How to Find Internal Resistance from a Voltage-Current Graph

A battery reads 5.80 V while supplying 0.40 A. You connect a lower-resistance load, the current rises to 1.40 A, and the voltage falls to 5.30 V. The battery hasn’t suddenly lost half a volt of electromotive force. More energy is being transferred inside it as the current increases.

How can you measure that hidden internal resistance without opening the battery?

Plot terminal voltage, \(V\), on the vertical axis against current, \(I\), on the horizontal axis. The internal resistance is the magnitude of the negative gradient:

\[ \begin{aligned} r&=-\frac{\Delta V}{\Delta I}\\ &=-\frac{5.30-5.80}{1.40-0.40}\\ &=0.50\ \Omega \end{aligned} \]

The result is \(0.50\ \Omega\). The graph’s vertical intercept gives the electromotive force (EMF), provided the battery can be modelled with approximately constant EMF and internal resistance.

This is the central method for finding internal resistance from a terminal-voltage graph. The interesting part is understanding why it works, and recognising when a graph’s gradient doesn’t measure internal resistance alone.

01Why measuring the battery voltage once isn’t enough

Imagine beginning with the simplest possible experiment.

Connect a voltmeter directly across a battery’s positive and negative terminals. Leave the external load disconnected.

The voltmeter might display 6.00 V.

What have you learned about the battery’s internal resistance?

Almost nothing. You know its approximate open-circuit voltage, but you haven’t established how that voltage changes when the battery supplies current.

For an ideal voltmeter, no current flows through the circuit. There is therefore no voltage drop across the battery’s internal resistance.

The measured terminal voltage equals the EMF in this idealised condition.

An actual voltmeter draws some current, so its reading may be slightly lower. We’ll deal with that practical complication later.

The first design gives a useful starting measurement, but it cannot determine internal resistance. We need measurements at different currents.

02Building the terminal-voltage experiment

The revised apparatus contains a battery, a switch, an ammeter, a variable resistor, a voltmeter, and connecting wires.

Arrange the main current path as follows:

1. Connect the battery’s positive terminal to the switch.
2. Connect the switch to the ammeter.
3. Connect the ammeter to the variable resistor.
4. Connect the other end of the variable resistor to the battery’s negative terminal.
5. Connect the voltmeter directly across the battery’s positive and negative terminals.

The ammeter is in series with the adjustable load. The voltmeter measures the potential difference between the battery’s external terminals.

What should changing the resistance do?

Start with a relatively high external resistance. The current is small.

Now decrease that resistance.

You should predict two changes:

• The current supplied by the battery increases.
• The terminal voltage decreases if the battery has appreciable internal resistance.

The first change follows from reducing the total circuit resistance.

The second is what allows us to measure the internal resistance.

An ideal constant-voltage source would maintain the same voltage regardless of current. Its terminal-voltage graph would be horizontal.

A real battery usually shows a decrease in terminal voltage as discharge current increases.

How to collect useful measurements

For each setting of the variable resistor, record the ammeter reading and the corresponding terminal voltage.

Keep the switch closed only as long as needed to obtain a stable reading. Changing the load can warm the battery and alter its behaviour, particularly at high currents.

Repeat measurements where practical, and use a range of currents rather than several almost identical readings.

The important experimental controls are the battery’s temperature, state of charge, and measurement arrangement. We want the current to change because the external resistance changes, not because the battery itself is changing significantly during the experiment.

03Why the graph has a negative gradient

A simple model of a real battery contains two elements connected in series:

• An ideal source with electromotive force \(\varepsilon\), measured in volts (V).
• An internal resistance \(r\), measured in ohms (\(\Omega\)).

The EMF represents the energy supplied per coulomb of charge by the source.

The internal resistance represents an effective opposition to current inside the battery. In this model, some supplied energy is dissipated internally, leaving less energy per coulomb available to the external circuit.

You can picture the internal resistance as an ordinary resistor hidden inside the battery case.

That picture is useful for circuit calculations, but it isn’t a literal description of the battery’s chemistry. Real internal losses involve several electrochemical processes, and the effective resistance can depend on operating conditions.

When conventional current leaves the battery’s positive terminal, the internal resistance produces a voltage drop \(Ir\).

Here, \(I\) is the battery’s discharge current in amperes (A).

The energy balance per unit charge is:

\[ \varepsilon=V+Ir \]

Rearrange to make the measured terminal voltage the subject:

\[ V=\varepsilon-Ir \]

Now compare this with the equation of a straight line:

\[ y=c+mx \]

For a graph with terminal voltage on the vertical axis and current on the horizontal axis:

• \(y\) corresponds to \(V\).
• \(x\) corresponds to \(I\).
• The vertical intercept \(c\) is \(\varepsilon\).
• The gradient \(m\) is \(-r\).

Therefore:

\[ \boxed{r=-m} \]

This negative sign is important. A negative gradient doesn’t mean the battery has negative resistance. The resistance is a positive quantity; the negative gradient describes the direction in which terminal voltage changes as discharge current increases.

What the graph tells you

The gradient has units of volts per ampere:

\[ \frac{\mathrm{V}}{\mathrm{A}}=\Omega \]

That’s a useful check. If your gradient is measured in \(\mathrm{A/V}\), you’ve probably reversed the axes.

The vertical intercept also needs care. It represents the extrapolated voltage at zero discharge current. It isn’t necessarily one of the measurements you actually collected.

For a more detailed explanation of what EMF means compared with terminal potential difference, see How to Distinguish EMF from Potential Difference in Circuits.

04Finding internal resistance from experimental data

05When the graph can give a misleading answer

The gradient method is reliable only when the plotted quantities and the physical model match.

Three mistakes are especially worth recognising.

1. Reversing the axes

Suppose you plot current vertically and voltage horizontally.

The battery hasn’t changed. The mathematical gradient has.

Starting with:

\[ V=\varepsilon-Ir \]

Rearrange for current:

\[ I=\frac{\varepsilon}{r}-\frac{V}{r} \]

Now the gradient is \(-1/r\), not \(-r\).

So if current is vertical and voltage is horizontal:

\[ r=-\frac{1}{m} \]

The new gradient has units of \(\mathrm{A/V}\).

Always read the axis labels before applying a gradient formula.

2. Measuring the wrong voltage

The standard method requires terminal voltage across the battery itself.

Suppose there’s an additional \(0.30\ \Omega\) resistor in the external circuit, but the voltmeter only measures voltage across a variable load beyond that resistor.

As the current increases, the reading falls because of two voltage drops:

• The battery’s internal voltage drop.
• The voltage drop across the extra external resistor.

The measured relationship becomes:

\[ V_{\text{load}}=\varepsilon-I(r+R_s) \]

Here, \(V_{\text{load}}\) is the voltage across the variable load, and \(R_s\) is the extra series resistance.

The gradient is now \(-(r+R_s)\).

Mistaking its magnitude for \(r\) overestimates the internal resistance.

This problem can’t be fixed by drawing a more accurate best-fit line. The measurement location is wrong for the quantity you’re trying to find.

3. Assuming every battery produces a straight line

A straight-line graph is expected only when the battery behaves approximately like a constant EMF in series with a constant resistance.

Real data may curve because the effective internal resistance changes with current, temperature, state of charge, or other operating conditions.

A few scattered points don’t necessarily disprove the model. Ordinary measurement uncertainty can make an otherwise linear relationship look imperfect.

But a repeated, systematic curve is different.

If the gradient becomes progressively steeper, a single resistance value may not describe the entire data range.

A useful experimental decision is to ask whether one straight line represents the points within their uncertainty.

If it does, report the effective resistance from its gradient over the tested conditions.

If it doesn’t, explain why a constant-resistance model may be insufficient. A local gradient can describe behaviour near one operating current, but it is not necessarily a fixed property of the battery.

06A contrasting case: charging reverses the voltage change

So far, we’ve treated current leaving the battery’s positive terminal as positive. That’s our discharge-current sign convention.

What if a charger forces current into the battery’s positive terminal instead?

The terminal voltage can now be greater than the battery’s EMF.

Using \(I_c\) for the positive magnitude of the charging current:

\[ V=\varepsilon+I_c r \]

The charger must supply enough voltage to overcome the battery’s EMF and the internal voltage drop.

This doesn’t require negative internal resistance.

If we keep one signed current axis, with discharge current positive and charging current negative, both behaviours follow the same equation:

\[ V=\varepsilon-Ir \]

A charging measurement appears to the left of \(I=0\), with terminal voltage above the EMF.

This is why the sign convention matters. If you plot charging-current magnitude as positive on a separate graph, its terminal voltage increases with that magnitude, producing a positive gradient.

The physical resistance remains positive.

07Practical extension: what if the meters aren’t ideal?

For ordinary HSC-style circuit calculations, meters are generally treated as ideal unless additional resistance information is supplied.

A real voltmeter has a finite input resistance, and a real ammeter adds some resistance to the path containing it.

This section is a practical measurement extension. The purpose is to understand how the instruments can affect the graph, not to suggest that the full non-ideal-meter analysis is required for every HSC question.

The voltmeter may draw current that the ammeter misses

Consider the usual arrangement: a voltmeter is connected directly across the battery terminals, while an ammeter is in the adjustable-load branch.

An ideal voltmeter draws no current.

A real voltmeter of resistance \(R_V\) draws:

\[ I_V=\frac{V}{R_V} \]

Here, \(I_V\) is the voltmeter current in amperes, and \(R_V\) is the voltmeter’s input resistance in ohms.

If the load-branch ammeter reads \(I_A\), the total battery current is:

\[ I_b=I_A+I_V \]

The battery’s internal voltage drop depends on \(I_b\), not just the ammeter’s branch reading.

Consequently:

\[ V=\varepsilon-r\left(I_A+\frac{V}{R_V}\right) \]

Rearranging gives:

\[ V=\frac{\varepsilon-rI_A}{1+r/R_V} \]

This is still a straight-line relationship between \(V\) and \(I_A\).

However, its vertical intercept is now:

\[ \frac{\varepsilon}{1+r/R_V} \]

and its gradient is:

\[ -\frac{r}{1+r/R_V} \]

Both are smaller in magnitude than the corresponding ideal-meter quantities.

That creates a surprisingly subtle problem: an excellent straight-line fit can still produce the wrong internal resistance if the horizontal axis isn’t the total battery current.

Moving the ammeter changes what the graph measures

If the ammeter measures the total current leaving the battery before the current splits between the voltmeter and the load, its reading is \(I_b\).

Provided terminal voltage is measured directly across the battery, we recover:

\[ V=\varepsilon-rI_b \]

The gradient is now \(-r\).

A non-zero ammeter resistance can still change the currents reached with particular load settings. But it doesn’t change this relationship between terminal voltage and total battery current under the constant-\(\varepsilon\), constant-\(r\) model.

In a typical digital-voltmeter measurement, the input resistance is large enough that voltmeter current is often negligible compared with the load current.

But when the battery resistance is unusually large, the meter input resistance is low, or high measurement precision is needed, instrument loading deserves attention.

08Questions and solutions

These five questions progress from reading a straightforward gradient to testing whether a perfectly straight graph really identifies the battery’s internal resistance.

Unless otherwise stated, assume steady DC conditions, negligible wire resistance, and constant battery EMF and internal resistance. Treat meters as ideal except where finite resistance is explicitly introduced.

Question 1

A battery’s terminal voltage is plotted against discharge current.

A straight line passes through the points \((0.25\ \mathrm{A},\,4.90\ \mathrm{V})\) and \((1.25\ \mathrm{A},\,4.10\ \mathrm{V})\).

Determine:

a. The battery’s internal resistance.

b. Its EMF.

c. The terminal voltage at a discharge current of 0.75 A.

Solution 1

The internal resistance is \(0.80\ \Omega\), the EMF is \(5.10\ \mathrm{V}\), and the terminal voltage at 0.75 A is \(4.50\ \mathrm{V}\).

The battery is discharging, so the relationship is \(V=\varepsilon-Ir\).

Part a: Find the gradient.

\[ \begin{aligned} m&=\frac{4.10-4.90}{1.25-0.25}\\ &=\frac{-0.80\ \mathrm{V}}{1.00\ \mathrm{A}}\\ &=-0.80\ \mathrm{V/A} \end{aligned} \]

Therefore:

\[ r=-m=0.80\ \Omega \]

Part b: Calculate the EMF.

\[ \begin{aligned} \varepsilon&=V+Ir\\ &=4.90+(0.25)(0.80)\\ &=5.10\ \mathrm{V} \end{aligned} \]

Part c: Predict the new terminal voltage.

\[ \begin{aligned} V&=\varepsilon-Ir\\ &=5.10-(0.75)(0.80)\\ &=4.50\ \mathrm{V} \end{aligned} \]

A tempting mistake is to report the gradient as \(-0.80\ \Omega\). The gradient is negative because terminal voltage decreases with increasing discharge current. The physical internal resistance is positive.

Question 2

Two battery packs, A and B, are tested separately. Their terminal-voltage graphs are straight lines with the following points:

Each battery is used separately to supply the same ohmic resistor \(R\).

a. Determine the EMF and internal resistance of each battery.

b. For \(R=2.0\ \Omega\), calculate the current and terminal voltage with each battery. Which delivers greater power to the resistor?

c. Find the positive value of \(R\) for which both batteries deliver the same power to that resistor.

d. A student argues that battery B must always supply more power because its EMF is higher. Explain why this prediction fails, and identify which battery supplies more power for resistances above the value found in part c.

Solution 2

Battery A delivers more power to a \(2.0\ \Omega\) load, despite having the lower EMF. The batteries deliver equal power when \(R=5.6\ \Omega\). For larger positive load resistances, battery B delivers more power.

Part a: Find the characteristics of each battery.

The EMFs are the vertical intercepts:

\[ \varepsilon_A=6.0\ \mathrm{V} \]

\[ \varepsilon_B=6.6\ \mathrm{V} \]

Their internal resistances come from the gradients:

\[ \begin{aligned} r_A&=-\frac{4.8-6.0}{3.0-0}\\ &=0.40\ \Omega \end{aligned} \]

\[ \begin{aligned} r_B&=-\frac{3.6-6.6}{3.0-0}\\ &=1.0\ \Omega \end{aligned} \]

Battery B has a higher EMF but also a substantially higher internal resistance.

Part b: Compare the batteries using the same load.

For a battery supplying an external resistor:

\[ I=\frac{\varepsilon}{R+r} \]

Battery A:

\[ \begin{aligned} I_A&=\frac{6.0}{2.0+0.40}\\ &=2.5\ \mathrm{A}\\ V_A&=(2.5)(2.0)\\ &=5.0\ \mathrm{V} \end{aligned} \]

Battery B:

\[ \begin{aligned} I_B&=\frac{6.6}{2.0+1.0}\\ &=2.2\ \mathrm{A}\\ V_B&=(2.2)(2.0)\\ &=4.4\ \mathrm{V} \end{aligned} \]

Power delivered to an ohmic resistor is \(P=I^2R\).

For A:

\[ P_A=(2.5)^2(2.0)=12.5\ \mathrm{W} \]

For B:

\[ P_B=(2.2)^2(2.0)=9.68\ \mathrm{W} \]

Battery A supplies more power.

Part c: Find the equal-power load.

Because both batteries are compared using the same positive resistance, equal power means equal current magnitude, and therefore equal terminal voltage.

Set their current expressions equal:

\[ \frac{6.0}{R+0.40} = \frac{6.6}{R+1.0} \]

Cross-multiply:

\[ \begin{aligned} 6.0(R+1.0)&=6.6(R+0.40)\\ 6.0R+6.0&=6.6R+2.64\\ 3.36&=0.60R\\ R&=5.6\ \Omega \end{aligned} \]

At this resistance:

\[ \begin{aligned} I_A&=\frac{6.0}{5.6+0.40}\\ &=1.0\ \mathrm{A} \end{aligned} \]

and:

\[ \begin{aligned} I_B&=\frac{6.6}{5.6+1.0}\\ &=1.0\ \mathrm{A} \end{aligned} \]

Both batteries have terminal voltage 5.6 V and deliver 5.6 W.

Part d: Explain the reversal.

Battery B begins with 0.6 V more EMF, but loses voltage internally five-halves times as quickly with increasing current because its resistance is \(1.0\ \Omega\) instead of \(0.40\ \Omega\).

For low external resistance, high current makes B’s internal voltage drop especially costly. Battery A performs better.

For large external resistance, current becomes small and the internal voltage drops become less important. Battery B’s higher EMF then gives it the advantage.

The two terminal-voltage lines intersect at 1.0 A, but that doesn’t mean we should compare the batteries at equal arbitrary currents when supplying a fixed load. Each battery finds its own operating current through the same resistor.

The intersection is relevant here because it also corresponds to a common load of \(5.6\ \Omega\).

The trap is treating EMF as though it were the voltage automatically delivered to the external resistor.

Question 3

A student investigates a battery using a variable resistor, an ideal ammeter, and an ideal voltmeter.

A fixed protective resistor of \(0.30\ \Omega\) is connected in series with the variable resistor.

Unfortunately, the voltmeter is connected only across the variable resistor, not across the battery terminals.

The student obtains these two points on a straight graph of measured load voltage against current:

a. Determine the battery’s actual internal resistance and EMF.

b. Calculate the true battery terminal voltage when the current is 2.00 A.

c. The student moves the ideal voltmeter directly across the battery terminals without changing anything else. Determine the new graph’s gradient and vertical intercept. Explain whether moving the voltmeter changes the current.

Solution 3

The battery’s actual internal resistance is \(0.50\ \Omega\), not \(0.80\ \Omega\). Its EMF is 8.00 V. At 2.00 A, the true terminal voltage is 7.00 V.

Part a: Interpret the measured gradient.

First calculate the slope of the student’s graph:

\[ \begin{aligned} m&=\frac{6.40-7.60}{2.00-0.50}\\ &=\frac{-1.20}{1.50}\\ &=-0.80\ \mathrm{V/A} \end{aligned} \]

The voltmeter measures only the variable resistor.

The battery’s EMF is shared between the internal resistance, the protective resistor, and the variable resistor.

Therefore:

\[ V_{\text{load}}=\varepsilon-I(r+R_s) \]

where \(R_s=0.30\ \Omega\).

The magnitude of the measured gradient is:

\[ \begin{aligned} 0.80&=r+0.30\\ r&=0.50\ \Omega \end{aligned} \]

To find the EMF, use either measured point:

\[ \begin{aligned} \varepsilon&=V_{\text{load}}+I(r+R_s)\\ &=7.60+(0.50)(0.80)\\ &=8.00\ \mathrm{V} \end{aligned} \]

Part b: Calculate the actual terminal voltage.

Terminal voltage includes the voltage supplied to both external resistors.

\[ \begin{aligned} V&=\varepsilon-Ir\\ &=8.00-(2.00)(0.50)\\ &=7.00\ \mathrm{V} \end{aligned} \]

The student measured only 6.40 V because the protective resistor also drops:

\[ \begin{aligned} V_s&=IR_s\\ &=(2.00)(0.30)\\ &=0.60\ \mathrm{V} \end{aligned} \]

And \(6.40+0.60=7.00\ \mathrm{V}\), so the energy-per-charge balance is consistent.

Part c: Correct the measurement arrangement.

With the voltmeter directly across the battery terminals:

\[ V=8.00-0.50I \]

The new gradient is \(-0.50\ \mathrm{V/A}\), and the intercept remains 8.00 V.

An ideal voltmeter draws zero current. Moving it does not change the external circuit’s current.

The original graph wasn’t mathematically wrong. It was a valid measurement of a different voltage. Its gradient included an additional external resistance.

Question 4

An electrochemical cell is measured twice in quick succession.

During the first measurement, an external charger drives a charging current of 0.80 A into the cell’s positive terminal. The voltmeter across the cell reads 6.40 V.

The charger is an ideal voltage source connected through a \(2.0\ \Omega\) series resistor.

The charger is then disconnected. A separate \(2.0\ \Omega\) resistor is connected directly across the cell, producing a discharge current of 2.40 A and a measured terminal voltage of 4.80 V.

Assume the cell’s EMF and internal resistance are constant across both measurements.

a. Treat discharge current as positive and charging current as negative. Determine the cell’s EMF and internal resistance using a single terminal-voltage relationship.

b. Determine the ideal charger’s EMF.

c. Explain why plotting both current magnitudes as positive would give a misleading gradient. Identify one physical reason the two measurements might not follow the same straight line in a real experiment.

Solution 4

The cell has an EMF of 6.00 V and an internal resistance of \(0.50\ \Omega\). The charger’s EMF is 8.00 V.

Both voltage readings fit one straight-line model once the direction of current is represented correctly.

Part a: Use signed current rather than current magnitude.

Our convention is:

• Positive \(I\): discharge current leaving the cell’s positive terminal.
• Negative \(I\): charging current entering the cell’s positive terminal.

Therefore the two graph coordinates are:

\[ (-0.80\ \mathrm{A},\ 6.40\ \mathrm{V}) \]

and:

\[ (2.40\ \mathrm{A},\ 4.80\ \mathrm{V}) \]

Calculate the gradient:

\[ \begin{aligned} m&=\frac{4.80-6.40}{2.40-(-0.80)}\\ &=\frac{-1.60}{3.20}\\ &=-0.50\ \mathrm{V/A} \end{aligned} \]

So:

\[ r=0.50\ \Omega \]

Use the discharging measurement to find EMF:

\[ \begin{aligned} \varepsilon&=V+Ir\\ &=4.80+(2.40)(0.50)\\ &=6.00\ \mathrm{V} \end{aligned} \]

Check the charging measurement:

\[ \begin{aligned} V&=\varepsilon-Ir\\ &=6.00-(-0.80)(0.50)\\ &=6.40\ \mathrm{V} \end{aligned} \]

It agrees exactly.

Part b: Calculate the charger EMF.

The charger must overcome both the cell’s terminal voltage while charging and the voltage drop across the external \(2.0\ \Omega\) resistor.

That resistor’s voltage drop is:

\[ \begin{aligned} V_R&=I_cR\\ &=(0.80)(2.0)\\ &=1.60\ \mathrm{V} \end{aligned} \]

Therefore, with \(E_c\) representing the charger’s EMF:

\[ \begin{aligned} E_c&=V+V_R\\ &=6.40+1.60\\ &=8.00\ \mathrm{V} \end{aligned} \]

Part c: Explain why the sign matters.

If both current magnitudes were plotted as positive, the points would be \((0.80,6.40)\) and \((2.40,4.80)\).

Their apparent gradient would be:

\[ \frac{-1.60}{1.60} =-1.0\ \mathrm{V/A} \]

That would incorrectly suggest an internal resistance of \(1.0\ \Omega\).

The mistake comes from treating two opposite current directions as though they were the same physical condition.

During discharge, the internal voltage drop reduces terminal voltage below EMF. During charging, the required terminal voltage lies above EMF.

In a real experiment, rapid charging and discharging can also change the cell’s internal electrochemical state. Heating, polarisation, and other transient effects may cause the effective parameters to differ between measurements.

The straight-line construction is valid here because the question explicitly assumes constant EMF and internal resistance.

Question 5

A battery is tested using two different ammeter positions.

A voltmeter with finite input resistance \(R_V\) stays connected directly across the battery’s terminals throughout both experiments.

The external load is varied, and an ideal ammeter is placed either:

• In the shared battery lead, where it measures total current before the current divides between the voltmeter and load.
• In the adjustable-load branch alone, where it does not measure voltmeter current.

The battery has constant EMF \(\varepsilon\) and internal resistance \(r\).

Both experiments produce excellent straight-line fits of terminal voltage against the ammeter reading \(I_A\):

Graph X:

\[ V=9.0-3.0I_A \]

Graph Y:

\[ V=7.2-2.4I_A \]

Here, voltage is measured in volts and current in amperes. Both experiments involve discharge, the same voltmeter is used throughout, and all instruments except the finite-resistance voltmeter are ideal.

a. Identify which ammeter position produced each graph. Determine the actual battery EMF, internal resistance, and voltmeter input resistance.

b. During the experiment represented by graph Y, the ammeter reads 1.0 A. Determine the terminal voltage, voltmeter current, and total battery current. Show that these results are also consistent with graph X.

c. Suppose graph Y were the only available experimental evidence. Could the battery EMF, internal resistance, and voltmeter resistance all be determined uniquely? Demonstrate your conclusion by constructing one different physically possible combination of these three quantities that produces exactly graph Y.

d. The voltmeter’s input resistance is doubled, and the ammeter remains in the position used for graph Y. Predict the new vertical intercept and gradient. Explain why the magnitude of the gradient changes in that direction, even though the voltmeter is now less disruptive to the circuit.

Solution 5

Graph X comes from the ammeter measuring total battery current. Graph Y comes from the ammeter measuring only load-branch current.

The battery’s actual EMF is 9.0 V, its internal resistance is \(3.0\ \Omega\), and the voltmeter resistance is \(12\ \Omega\).

Graph Y alone cannot determine these three quantities uniquely. Doubling the voltmeter resistance changes its fitted intercept to 8.0 V and its gradient to approximately \(-2.67\ \mathrm{V/A}\).

The reason is that voltmeter current affects both the apparent slope and the intercept.

Part a: Identify the true battery characteristic.

When the ammeter measures total battery current, its reading includes any current drawn by the voltmeter.

The battery obeys:

\[ V=\varepsilon-rI_A \]

Thus graph X gives:

\[ \varepsilon=9.0\ \mathrm{V} \]

and:

\[ r=3.0\ \Omega \]

Now consider the load-branch arrangement.

The battery supplies both load current and voltmeter current:

\[ I_b=I_A+\frac{V}{R_V} \]

Substitute into the battery equation:

\[ V=9.0-3.0\left(I_A+\frac{V}{R_V}\right) \]

Rearrange:

\[ V\left(1+\frac{3.0}{R_V}\right) =9.0-3.0I_A \]

Therefore:

\[ V=\frac{9.0-3.0I_A}{1+3.0/R_V} \]

The vertical intercept of graph Y is 7.2 V, so:

\[ \begin{aligned} 7.2&=\frac{9.0}{1+3.0/R_V}\\ 1+\frac{3.0}{R_V}&=1.25\\ \frac{3.0}{R_V}&=0.25\\ R_V&=12\ \Omega \end{aligned} \]

Check that this also produces graph Y’s gradient:

\[ \begin{aligned} m_Y&=-\frac{3.0}{1+3.0/12}\\ &=-\frac{3.0}{1.25}\\ &=-2.4\ \mathrm{V/A} \end{aligned} \]

Both its gradient and intercept are explained by the same finite voltmeter resistance.

Graph Y therefore represents the ammeter in the load branch.

Part b: Check the currents and voltage.

For graph Y, let \(I_A=1.0\ \mathrm{A}\).

The terminal voltage is:

\[ \begin{aligned} V&=7.2-2.4(1.0)\\ &=4.8\ \mathrm{V} \end{aligned} \]

The voltmeter current is:

\[ \begin{aligned} I_V&=\frac{V}{R_V}\\ &=\frac{4.8}{12}\\ &=0.40\ \mathrm{A} \end{aligned} \]

Total battery current is:

\[ \begin{aligned} I_b&=I_A+I_V\\ &=1.0+0.40\\ &=1.40\ \mathrm{A} \end{aligned} \]

Now use graph X with that total battery current:

\[ \begin{aligned} V&=9.0-3.0(1.40)\\ &=9.0-4.2\\ &=4.8\ \mathrm{V} \end{aligned} \]

The results agree.

The battery hasn’t changed its internal resistance between runs. The two ammeter positions measure different currents in the same physical circuit.

Part c: Test whether graph Y alone identifies the battery.

No. Graph Y alone does not uniquely determine \(\varepsilon\), \(r\), and \(R_V\).

Its general equation is:

\[ V=\frac{\varepsilon-rI_A}{1+r/R_V} \]

The observed intercept gives:

\[ 7.2=\frac{\varepsilon}{1+r/R_V} \]

The observed gradient gives:

\[ 2.4=\frac{r}{1+r/R_V} \]

These are two relationships involving three unknown quantities.

To prove the ambiguity, construct a second model.

Suppose:

\[ \begin{aligned} \varepsilon&=12\ \mathrm{V}\\ r&=4.0\ \Omega\\ R_V&=6.0\ \Omega \end{aligned} \]

Its apparent vertical intercept is:

\[ \begin{aligned} c&=\frac{12}{1+4/6}\\ &=7.2\ \mathrm{V} \end{aligned} \]

Its apparent gradient is:

\[ \begin{aligned} m&=-\frac{4.0}{1+4/6}\\ &=-2.4\ \mathrm{V/A} \end{aligned} \]

So this completely different battery-and-voltmeter combination produces exactly the same graph Y:

\[ V=7.2-2.4I_A \]

There is no measurement noise to blame. The two different models are mathematically indistinguishable from this one graph.

Graph X supplies the missing information because measuring total battery current removes the unknown contribution of the voltmeter from the horizontal-axis reading.

Part d: Double the voltmeter resistance.

The original input resistance is \(12\ \Omega\).

Doubling it gives:

\[ R_V'=24\ \Omega \]

The battery itself still has:

\[ \varepsilon=9.0\ \mathrm{V}, \qquad r=3.0\ \Omega \]

For the load-branch ammeter arrangement:

\[ V=\frac{9.0-3.0I_A}{1+3.0/24} \]

The denominator is:

\[ 1+\frac{3.0}{24}=1.125 \]

So the new vertical intercept is:

\[ \begin{aligned} c'&=\frac{9.0}{1.125}\\ &=8.0\ \mathrm{V} \end{aligned} \]

The new gradient is:

\[ \begin{aligned} m'&=-\frac{3.0}{1.125}\\ &\approx-2.67\ \mathrm{V/A} \end{aligned} \]

The new graph is approximately:

\[ V=8.0-2.67I_A \]

There are two connected effects.

First, increasing the voltmeter resistance reduces the current drawn by the voltmeter. Even when the load-branch ammeter reads zero, the voltmeter still draws some current. Reducing that current moves the apparent vertical intercept closer to the true EMF of 9.0 V.

Second, because less current is hidden in the voltmeter branch, the load-branch ammeter becomes a more accurate measure of total battery current.

The apparent gradient therefore moves from \(-2.4\ \mathrm{V/A}\) towards the true gradient of \(-3.0\ \mathrm{V/A}\).

Its magnitude increases even though measurement interference decreases.

That is the conceptual trap. A gentler slope isn’t always evidence of a smaller internal resistance, and a perfectly straight graph isn’t proof that the measured current is the correct one.

09What to do with the gradient after you’ve found it

The internal resistance measured from a terminal-voltage graph tells you how much extra terminal voltage the battery loses for each additional ampere of discharge current.

But that number is useful only alongside the conditions under which it was obtained.

A good experimental conclusion should state the fitted resistance in ohms, the EMF in volts, the current range tested, and whether the constant-resistance model adequately describes the measurements.

If the graph is approximately linear and the voltmeter measures the actual battery terminals, you can use:

\[ r=-\frac{\Delta V}{\Delta I} \]

You can then predict how terminal voltage changes when the external load changes.

The next useful question is where the unavailable electrical energy goes. Once the internal resistance and current are known, you can calculate the power dissipated internally using \(P_{\text{internal}}=I^2r\), where \(P_{\text{internal}}\) is measured in watts (W).

That leads directly to How to Calculate Energy Dissipated Inside a Battery, where the voltage drop becomes a measurable energy loss over time.