading displayed by each voltmeter when both are connected.

Explain why the student’s prediction is incorrect.

Solution 5

The student’s prediction is incorrect. Both meters read approximately 3.93 V when connected simultaneously. The individual readings are different because each meter changes the circuit’s loading, and connecting both meters increases that loading further.

This problem requires us to work backwards from two measurements to determine the circuit’s original equivalent source.

Part a: Determine the equivalent source EMF

When a voltmeter with resistance \(R_v\) is connected across a source with EMF \(E\) and series resistance \(R_s\), the measured voltage is:

\[
V=E\frac{R_v}{R_s+R_v}
\]

Use the 1.00 MΩ voltmeter reading first.

\[
4.00=E\frac{1.00}{R_s+1.00}
\]

Rearranging:

\[
E=4.00(R_s+1.00)
\]

Here, resistance is expressed in MΩ.

Now use the 10.0 MΩ voltmeter reading:

\[
4.80=E\frac{10.0}{R_s+10.0}
\]

Substitute the expression for \(E\):

\[
4.80=
\frac{40.0(R_s+1.00)}
{R_s+10.0}
\]

Cross-multiplying:

\[
4.80(R_s+10.0)=40.0(R_s+1.00)
\]

Expanding:

\[
4.80R_s+48.0=40.0R_s+40.0
\]

\[
8.0=35.2R_s
\]

\[
R_s=0.2273\text{ M}\Omega
\]

Part b: Determine the effective source resistance

Therefore:

\[
\boxed{R_s=0.227\text{ M}\Omega}
\]

Substitute this into the earlier expression for \(E\):

\[
E=4.00(0.2273+1.00)
\]

\[
\boxed{E=4.91\text{ V}}
\]

The equivalent source has an EMF of approximately 4.91 V and an effective series resistance of 227 kΩ.

Part c: Calculate the equivalent resistance of both voltmeters

When two resistors are connected in parallel, their equivalent resistance is:

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

For the two voltmeters:

\[
R_{\parallel}=
\frac{10.0\times1.00}{10.0+1.00}
\]

\[
\boxed{R_{\parallel}=0.909\text{ M}\Omega}
\]

Notice that this resistance is smaller than either individual meter resistance.

This is the crucial part of the problem.

The two meters do not act independently. They create a single parallel combination that draws more current from the source than either meter alone.

Part d: Calculate the simultaneous reading

The two meters are connected across the same terminals, so they must measure the same potential difference.

Use the equivalent resistance of 0.909 MΩ:

\[
V_{\text{measured}}=
E\frac{R_{\parallel}}{R_s+R_{\parallel}}
\]

Substituting:

\[
V_{\text{measured}}=
4.91\frac{0.909}{0.227+0.909}
\]

\[
\boxed{V_{\text{measured}}\approx3.93\text{ V}}
\]

Both voltmeters therefore display approximately 3.93 V.

Why was the student’s prediction wrong?

The student assumed that the higher-resistance meter would continue to measure approximately the same voltage it measured when connected alone.

However, the 1.00 MΩ meter creates an additional parallel path for current. This reduces the combined meter resistance to 0.909 MΩ.

The effective resistance of the measuring instruments is now lower than either individual resistance. More current is drawn from the source, increasing the voltage drop across its 0.227 MΩ series resistance.

Consequently, the terminal voltage falls.

There is another important detail: both voltmeters must display the same reading when connected across the same two terminals. They cannot display different potential differences across identical connection points.

The original readings of 4.80 V and 4.00 V do not necessarily mean that one meter is inaccurate. They are different readings because the meters load the circuit differently.

This is a much more subtle measurement problem than simply connecting a voltmeter incorrectly.

08What to remember about voltmeter loading

The key ideas are:

• An ideal voltmeter has infinite resistance and draws no current.
• A practical voltmeter has finite input resistance and draws some current.
• A voltmeter is connected in parallel with the component being measured.
• Its resistance creates an additional parallel branch, reducing the equivalent resistance of that branch.
• In circuits with series resistance, the extra current can change voltage drops and reduce the measured potential difference.
• An ideal voltage source can maintain its voltage even when the voltmeter draws additional current.
• The loading effect becomes smaller as the voltmeter’s input resistance becomes much larger than the effective source resistance.

Finally, remember that voltmeter loading is different from an incorrect meter connection. A voltmeter connected in series instead of parallel can produce an unexpected or zero reading for a different reason. That problem is explained in How to Explain a Zero Reading from an Incorrect Meter Connection.

Once you understand loading, you can make better decisions about measurement accuracy, circuit resistance, and the assumptions behind ideal electrical instruments. Those ideas become especially useful when analysing voltage dividers and batteries with internal resistance.