Why Battery Terminal Voltage Falls When Current Increases

A battery reads 6.0 V when nothing is connected to it. Switch on a lamp, and the voltage measured across the same battery drops to 5.0 V. Connect a device that draws even more current, and the voltage falls further. Why should the battery deliver less voltage just when the circuit demands more current?

The answer is internal resistance. A real battery transfers some of its energy inside itself as current flows. The greater the current, the greater the internal voltage drop, leaving a smaller potential difference between the battery’s external terminals.

This is called terminal voltage drop, or voltage sag. It explains why a battery can appear healthy when disconnected but struggle to power a demanding device.

01Why the battery’s voltage depends on current

A battery supplies energy, not a fixed voltage under every condition

Imagine a battery connected to a small motor.

Inside the battery, chemical reactions provide energy to electric charges. Those charges transfer energy to the motor as they move through the external circuit.

Voltage measures energy transferred per unit charge. A battery with an electromotive force (EMF) of 6.0 V supplies 6.0 joules of energy per coulomb of charge under the conditions described by its EMF.

But does all that energy reach the motor?

Not necessarily.

A real battery contains electrodes, electrolyte, and other materials involved in moving charge. These processes are not perfectly efficient. Some energy is transferred internally rather than being delivered to the external circuit.

One useful comparison is a long charging cable.

When a phone draws current through a cable with electrical resistance, some voltage is lost along the cable. Increase the current, and the voltage drop across that resistance increases.

A battery behaves similarly, except the effective resistance is associated with processes inside the battery rather than an external cable.

The analogy has a limit: internal resistance isn’t necessarily a separate physical resistor hidden inside the battery. It is a simplified way of representing several electrochemical and electrical effects.

The equivalent circuit of a real battery

We can represent a real battery using two components connected in series:

• An ideal source of EMF, which supplies energy to charges.
• An internal resistance, which accounts for voltage lost inside the battery.

Suppose conventional current flows out of the battery’s positive terminal.

The chemical source provides energy per coulomb. Some is transferred internally, and the rest is available to the external circuit.

Conservation of energy gives:

\[ \varepsilon=V_t+Ir \]

Rearranging:

\[ V_t=\varepsilon-Ir \]

Here:

• \(\varepsilon\) is the battery’s EMF, measured in volts (V).
• \(V_t\) is its terminal potential difference, measured in volts (V).
• \(I\) is the current leaving the positive terminal, measured in amperes (A).
• \(r\) is the internal resistance, measured in ohms (\(\Omega\)).
• \(Ir\) is the voltage drop associated with internal resistance, measured in volts (V).

The equation assumes a steady discharging current, with approximately constant EMF and internal resistance.

Notice what happens when current increases.

If \(r\) stays constant, the internal voltage drop \(Ir\) becomes larger. Because the source still supplies approximately the same EMF, less voltage remains across the external terminals.

For example, if \(r=0.40\ \Omega\), increasing the current from 1.0 A to 2.0 A increases the internal voltage drop from 0.40 V to 0.80 V.

The terminal voltage therefore falls by 0.40 V.

Crucially, current isn’t being used up inside the battery. In the simple series model, the same current passes through the internal resistance and the external load. What changes is the energy transferred per coulomb.

For a fuller distinction between the source’s EMF and a measured voltage, see How to Distinguish EMF from Potential Difference in Circuits.

02The experiment: why does the voltmeter reading fall?

Suppose we’re investigating a battery whose voltage appears to decrease when a lamp is connected.

The first question is whether the battery really produces a smaller terminal potential difference at higher currents, or whether something else in the circuit is responsible.

Step 1: Build a circuit that changes the current

Start with a battery, a switch, an ammeter, and an adjustable resistance.

Connect the battery’s positive terminal to the switch, then to the ammeter, then to the adjustable resistor. Connect the other end of the resistor back to the negative terminal.

This forms one complete series path when the switch is closed.

Now connect a voltmeter directly across the battery’s two external posts.

The meters have different jobs.

The ammeter measures the current in the external load branch. The voltmeter measures the potential difference between the battery terminals.

For the main investigation, assume ideal meters: the ammeter has zero resistance, and the voltmeter draws no current.

With the switch open, no current flows in the ideal-meter model.

Therefore:

\[ V_t=\varepsilon \]

The open-circuit reading gives the battery’s EMF in this simplified model.

Step 2: Predict what an ideal battery would do

Imagine the battery had no internal resistance.

You close the switch and gradually decrease the external resistance.

What should happen to the voltage across the battery?

For an ideal source of constant EMF, the terminal voltage would remain constant, even as current increased.

Current would change because the external resistance changed, but the battery’s terminal potential difference would not.

Now replace the ideal battery with a real one.

The internal-resistance model predicts a different pattern: decreasing the external resistance increases the current, which increases the internal voltage drop and reduces the terminal voltage.

Step 3: Compare current and terminal voltage

The following table contains illustrative data, calculated for a hypothetical battery with an EMF of 6.00 V and an internal resistance of 0.50 \(\Omega\). These are not measurements from a historical or performed experiment.

The resistance values represent the entire external load, with ideal meters and negligible connecting-wire resistance.

As the current rises from 0.50 A to 1.50 A, the terminal voltage falls from 5.75 V to 5.25 V.

The battery hasn’t necessarily lost a significant amount of charge during these brief readings. Its terminal voltage has changed because the current through its internal resistance has changed.

The final reading is important.

When the external resistance returns to 11.5 \(\Omega\), the current returns to 0.50 A, and the terminal voltage returns to 5.75 V.

This is exactly what the constant-EMF, constant-internal-resistance model predicts.

In a real experiment, returning to an earlier load is a useful check for drift, heating, and changes in the battery during the measurements.

Step 4: Turn the measurements into a graph

Plot terminal voltage \(V_t\) on the vertical axis and current \(I\) on the horizontal axis.

The equation is:

\[ V_t=\varepsilon-rI \]

Compare it with the straight-line relationship \(y=c+mx\).

The vertical intercept represents the EMF, while the gradient represents the negative of the internal resistance.

Calculate the gradient using two points on the line:

\[ \begin{aligned} m&=\frac{\Delta V_t}{\Delta I}\\ &=\frac{5.25-5.75}{1.50-0.50}\\ &=-0.50\ \Omega \end{aligned} \]

The negative sign tells us that voltage decreases as current increases.

Since \(m=-r\):

\[ r=0.50\ \Omega \]

The vertical intercept gives:

\[ \varepsilon=6.00\ \mathrm{V} \]

We have used the relationship between current and terminal voltage to estimate a property of the battery that cannot be read directly from a voltmeter: its effective internal resistance.

Why a single reading isn’t enough

Suppose we had measured only the voltage across a lamp and found it was lower than expected.

That observation alone wouldn’t establish that the missing voltage was lost inside the battery.

The connecting wires, switches, and electrical contacts may also have resistance. A voltmeter connected across a distant lamp measures the voltage at the lamp, not necessarily the voltage at the battery posts.

Likewise, if the battery changes during a long experiment, a falling voltage might partly reflect a change in its EMF or effective internal resistance.

A stronger experimental design therefore uses several current settings, measures directly at the battery terminals, and repeats an earlier setting to test whether the readings are reversible.

In an actual investigation, keep the loads within the battery’s and equipment’s safe operating limits, avoid short circuits, and minimise unnecessary discharge and heating.

Even a straight-line graph doesn’t prove that internal resistance is perfectly constant under every condition. It shows that a constant-\(r\) model describes the measurements over the tested range.

05When the terminal voltage does not follow the usual prediction

The rule that terminal voltage falls as current increases assumes the battery is discharging and its EMF and internal resistance remain approximately constant.

There are three useful contrasting cases.

Case 1: An ideal battery

For an ideal battery:

\[ r=0 \]

Therefore:

\[ V_t=\varepsilon \]

Its terminal voltage is independent of current, provided its EMF is constant.

This ideal model is useful when internal resistance is negligible compared with the external circuit resistance.

Case 2: A battery being charged

Suppose an external power supply forces current into the positive terminal of a rechargeable battery.

The direction of current through the battery has reversed.

We defined positive current as current leaving the battery’s positive terminal. During charging, that signed current is negative.

If \(I_c\) is the positive magnitude of the charging current, the same simple battery model gives:

\[ V_t=\varepsilon+I_cr \]

The terminal voltage is now greater than the EMF.

For example, consider a battery with an EMF of 6.0 V and internal resistance of 0.50 \(\Omega\).

If it is being charged at 1.0 A:

\[ \begin{aligned} V_t&=6.0+(1.0)(0.50)\\ &=6.5\ \mathrm{V} \end{aligned} \]

The external supply must provide sufficient terminal potential difference to drive current into the battery.

So the statement “a real battery’s terminal voltage is always lower than its EMF” is false.

It is lower during discharge and higher during charging in this constant-\(\varepsilon\), constant-\(r\) model.

Case 3: The battery itself changes during use

An actual battery doesn’t necessarily have constant EMF or internal resistance.

Its terminal response can depend on temperature, state of charge, current history, and electrochemical processes.

For instance, a battery may show an immediate voltage drop when a heavy load is connected, then a partial voltage recovery after the load is removed.

Some of this behaviour may be represented by internal resistance. Slower recovery and time-dependent behaviour can require a more detailed model.

An immediate drop under load and a gradual decrease during prolonged use aren’t necessarily the same phenomenon.

You should therefore avoid treating every falling voltage reading as evidence that internal resistance has increased.

06Practical extension: real voltmeters and ammeters

The experiment above assumed ideal measuring instruments.

That is usually an appropriate first model for standard circuit calculations. In a careful practical investigation, however, the meters themselves can affect the circuit.

A voltmeter draws some current

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

A real voltmeter has a finite input resistance, \(R_V\), measured in ohms.

When connected across the battery posts, it provides another current path.

Even when the external load switch is open, current may flow through the voltmeter.

Consider an illustrative battery with:

\[ \varepsilon=6.0\ \mathrm{V}, \qquad r=0.50\ \Omega \]

Connect a voltmeter with an intentionally low input resistance of 100 \(\Omega\), and leave every other external load disconnected.

The voltmeter current is:

\[ \begin{aligned} I_V&=\frac{\varepsilon}{R_V+r}\\ &=\frac{6.0}{100+0.50}\\ &=0.0597\ \mathrm{A} \end{aligned} \]

The voltage drop inside the battery is:

\[ \begin{aligned} I_Vr&=(0.0597)(0.50)\\ &=0.0299\ \mathrm{V} \end{aligned} \]

The voltmeter therefore reads approximately:

\[ V_t=6.0-0.0299 =5.97\ \mathrm{V} \]

It doesn’t read exactly 6.00 V because it draws current.

A voltmeter with much greater input resistance would produce a far smaller effect.

An ammeter also changes the circuit

An ideal ammeter has zero resistance.

A real ammeter has some resistance, \(R_A\), which contributes to the resistance of the branch where it is connected.

This can reduce the current compared with the circuit before the ammeter was inserted.

A particularly important distinction is that an ammeter placed only in the load branch does not measure current passing through a separate voltmeter branch.

At the battery’s positive terminal:

\[ I_{\text{battery}}=I_A+I_V \]

where \(I_A\) is the load-branch current and \(I_V\) is the voltmeter current.

Using \(I_A\) in the internal-resistance equation when it doesn’t represent the total battery current can give incorrect estimates of EMF and internal resistance.

These nonideal-meter effects are practical extensions. Unless the problem supplies instrument resistances or explicitly asks about measurement limitations, start with ideal meters.

07A decision process for falling voltage readings

When you see a battery voltage decrease in an HSC-style question, don’t immediately assume the battery is running flat.

First identify exactly what changed and what was measured.

One final relationship ties the experiment to the external circuit.

For a battery supplying a single resistor \(R\), with negligible other external resistance:

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

and:

\[ V_t=IR \]

Therefore:

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

This expression explains two observations at once.

When \(R\) is much larger than \(r\), terminal voltage is close to the EMF.

When \(R\) becomes comparable with \(r\), a larger fraction of the source’s voltage is lost internally.

The important comparison is not simply whether current is “large”. It is how significant the internal resistance is relative to the rest of the circuit.

08Questions and solutions

The questions increase in difficulty. The final two require you to distinguish external load current from the current through an individual battery.

Question 1

A cell has an EMF of 7.2 V and internal resistance of 0.80 \(\Omega\).

It supplies a steady current of 1.5 A to a single external resistor.

Assume the connecting wires have negligible resistance.

(a) Calculate the terminal potential difference.

(b) Determine the resistance of the external resistor.

(c) Calculate the energy transferred internally per coulomb of charge.

Solution 1

The terminal potential difference is 6.0 V, the external resistance is 4.0 \(\Omega\), and the internal energy transfer is 1.2 J per coulomb.

For part (a), use the terminal-voltage equation:

\[ \begin{aligned} V_t&=\varepsilon-Ir\\ &=7.2-(1.5)(0.80)\\ &=7.2-1.2\\ &=6.0\ \mathrm{V} \end{aligned} \]

For part (b), the external resistor receives the entire terminal potential difference.

Using Ohm’s law:

\[ \begin{aligned} R&=\frac{V_t}{I}\\ &=\frac{6.0}{1.5}\\ &=4.0\ \Omega \end{aligned} \]

For part (c), the internal voltage drop is:

\[ Ir=(1.5)(0.80)=1.2\ \mathrm{V} \]

Because one volt represents one joule per coulomb, the internal energy transfer is:

\[ 1.2\ \mathrm{J/C} \]

The tempting mistake is to calculate the external resistance using \(7.2/1.5\). That would incorrectly assume the load receives the full EMF. The correct external voltage is 6.0 V.

Question 2

A battery with an EMF of 12.0 V and internal resistance of 2.0 \(\Omega\) powers a lamp.

The lamp is approximated as an ohmic resistor of constant resistance 10.0 \(\Omega\).

A second resistor of 5.0 \(\Omega\) is connected in parallel with the lamp using a switch. All wires and switches have negligible resistance.

(a) Determine the battery current, terminal voltage, and lamp power before the second resistor is connected.

(b) Determine the total battery current, terminal voltage, current through the original lamp, and lamp power after the switch closes.

(c) A student argues that adding a parallel branch shouldn’t change the original lamp because parallel components share the same voltage. Explain the flaw in this prediction.

(d) State what would happen to the original lamp’s power before and after adding the branch if the battery were replaced by an ideal 12.0 V source.

Solution 2

The original lamp’s power falls from 10.0 W to 5.63 W when the second branch is connected, despite the battery supplying more total current. An ideal 12.0 V source would not cause this decrease.

For part (a), initially only the 10.0 \(\Omega\) lamp is connected.

The total resistance is:

\[ R_{\text{total}}=10.0+2.0 =12.0\ \Omega \]

Battery current:

\[ \begin{aligned} I&=\frac{12.0}{12.0}\\ &=1.0\ \mathrm{A} \end{aligned} \]

Terminal voltage:

\[ \begin{aligned} V_t&=12.0-(1.0)(2.0)\\ &=10.0\ \mathrm{V} \end{aligned} \]

Lamp power is:

\[ \begin{aligned} P&=\frac{V_t^2}{R}\\ &=\frac{10.0^2}{10.0}\\ &=10.0\ \mathrm{W} \end{aligned} \]

Here \(P\) is the rate of energy transfer, measured in watts (W).

For part (b), closing the switch connects the 10.0 \(\Omega\) and 5.0 \(\Omega\) resistors in parallel.

Their equivalent resistance \(R_{\text{eq}}\) is:

\[ \begin{aligned} \frac{1}{R_{\text{eq}}} &=\frac{1}{10.0}+\frac{1}{5.0}\\ &=0.30\ \Omega^{-1} \end{aligned} \]

Therefore:

\[ R_{\text{eq}}=\frac{10}{3}\ \Omega \]

This equivalent resistance is in series with the battery’s internal resistance.

The new battery current is:

\[ \begin{aligned} I_{\text{total}} &=\frac{12.0}{2.0+10/3}\\ &=2.25\ \mathrm{A} \end{aligned} \]

The terminal voltage becomes:

\[ \begin{aligned} V_t&=12.0-(2.25)(2.0)\\ &=7.50\ \mathrm{V} \end{aligned} \]

Both parallel branches receive this voltage.

Current through the original lamp:

\[ \begin{aligned} I_{\text{lamp}}&=\frac{7.50}{10.0}\\ &=0.750\ \mathrm{A} \end{aligned} \]

Current through the additional resistor:

\[ I_{\text{new}}=\frac{7.50}{5.0} =1.50\ \mathrm{A} \]

Their sum is 2.25 A, confirming the total current.

The original lamp’s new power is:

\[ \begin{aligned} P_{\text{lamp}} &=\frac{7.50^2}{10.0}\\ &=5.625\ \mathrm{W}\\ &\approx5.63\ \mathrm{W} \end{aligned} \]

For part (c), the student’s statement about parallel components sharing voltage is correct, but the prediction assumes the source voltage is fixed.

The battery current has increased from 1.0 A to 2.25 A.

The internal voltage drop has consequently increased from 2.0 V to 4.5 V.

The terminal voltage falls, so both parallel branches receive less voltage than the original lamp received before the switch closed.

The second branch hasn’t simply taken some of a fixed current away from the lamp. It has changed the total load, causing the nonideal battery’s terminal voltage to fall.

For part (d), an ideal source with zero internal resistance would maintain 12.0 V across the lamp in both arrangements.

The lamp would therefore have constant power:

\[ \begin{aligned} P_{\text{ideal}} &=\frac{12.0^2}{10.0}\\ &=14.4\ \mathrm{W} \end{aligned} \]

Its power would not change when the second branch was connected.

The contrast isolates the cause: the original lamp becomes less powerful because of the battery’s internal resistance, not because parallel circuits inherently reduce the voltage of existing branches.

Question 3

A student investigates a battery using a variable external resistor.

The student first records readings quickly, changing from a low current to a high current and immediately returning to the original low current.

After the battery has supplied current for an extended period, the student takes another pair of readings.

The following data are illustrative.

Assume the meters are ideal and connected correctly, and that EMF and internal resistance remain approximately constant during each short set of readings, although they may differ between the early and later sets.

(a) Determine the EMF and internal resistance during the early measurements.

(b) Determine the EMF and internal resistance during the later measurements.

(c) A student concludes that the later voltage readings are lower because the battery’s internal resistance has increased. Evaluate the conclusion.

(d) Explain why returning to the original current during the early measurements was a useful experimental control.

Solution 3

The early measurements give \(\varepsilon=6.00\ \mathrm{V}\) and \(r=0.40\ \Omega\). The later measurements give \(\varepsilon=5.80\ \mathrm{V}\) and the same \(r=0.40\ \Omega\). Under the stated model, the later voltage shift is caused by a lower EMF, not an increase in internal resistance.

For part (a), calculate the internal resistance from the early pair:

\[ \begin{aligned} r&=\frac{5.80-5.40}{1.50-0.50}\\ &=\frac{0.40}{1.00}\\ &=0.40\ \Omega \end{aligned} \]

Then find the EMF:

\[ \begin{aligned} \varepsilon&=V_t+Ir\\ &=5.80+(0.50)(0.40)\\ &=6.00\ \mathrm{V} \end{aligned} \]

For part (b), use the later measurements:

\[ \begin{aligned} r&=\frac{5.60-5.20}{1.50-0.50}\\ &=\frac{0.40}{1.00}\\ &=0.40\ \Omega \end{aligned} \]

The later EMF is:

\[ \begin{aligned} \varepsilon&=5.60+(0.50)(0.40)\\ &=5.80\ \mathrm{V} \end{aligned} \]

For part (c), the conclusion that internal resistance increased is not supported.

Both measurement sets show a terminal-voltage decrease of 0.40 V for a current increase of 1.00 A.

Their voltage-current gradients are identical.

The entire later relationship is shifted down by 0.20 V, meaning the inferred EMF is 0.20 V smaller.

If internal resistance alone had increased while EMF remained constant, the voltage-current graph would have become steeper rather than simply shifting downward.

This conclusion depends on the constant-parameter model within each measurement set. A real battery could show more complicated behaviour outside the measured range.

For part (d), the early return to 0.50 A produces the original 5.80 V reading.

This supports the interpretation that the short-term voltage change was caused by changing current, rather than a continuing loss of EMF during those particular measurements.

Without the repeated reading, the falling voltage might have been confused with a change occurring over time.

The important lesson is that a single low-voltage reading cannot tell you whether the cause is a lower EMF, a larger internal resistance, or another measurement condition.

Question 4

A real battery has an EMF of 6.0 V and internal resistance of 1.0 \(\Omega\).

A voltmeter with input resistance \(R_V=5.0\ \Omega\) is connected directly across the battery terminals.

A separate external branch contains a 5.0 \(\Omega\) resistor, an ideal ammeter, and a switch, all in series. This branch is connected across the same battery terminals.

The ammeter measures current only through the external resistor branch.

Initially, the switch is open.

(a) Determine the voltmeter and ammeter readings, and the total current supplied by the battery.

The switch is then closed.

(b) Determine the new terminal voltage, the ammeter reading, and the total battery current.

(c) A student plots the two voltmeter readings against the corresponding ammeter readings. They use the intercept and gradient to estimate the battery’s EMF and internal resistance. Determine their apparent estimates.

(d) Explain precisely why those estimates differ from the battery’s actual EMF and internal resistance.

Assume all resistances and the EMF remain constant.

Solution 4

With the switch open, the voltmeter reads 5.00 V while the ammeter reads zero, even though the battery supplies 1.00 A. With the switch closed, the voltmeter reads approximately 4.29 V, the ammeter reads 0.857 A, and the battery supplies 1.71 A. Using the load-branch ammeter readings gives incorrect estimates of 5.00 V for EMF and 0.833 \(\Omega\) for internal resistance.

For part (a), when the switch is open, no current can flow through the 5.0 \(\Omega\) external resistor branch.

Therefore:

\[ I_A=0\ \mathrm{A} \]

But the voltmeter is still connected across the battery and provides a complete current path.

The total resistance seen by the EMF source is:

\[ R_{\text{total}}=1.0+5.0 =6.0\ \Omega \]

So battery current is:

\[ \begin{aligned} I_{\text{battery}} &=\frac{6.0}{6.0}\\ &=1.00\ \mathrm{A} \end{aligned} \]

The voltmeter reading is the potential difference across its own 5.0 \(\Omega\) resistance:

\[ \begin{aligned} V_t&=(1.00)(5.0)\\ &=5.00\ \mathrm{V} \end{aligned} \]

For part (b), closing the switch creates two external parallel branches, each with resistance 5.0 \(\Omega\).

Their equivalent resistance is:

\[ \begin{aligned} R_{\text{eq}} &=\frac{(5.0)(5.0)}{5.0+5.0}\\ &=2.50\ \Omega \end{aligned} \]

The battery supplies current through its 1.0 \(\Omega\) internal resistance and this equivalent external resistance.

Therefore:

\[ \begin{aligned} I_{\text{battery}} &=\frac{6.0}{1.0+2.50}\\ &=1.714\ \mathrm{A} \end{aligned} \]

The terminal potential difference becomes:

\[ \begin{aligned} V_t&=(1.714)(2.50)\\ &=4.286\ \mathrm{V} \end{aligned} \]

The two parallel external branches have the same voltage.

Current through the load branch is:

\[ \begin{aligned} I_A&=\frac{4.286}{5.0}\\ &=0.857\ \mathrm{A} \end{aligned} \]

The voltmeter branch also carries approximately 0.857 A.

These currents add to give the battery current:

\[ I_{\text{battery}}=1.714\ \mathrm{A} \]

For part (c), the student’s two plotted points are approximately:

\[ (0,\ 5.00) \quad\text{and}\quad (0.857,\ 4.286) \]

where the first coordinate is the ammeter reading in amperes and the second is the terminal voltage in volts.

The extrapolated vertical intercept is 5.00 V.

The apparent magnitude of the gradient is:

\[ \begin{aligned} r_{\text{apparent}} &=\frac{5.00-4.286}{0.857-0}\\ &\approx0.833\ \Omega \end{aligned} \]

So the student incorrectly estimates:

\[ \varepsilon_{\text{apparent}}=5.00\ \mathrm{V} \]

and:

\[ r_{\text{apparent}}=0.833\ \Omega \]

For part (d), the mistake is treating the ammeter reading as though it were the total current through the battery.

The correct battery equation is:

\[ V_t=\varepsilon-I_{\text{battery}}r \]

But the ammeter measures only the external resistor current.

Even when the ammeter reads zero, current still flows through the voltmeter. This causes a voltage drop inside the battery.

Consequently, the apparent zero-current intercept is not the true EMF.

As the load current increases, the voltmeter current changes too. The measured load-branch current therefore does not capture the full change in battery current.

The surprising result is that the measured voltage-current line can look perfectly straight while still giving incorrect battery parameters.

A straight line is evidence for a model only if the correct quantities have been measured.

Question 5

Two rechargeable battery packs, A and B, are represented by constant-EMF, constant-internal-resistance models.

They are connected in parallel, positive terminal to positive terminal and negative terminal to negative terminal.

A variable external resistor \(R\) is connected across their common terminals.

Pack A has:

\[ \varepsilon_A=10.0\ \mathrm{V}, \qquad r_A=1.0\ \Omega \]

Pack B has:

\[ \varepsilon_B=8.0\ \mathrm{V}, \qquad r_B=1.0\ \Omega \]

Assume ideal connecting wires, no other components, and steady-state conditions. Treat this as a theoretical circuit; real battery packs with unequal EMFs should not simply be connected in parallel because substantial circulating currents may result.

Take current leaving either pack’s positive terminal as positive. A negative current therefore indicates that the pack is being charged.

The external load is tested in three states:

• State 1: The external load is disconnected, but A and B remain connected in parallel.
• State 2: \(R=8.5\ \Omega\).
• State 3: \(R=2.5\ \Omega\).

(a) Determine the common terminal voltage, the signed current through each battery pack, and the external load current in all three states.

(b) Explain why Pack B can have a terminal voltage greater than its EMF in States 1 and 2.

(c) Determine the external load resistance at which Pack B changes from charging to discharging.

(d) As the external load current increases from zero, does the power dissipated in Pack B’s internal resistance necessarily increase? Calculate its internal power dissipation in the three states and at the transition.

(e) An experimenter plots the common terminal voltage against the external load current and obtains a straight line. Find the line’s intercept and gradient. Explain why neither gives the EMF or internal resistance of either individual battery pack.

Solution 5

The common terminal voltage falls from 9.0 V to 8.5 V and then 7.5 V as the external load increases. However, Pack B initially absorbs current rather than supplying it. Its charging current decreases before reversing direction, and its internal heating initially decreases even though external load current increases.

This is the central complication: the current through an individual battery need not be the same as the current through the external load when multiple sources are connected.

Part (a): Find the currents and terminal voltages

Let \(I_A\) and \(I_B\) represent signed currents leaving the positive terminals of Packs A and B.

Let \(I_L\) represent the current flowing through the external load resistor.

For Pack A:

\[ V_t=\varepsilon_A-I_Ar_A \]

Therefore:

\[ I_A=\frac{\varepsilon_A-V_t}{r_A} \]

Similarly, for Pack B:

\[ I_B=\frac{\varepsilon_B-V_t}{r_B} \]

Conservation of charge at the positive junction gives:

\[ I_L=I_A+I_B \]

The external load obeys:

\[ I_L=\frac{V_t}{R} \]

Because both internal resistances are 1.0 \(\Omega\), combining the two battery equations gives:

\[ V_t=9.0\ \mathrm{V}-I_L(0.50\ \Omega) \]

This is the terminal-voltage relationship for the two-pack combination.

For State 1, the external load is disconnected.

Therefore:

\[ I_L=0\ \mathrm{A} \]

The common terminal voltage is:

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

Current from Pack A is:

\[ \begin{aligned} I_A&=\frac{10.0-9.0}{1.0}\\ &=+1.0\ \mathrm{A} \end{aligned} \]

The numerator represents volts and the denominator represents ohms, giving amperes.

For Pack B:

\[ \begin{aligned} I_B&=\frac{8.0-9.0}{1.0}\\ &=-1.0\ \mathrm{A} \end{aligned} \]

Pack A supplies 1.0 A, which flows into Pack B.

No current flows through the external load, but current still flows between the two packs.

For State 2, the external resistance is 8.5 \(\Omega\).

Using the equivalent two-pack voltage relationship:

\[ \begin{aligned} I_L&=\frac{9.0}{8.5+0.50}\\ &=1.00\ \mathrm{A} \end{aligned} \]

Terminal voltage:

\[ \begin{aligned} V_t&=I_LR\\ &=(1.00)(8.5)\\ &=8.50\ \mathrm{V} \end{aligned} \]

Current from Pack A:

\[ \begin{aligned} I_A&=\frac{10.0-8.50}{1.0}\\ &=+1.50\ \mathrm{A} \end{aligned} \]

Current from Pack B:

\[ \begin{aligned} I_B&=\frac{8.0-8.50}{1.0}\\ &=-0.50\ \mathrm{A} \end{aligned} \]

Pack A supplies 1.50 A.

Of that, 1.00 A goes through the external resistor, while 0.50 A enters Pack B.

For State 3, \(R=2.5\ \Omega\).

\[ \begin{aligned} I_L&=\frac{9.0}{2.5+0.50}\\ &=3.00\ \mathrm{A} \end{aligned} \]

Terminal voltage:

\[ V_t=(3.00)(2.5)=7.50\ \mathrm{V} \]

Pack currents:

\[ \begin{aligned} I_A&=\frac{10.0-7.50}{1.0}\\ &=+2.50\ \mathrm{A}\\[4pt] I_B&=\frac{8.0-7.50}{1.0}\\ &=+0.50\ \mathrm{A} \end{aligned} \]

Both packs now supply current to the external resistor.

The results are:

Part (b): Why is Pack B’s terminal voltage greater than its EMF?

Pack B is being charged in States 1 and 2.

Conventional current enters its positive terminal, so its signed discharge current is negative.

For a charging battery:

\[ V_t=\varepsilon_B+I_cr_B \]

where \(I_c\) is the positive magnitude of the charging current.

In State 1:

\[ V_t=8.0+(1.0)(1.0) =9.0\ \mathrm{V} \]

In State 2:

\[ V_t=8.0+(0.50)(1.0) =8.5\ \mathrm{V} \]

There is no contradiction with the internal-resistance model.

The terminal voltage is higher than Pack B’s EMF precisely because current is being driven into it.

Part (c): Find where Pack B changes direction

At the instant Pack B changes from charging to discharging, its current is zero.

Therefore:

\[ I_B=0 \]

For Pack B:

\[ V_t=\varepsilon_B-I_Br_B \]

So at the transition:

\[ V_t=8.0\ \mathrm{V} \]

Pack A must supply all the external load current.

Its current is:

\[ \begin{aligned} I_A&=\frac{10.0-8.0}{1.0}\\ &=2.0\ \mathrm{A} \end{aligned} \]

Since \(I_B=0\):

\[ I_L=2.0\ \mathrm{A} \]

The load resistance is:

\[ \begin{aligned} R&=\frac{V_t}{I_L}\\ &=\frac{8.0}{2.0}\\ &=4.0\ \Omega \end{aligned} \]

Thus, Pack B charges when the external resistance is greater than 4.0 \(\Omega\), carries zero current at 4.0 \(\Omega\), and discharges when the resistance falls below 4.0 \(\Omega\), under the stated model.

Part (d): Does Pack B’s internal heating continually increase?

No. It initially decreases, reaches zero, and then increases.

The power dissipated in Pack B’s internal resistance is:

\[ P_B=I_B^2r_B \]

The square means the heating is positive regardless of current direction.

In State 1:

\[ P_B=(-1.0)^2(1.0) =1.00\ \mathrm{W} \]

In State 2:

\[ P_B=(-0.50)^2(1.0) =0.25\ \mathrm{W} \]

At the transition, when \(R=4.0\ \Omega\):

\[ P_B=(0)^2(1.0) =0\ \mathrm{W} \]

In State 3:

\[ P_B=(+0.50)^2(1.0) =0.25\ \mathrm{W} \]

The result is surprising because the total external load current increases throughout the sequence.

Yet the magnitude of current through Pack B initially decreases.

A student who assumes “more external current means more internal heating in every battery” would miss this behaviour.

The heating depends on the current through that particular battery, not the sum of currents delivered to the external circuit.

Part (e): Interpret the voltage-current graph

For the complete pair of packs, we obtained:

\[ V_t=9.0\ \mathrm{V}-I_L(0.50\ \Omega) \]

The graph of common terminal voltage against external load current is a straight line with:

• Vertical intercept: 9.0 V.
• Gradient: \(-0.50\ \Omega\).

These are the parameters of an equivalent source representing the combined battery network as viewed by the external load.

The equivalent EMF is 9.0 V, and the equivalent internal resistance is 0.50 \(\Omega\).

Neither value belongs to an individual pack.

Pack A has EMF 10.0 V and internal resistance 1.0 \(\Omega\).

Pack B has EMF 8.0 V and internal resistance 1.0 \(\Omega\).

The effective EMF lies between the two individual EMFs, while the effective internal resistance is smaller than either individual resistance because the two internal resistances contribute through parallel current paths.

The crucial trap is assuming that zero external load current means zero current through every battery.

With unequal EMFs, the packs can exchange current even when no external device is connected.

That is why the common terminal voltage can be 9.0 V despite neither battery having an EMF of 9.0 V.

This is also the deeper lesson behind ordinary terminal-voltage measurements: before interpreting a falling voltage, identify which current actually passes through the source or internal resistance you’re analysing.

Once that current is known, the next useful question is how quickly energy is being transferred internally. Voltage sag gives the internal energy loss per coulomb in the simple model, while \(P=I^2r\) gives the rate of internal energy dissipation. That connection is developed in How to Calculate Energy Dissipated Inside a Battery.