How to Calculate Energy Dissipated Inside a Battery

A battery reads 6.0 V when nothing is connected to it. Connect a small lamp, and the voltage across the battery’s terminals falls to 5.4 V while a current of 0.60 A flows.

If the lamp runs for 10 minutes, how much energy is dissipated inside the battery instead of reaching the lamp?

First, make a prediction. Does each coulomb of charge deliver 6.0 J or 5.4 J to the lamp?

It delivers 5.4 J. The remaining 0.60 J per coulomb is transferred inside the battery. In our simplified model, that energy becomes heat because the battery has internal resistance.

The current is 0.60 A, meaning 0.60 coulombs of charge pass each second. Over 600 seconds, the total charge is:

\[ Q=It=0.60\times600=360\ \mathrm{C} \]

The energy dissipated inside the battery is therefore:

\[ E_{\mathrm{int}}=0.60\times360 =216\ \mathrm{J} \]

So the battery dissipates 216 J internally during those 10 minutes.

This is the central method. Find the energy lost per coulomb inside the battery, then multiply by the charge that passes through it.

There are two useful equations for doing this:

\[ E_{\mathrm{int}}=I(\varepsilon-V)t \]

and

\[ E_{\mathrm{int}}=I^2rt \]

The first uses the difference between the battery’s electromotive force and its terminal voltage. The second uses its internal resistance.

Let’s establish what those quantities mean, how to measure them, and which equation to use when the information in a question changes.

01Why does a battery lose energy internally?

An ideal battery would transfer all the electrical energy it supplies to the external circuit.

Real batteries aren’t ideal. Current must move through materials inside the cell, including electrodes and an electrolyte. The processes involved in transporting charge and sustaining the chemical reactions produce internal energy losses.

For a simple circuit calculation, we model those losses using a small resistance inside the battery.

Imagine a battery containing an ideal energy source followed by an ordinary resistor. The resistor is hidden inside the battery case.

That’s a useful picture, but it’s not a literal description of the chemistry. The internal resistance represents the combined effect of processes occurring throughout the cell. It may change with temperature, current, and the battery’s state of charge.

We need three quantities:

• Electromotive force, \(\varepsilon\), measured in volts (V): the energy supplied by the battery per coulomb of charge in the ideal-source model.
• Terminal voltage, \(V\), measured in volts (V): the potential difference across the battery’s external terminals while it operates.
• Internal resistance, \(r\), measured in ohms (\(\Omega\)): the effective resistance representing energy dissipation inside the battery.

Despite its name, electromotive force (emf) is not a mechanical force. It is energy supplied per unit charge.

Why is terminal voltage lower than emf?

Suppose each coulomb receives 12 J of energy from the battery’s chemical processes.

If 1 J per coulomb is dissipated internally, only 11 J per coulomb remains available to the external circuit.

The energy balance is:

\[ \varepsilon=V+Ir \]

Rearranging gives:

\[ V=\varepsilon-Ir \]

Here:

• \(I\) is the current in amperes (A).
• \(Ir\) is the internal voltage drop in volts.

This equation assumes the battery is discharging, with conventional current leaving its positive terminal.

Notice what happens when current increases. If \(\varepsilon\) and \(r\) remain constant, the internal voltage drop increases and the terminal voltage decreases.

That’s why a battery can show a healthy voltage with nothing connected but a much lower voltage when supplying a large current.

02The two methods for calculating internal energy loss

We can derive both methods from one simple idea: energy transferred equals charge multiplied by potential difference.

Method 1: When emf and terminal voltage are known

The voltage difference inside the battery is:

\[ \Delta V=\varepsilon-V \]

The charge transferred over a time \(t\) is:

\[ Q=It \]

Multiplying gives:

\[ E_{\mathrm{int}}=I(\varepsilon-V)t \]

Use this when you’re given the battery’s emf, measured terminal voltage, current, and operating time.

Method 2: When internal resistance is known

Ohm’s law gives:

\[ \varepsilon-V=Ir \]

Substituting into the previous energy equation:

\[ E_{\mathrm{int}}=I(Ir)t \]

Therefore:

\[ E_{\mathrm{int}}=I^2rt \]

This is the most direct method when internal resistance and current are known.

It also tells us something important: internal heating depends on the square of the current.

If the current doubles while resistance stays constant, the rate of energy dissipation increases by a factor of four.

Energy versus power

Power is the rate at which energy is transferred.

For internal resistance:

\[ P_{\mathrm{int}}=I^2r \]

where \(P_{\mathrm{int}}\) is the internal power dissipation in watts (W).

If that power remains constant for \(t\) seconds:

\[ E_{\mathrm{int}}=P_{\mathrm{int}}t \]

Power is measured in watts. Energy is measured in joules.

One watt sustained for one second transfers one joule.

Which equation should you use?

In the third row, \(R\) is the total external resistance in ohms, assuming the circuit can be represented by one equivalent resistor.

03How can we measure a battery’s internal resistance?

In an exam, you might be given the internal resistance directly.

In an experiment, you usually need to determine it from voltage and current measurements.

The challenge is that internal resistance is hidden inside the battery. We cannot simply place a multimeter across it as though it were a separate, accessible resistor.

We need to infer it from how the battery behaves when current flows.

The first approach: Measure the battery without a load

Connect a voltmeter across the battery’s positive and negative terminals, leaving the external circuit disconnected.

What should it read?

If the voltmeter draws negligible current, almost no current passes through the internal resistance.

Therefore:

\[ Ir\approx0 \]

and:

\[ V_{\mathrm{open}}\approx\varepsilon \]

This gives an estimate of the emf.

But there is a problem. A single open-circuit voltage reading cannot tell us the internal resistance or the energy dissipated when the battery supplies current.

We need a way to vary the current and measure what happens to the terminal voltage.

The revised experiment: Change the external resistance

Use a battery, switch, ammeter, variable resistor, voltmeter, and connecting wires.

The connections matter.

Starting at the battery’s positive terminal, connect the switch, ammeter, and variable resistor in series, then return to the negative terminal.

Connect the voltmeter directly across the battery’s two external terminals.

This means:

• The ammeter measures current in the external load branch.
• The voltmeter measures the battery’s terminal potential difference.
• The variable resistor changes the current drawn from the battery.
• The switch allows the circuit to be opened between measurements.

In the basic experiment, we assume an ideal ammeter with zero resistance and an ideal voltmeter with infinite resistance.

Begin with a high external resistance to limit the current.

Close the switch briefly, measure the current and terminal voltage, and open the switch again. Then decrease the external resistance slightly and repeat.

For a school experiment, use a suitable current-limited circuit and keep currents within the ratings of the cell and equipment. Do not deliberately short-circuit the battery.

Opening the switch between readings helps reduce heating and changes in the battery’s state of charge.

What should the measurements show?

Before collecting any data, predict the trend.

If reducing the external resistance increases current, will the terminal voltage rise or fall?

It should fall.

The model predicts:

\[ V=\varepsilon-rI \]

That has the same form as the equation of a straight line:

\[ y=c+mx \]

If we plot terminal voltage vertically and current horizontally:

• The vertical intercept is the emf, \(\varepsilon\).
• The gradient is \(-r\).
• The magnitude of the negative gradient is the internal resistance.

The graph also tests our model. If the measurements don’t approximately follow a straight line, a constant internal resistance may not describe the battery accurately over that operating range.

What might make the experiment unreliable?

The straight-line model requires \(\varepsilon\) and \(r\) to remain approximately constant.

A real battery may warm up, discharge slightly between readings, or behave differently at higher currents.

Contact resistance and measurement uncertainty can also affect the results.

A useful experimental improvement is to repeat each measurement, keep the operating temperature as stable as practical, and use a fitted line through several data points rather than calculating resistance from one pair alone.

If the graph curves noticeably, don’t force the data into a constant-resistance model without explaining the limitation.

04Practical extension: What if the meters aren’t ideal?

The usual school model assumes that the voltmeter draws no current and the ammeter adds no resistance.

Real instruments affect a circuit to some extent.

This is a practical measurement extension, rather than a claim about additional prescribed HSC examination content.

A voltmeter can draw current

A real voltmeter has a large but finite resistance, \(R_V\).

Because it is connected across the battery terminals, some current flows through the voltmeter itself.

If the ammeter is in the separate load branch, it measures only that branch’s current.

The battery must supply both currents:

\[ I_b=I_A+I_V \]

where:

• \(I_b\) is the total battery current.
• \(I_A\) is the current measured in the load branch.
• \(I_V\) is the current through the voltmeter.

The voltmeter current is:

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

Therefore, internal heating must be calculated using \(I_b\), not automatically \(I_A\).

A non-ideal ammeter also has a small resistance, \(R_A\), in series with the branch containing it. That changes the circuit current and dissipates additional energy outside the battery.

It does not change the basic internal-loss formula:

\[ E_{\mathrm{int}}=I_b^2rt \]

The important question is which current actually passes through the battery’s internal resistance.

05Two situations where the usual method needs care

Changing current

Suppose a device draws 1 A for a while, then 3 A.

You cannot generally calculate internal energy loss by taking the average current and substituting it into \(I^2rt\).

The current is squared before we calculate energy.

For separate intervals, use:

\[ E_{\mathrm{int}} =\sum_k I_k^2r\Delta t_k \]

Here, \(I_k\) is the current during interval \(k\), and \(\Delta t_k\) is that interval’s duration.

This assumes internal resistance stays constant and current is approximately constant within each interval.

Charging instead of discharging

For a discharging battery, conventional current leaves its positive terminal:

\[ V=\varepsilon-Ir \]

During charging, an external source forces current into the battery’s positive terminal.

If we call the positive magnitude of that charging current \(I_{\mathrm{ch}}\), the terminal voltage becomes:

\[ V=\varepsilon+I_{\mathrm{ch}}r \]

The terminal voltage must exceed the emf to drive charging current through the internal resistance.

Yet internal resistive heating remains positive:

\[ E_{\mathrm{int}}=I_{\mathrm{ch}}^2rt \]

Do not interpret the higher charging voltage as negative energy dissipation.

The energy supplied during charging is divided between energy stored by the battery’s chemistry and internal losses, within this simplified model.

This charging model assumes a rechargeable cell that can accept current. Actual charging behaviour involves additional chemical limitations and control systems.

06Questions and solutions

Question 1

A battery has an emf of 4.50 V. When connected to a small lamp, its terminal voltage is 4.20 V and it supplies a constant current of 0.30 A.

Calculate:

(a) The battery’s internal resistance.

(b) The energy dissipated inside the battery over two minutes.

Assume constant emf and internal resistance.

Solution 1

The battery has an internal resistance of \(1.0\ \Omega\) and dissipates 10.8 J internally over two minutes.

(a) Internal resistance

The voltage difference is caused by the internal resistance in the model.

\[ \begin{aligned} r&=\frac{\varepsilon-V}{I}\\ &=\frac{4.50-4.20}{0.30}\\ &=1.0\ \Omega \end{aligned} \]

(b) Internal energy

Convert two minutes into seconds:

\[ t=120\ \mathrm{s} \]

Then:

\[ \begin{aligned} E_{\mathrm{int}}&=I^2rt\\ &=(0.30)^2(1.0)(120)\\ &=10.8\ \mathrm{J} \end{aligned} \]

The tempting mistake is to use \(VIt\), which calculates energy delivered to the external circuit, not energy dissipated inside the battery.

Here, the battery loses 0.30 J internally for every coulomb passing through it.

Question 2

An illustrative battery has an emf of 6.0 V and an unknown internal resistance.

It powers lamp A, modelled as a fixed \(9.5\ \Omega\) resistor. With only lamp A connected, the current is 0.60 A.

A switch is then closed, connecting a second lamp, B, in parallel with lamp A. Both lamps are modelled as fixed resistors. The ammeter in the common supply wire now measures a total current of 1.00 A.

Assume ideal connecting wires, an ideal ammeter, and constant battery parameters.

(a) Determine the battery’s internal resistance.

(b) Calculate the new terminal voltage and the current through each lamp.

(c) Calculate the energy dissipated inside the battery during 120 seconds with both lamps connected.

(d) Explain why the current through lamp A changes even though its resistance stays constant.

Solution 2

The internal resistance is \(0.50\ \Omega\). With both lamps connected, the terminal voltage falls to 5.5 V, lamp A’s current decreases to approximately 0.579 A, and the battery dissipates 60 J internally over 120 seconds.

(a) Find internal resistance from the first arrangement.

With only lamp A connected:

\[ \begin{aligned} V_1&=I_1R_A\\ &=(0.60)(9.5)\\ &=5.70\ \mathrm{V} \end{aligned} \]

Here, \(R_A\) is lamp A’s resistance, not ammeter resistance.

The internal voltage drop is:

\[ 6.0-5.70=0.30\ \mathrm{V} \]

Therefore:

\[ r=\frac{0.30}{0.60}=0.50\ \Omega \]

(b) Find the new terminal voltage and branch currents.

With both lamps operating, the battery supplies 1.00 A in total.

\[ \begin{aligned} V_2&=\varepsilon-I_2r\\ &=6.0-(1.00)(0.50)\\ &=5.5\ \mathrm{V} \end{aligned} \]

Both parallel lamps have the same 5.5 V potential difference.

For lamp A:

\[ \begin{aligned} I_A&=\frac{V_2}{R_A}\\ &=\frac{5.5}{9.5}\\ &\approx0.579\ \mathrm{A} \end{aligned} \]

The remaining current passes through lamp B:

\[ \begin{aligned} I_B&=1.00-0.579\\ &\approx0.421\ \mathrm{A} \end{aligned} \]

(c) Calculate internal dissipation.

Use the total battery current:

\[ \begin{aligned} E_{\mathrm{int}}&=I_2^2rt\\ &=(1.00)^2(0.50)(120)\\ &=60\ \mathrm{J} \end{aligned} \]

(d) Explain the change in lamp A’s current.

A tempting prediction is that lamp A continues drawing 0.60 A because its resistance hasn’t changed.

But adding lamp B increases the total current taken from the battery. This increases the voltage drop across the battery’s internal resistance.

The battery’s terminal voltage falls from 5.70 V to 5.5 V.

Since lamp A’s resistance remains fixed, the lower voltage forces a slightly lower current through it.

This is an important interaction: changing one parallel branch can affect another branch through the battery’s internal resistance.

Question 3

A model battery has an emf of 10.0 V and an internal resistance of \(0.50\ \Omega\).

A device can operate under either of two discharge patterns:

• Pattern A: Draw a constant 2.0 A for 60 seconds.
• Pattern B: Draw 1.0 A for the first 30 seconds, then 3.0 A for the next 30 seconds.

Assume the battery’s emf and internal resistance remain constant, current changes instantaneously between intervals, and there are no other energy losses in the external connections.

(a) Show that both patterns transfer the same total charge.

(b) Calculate the internal energy dissipated during each pattern.

(c) Determine the total energy delivered through the battery’s terminals during each pattern.

(d) A student argues that both patterns must produce the same internal heating because their average current is identical. Explain the flaw in this reasoning.

Solution 3

Pattern B dissipates 150 J internally, compared with 120 J for Pattern A, despite transferring the same 120 C of charge. Consequently, Pattern A delivers 1080 J externally, while Pattern B delivers 1050 J.

(a) Total charge

For Pattern A:

\[ \begin{aligned} Q_A&=It\\ &=(2.0)(60)\\ &=120\ \mathrm{C} \end{aligned} \]

For Pattern B:

\[ \begin{aligned} Q_B&=(1.0)(30)+(3.0)(30)\\ &=30+90\\ &=120\ \mathrm{C} \end{aligned} \]

Both transfer 120 C.

(b) Internal energy dissipation

For Pattern A:

\[ \begin{aligned} E_A&=I^2rt\\ &=(2.0)^2(0.50)(60)\\ &=120\ \mathrm{J} \end{aligned} \]

For Pattern B, calculate each interval separately.

First 30 seconds:

\[ E_1=(1.0)^2(0.50)(30) =15\ \mathrm{J} \]

Second 30 seconds:

\[ E_2=(3.0)^2(0.50)(30) =135\ \mathrm{J} \]

Total:

\[ E_B=15+135=150\ \mathrm{J} \]

Pattern B dissipates 30 J more internally.

(c) Energy delivered through the terminals

The ideal source supplies:

\[ \begin{aligned} E_{\mathrm{source}}&=\varepsilon Q\\ &=(10.0)(120)\\ &=1200\ \mathrm{J} \end{aligned} \]

For Pattern A:

\[ E_{\mathrm{external},A} =1200-120 =1080\ \mathrm{J} \]

For Pattern B:

\[ E_{\mathrm{external},B} =1200-150 =1050\ \mathrm{J} \]

(d) Why the average-current argument fails

Both patterns have an average current of 2.0 A.

But power dissipated in internal resistance depends on \(I^2\), not directly on \(I\).

In Pattern B, the threefold increase from 1.0 A to 3.0 A produces nine times the internal heating power during the high-current interval.

That large increase is not cancelled by the smaller loss during the low-current interval.

Using the square of the average current would incorrectly predict equal energy loss. We must instead calculate the energy in each interval before adding the results.

Question 4

Practical extension: A voltmeter that changes the measurement

An illustrative battery has an emf of 6.0 V and internal resistance \(1.0\ \Omega\).

A \(5.0\ \Omega\) resistor is connected across its terminals through an ideal ammeter.

A non-ideal voltmeter, modelled as a \(10.0\ \Omega\) resistor, is also connected directly across the battery terminals. The voltmeter forms a separate parallel branch, so its current does not pass through the ammeter.

The voltmeter resistance is deliberately low to make the measurement-loading effect noticeable.

The arrangement operates steadily for 20 seconds.

(a) Calculate the current supplied by the battery.

(b) Determine the terminal voltage, ammeter reading, and voltmeter current.

(c) Calculate the energy dissipated inside the battery.

(d) Explain why using the ammeter reading directly in \(I^2rt\) gives the wrong result.

Solution 4

The battery supplies approximately 1.385 A and dissipates 38.3 J internally. The ammeter reads only 0.923 A because it does not measure the current flowing through the voltmeter.

(a) Calculate the equivalent external resistance and battery current.

The resistor and voltmeter are in parallel.

\[ \begin{aligned} R_{\mathrm{eq}} &=\frac{5.0\times10.0}{5.0+10.0}\\ &=3.33\ \Omega \end{aligned} \]

Including the battery’s internal resistance:

\[ \begin{aligned} I_b&=\frac{\varepsilon}{r+R_{\mathrm{eq}}}\\ &=\frac{6.0}{1.0+3.33}\\ &\approx1.385\ \mathrm{A} \end{aligned} \]

(b) Determine the terminal voltage and branch currents.

The internal voltage drop is approximately:

\[ I_br=(1.385)(1.0) =1.385\ \mathrm{V} \]

Therefore:

\[ V=6.0-1.385 \approx4.615\ \mathrm{V} \]

The ammeter measures the current through the \(5.0\ \Omega\) resistor:

\[ \begin{aligned} I_A&=\frac{4.615}{5.0}\\ &\approx0.923\ \mathrm{A} \end{aligned} \]

The voltmeter draws:

\[ \begin{aligned} I_V&=\frac{4.615}{10.0}\\ &\approx0.462\ \mathrm{A} \end{aligned} \]

Their sum is approximately 1.385 A, matching the battery current.

(c) Calculate internal dissipation.

\[ \begin{aligned} E_{\mathrm{int}}&=I_b^2rt\\ &=(1.385)^2(1.0)(20)\\ &\approx38.3\ \mathrm{J} \end{aligned} \]

(d) Explain the measurement error.

Using the ammeter reading would give:

\[ E_{\mathrm{wrong}} =(0.923)^2(1.0)(20) \approx17.0\ \mathrm{J} \]

That substantially underestimates the actual internal loss.

The mistake is assuming that every coulomb supplied by the battery passes through the ammeter.

It doesn’t. Current splits between the external resistor and voltmeter.

The governing constraint is conservation of charge at the junction:

\[ I_b=I_A+I_V \]

Internal heating depends on the total current through the battery, including the current consumed by the measuring instrument.

Question 5

Challenge extension: Two batteries, one external circuit

Two simplified rechargeable cells, A and B, are connected in parallel, with their positive terminals joined and their negative terminals joined.

Their properties are:

A switch allows an external \(2.0\ \Omega\) resistor to be connected across the shared terminals.

For the first 30 seconds, the external switch is open. The cells remain connected to each other.

For the next 30 seconds, the switch is closed, connecting the resistor.

Assume both cells obey the constant-emf, constant-resistance model, both can accept charging current, connecting wires are ideal, and switching is instantaneous. Their properties do not change over the 60 seconds.

This is a theoretical circuit. Connecting real batteries with different emfs directly in parallel can be unsafe.

(a) When the external switch is open, determine the shared terminal voltage and the current direction in each cell.

(b) Calculate the total energy dissipated inside the two cells during the first 30 seconds.

(c) When the \(2.0\ \Omega\) external resistor is connected, determine the new terminal voltage and currents in both cells.

(d) Calculate the total energy dissipated inside both cells over the complete 60 seconds.

(e) Determine the external resistance at which cell B changes between charging and discharging.

Explain why the statements “an open circuit means no internal heating” and “both batteries must supply current whenever the external load is connected” are not always correct.

Solution 5

The two cells dissipate approximately 143 J internally over the full 60 seconds. Cell B charges while the external resistor is disconnected but discharges after the \(2.0\ \Omega\) load is connected. It changes direction at an external resistance of \(4.0\ \Omega\).

The crucial idea is that the two cells are still connected to each other when the external switch is open.

(a) Find the voltage and current directions with no external load.

Both batteries share the same terminal voltage, which we’ll call \(V\).

Define each cell’s current as positive when it leaves that cell’s positive terminal.

For cell A:

\[ I_A=\frac{6.0-V}{2.0} \]

For cell B:

\[ I_B=\frac{4.0-V}{2.0} \]

With the external resistor disconnected, there is no net current leaving the two-cell combination.

Therefore:

\[ I_A+I_B=0 \]

Substituting:

\[ \frac{6.0-V}{2.0} +\frac{4.0-V}{2.0}=0 \]

This gives:

\[ V=5.0\ \mathrm{V} \]

Now calculate the cell currents:

\[ I_A=\frac{6.0-5.0}{2.0} =0.50\ \mathrm{A} \]

\[ I_B=\frac{4.0-5.0}{2.0} =-0.50\ \mathrm{A} \]

The negative sign means cell B is receiving current into its positive terminal.

Cell A is discharging at 0.50 A while cell B is charging at 0.50 A.

There is a complete current path between the two cells even though the external resistor is disconnected.

(b) Calculate internal heating during the first 30 seconds.

Each cell carries a current with magnitude 0.50 A.

For one cell:

\[ \begin{aligned} E&=I^2rt\\ &=(0.50)^2(2.0)(30)\\ &=15\ \mathrm{J} \end{aligned} \]

For two cells:

\[ E_{\mathrm{open}}=15+15 =30\ \mathrm{J} \]

This heat is not created from nothing.

During these 30 seconds, cell A supplies chemical energy equivalent to:

\[ (6.0)(0.50)(30)=90\ \mathrm{J} \]

Cell B stores, in the idealised model:

\[ (4.0)(0.50)(30)=60\ \mathrm{J} \]

The remaining 30 J is dissipated internally, exactly as calculated.

(c) Connect the external \(2.0\ \Omega\) resistor.

The shared terminal voltage is still \(V\), but current now also passes through the external resistor.

Its current is:

\[ I_R=\frac{V}{2.0} \]

Conservation of charge at the shared positive terminal requires:

\[ I_A+I_B=I_R \]

Substitute the cell equations:

\[ \frac{6.0-V}{2.0} +\frac{4.0-V}{2.0} =\frac{V}{2.0} \]

Multiplying by 2.0 and rearranging:

\[ 10.0-2V=V \]

Therefore:

\[ V=\frac{10.0}{3} \approx3.33\ \mathrm{V} \]

Now determine each current.

For cell A:

\[ \begin{aligned} I_A&=\frac{6.0-3.33}{2.0}\\ &\approx1.33\ \mathrm{A} \end{aligned} \]

For cell B:

\[ \begin{aligned} I_B&=\frac{4.0-3.33}{2.0}\\ &\approx0.333\ \mathrm{A} \end{aligned} \]

Both currents are positive.

Cell B is now discharging rather than charging.

The external resistor draws:

\[ I_R=\frac{3.33}{2.0} \approx1.67\ \mathrm{A} \]

This agrees with the sum of the two cell currents.

(d) Calculate internal dissipation over all 60 seconds.

During the second interval, use the unrounded currents \(4/3\ \mathrm{A}\) for cell A and \(1/3\ \mathrm{A}\) for cell B.

Cell A:

\[ \begin{aligned} E_A&=\left(\frac43\right)^2(2.0)(30)\\ &\approx106.7\ \mathrm{J} \end{aligned} \]

Cell B:

\[ \begin{aligned} E_B&=\left(\frac13\right)^2(2.0)(30)\\ &\approx6.67\ \mathrm{J} \end{aligned} \]

So internal energy lost during the loaded interval is:

\[ E_{\mathrm{loaded}} \approx113.3\ \mathrm{J} \]

Adding the 30 J from the open-load interval:

\[ \begin{aligned} E_{\mathrm{total}}&=30+113.3\\ &\approx143.3\ \mathrm{J} \end{aligned} \]

The two cells dissipate approximately 143 J internally over the entire minute.

(e) Determine the load resistance where cell B reverses current.

Cell B’s current is zero precisely when the shared terminal voltage equals its emf:

\[ V=4.0\ \mathrm{V} \]

At that voltage, cell A supplies:

\[ \begin{aligned} I_A&=\frac{6.0-4.0}{2.0}\\ &=1.0\ \mathrm{A} \end{aligned} \]

Because cell B supplies no current at this boundary, the entire 1.0 A must pass through the external resistor.

Therefore:

\[ \begin{aligned} R&=\frac{V}{I_R}\\ &=\frac{4.0}{1.0}\\ &=4.0\ \Omega \end{aligned} \]

This is the changeover resistance.

• For external resistance greater than \(4.0\ \Omega\), cell B receives charging current.
• At \(4.0\ \Omega\), cell B carries zero current.
• Below \(4.0\ \Omega\), cell B supplies current to the load.

The tempting mistake is to judge current direction using only the emf values or whether an external switch is open.

But current direction depends on the difference between each cell’s emf and the shared terminal voltage.

When the external load is disconnected, cell A can still drive current through cell B. When the load is connected, the shared voltage falls enough to reverse cell B’s current.

The governing principle hasn’t changed. Both cells obey their own terminal-voltage relationships, but they must also satisfy a shared terminal voltage and conservation of current at the junction.

That interaction is what makes the result surprising.

07What this calculation tells you about real batteries

A battery’s printed voltage doesn’t tell you how much energy a connected device will actually receive.

You also need to know the current it draws and how much terminal voltage the battery loses under that load.

For a short, approximately steady discharge, the useful calculation is:

\[ E_{\mathrm{int}}=I^2rt \]

For measured emf and terminal voltage, use:

\[ E_{\mathrm{int}}=I(\varepsilon-V)t \]

For changing currents, calculate the energy over separate intervals before adding them.

In each case, ask where the energy goes. The ideal source supplies it, the external circuit receives some of it, and the battery dissipates some internally.

The next useful distinction is between energy lost through internal resistance and energy stored or released by the battery’s chemical reactions. Internal resistance helps explain why terminal voltage falls under load, but it doesn’t explain what sets the emf or why some cells can be recharged.

Those questions are addressed through the redox chemistry of alkaline and lithium-ion batteries.