How to Distinguish EMF from Potential Difference in Circuits

A cell gives an open-circuit voltage reading of 1.50 V. Connect a lamp, and the voltmeter across the cell’s terminals now reads 1.20 V. Which value is the cell’s electromotive force (EMF)?

The EMF is approximately 1.50 V. The 1.20 V reading is the terminal potential difference while the cell is supplying current. Both readings can be correct because they describe different operating conditions.

The essential distinction is that EMF describes the energy supplied per coulomb by a source, while potential difference describes the energy transferred per coulomb between two points in a circuit.

01EMF and potential difference: what is the difference?

Both quantities measure energy transferred per unit charge. Both have the unit volt (V).

However, they describe different things.

Electromotive force (EMF)

The electromotive force, represented by \(\varepsilon\), is the energy supplied by a source per unit charge.

Inside a battery, chemical reactions transfer energy to electric charges. The EMF tells us how much energy the source supplies to each coulomb of charge.

Mathematically,

\[
\varepsilon=\frac{W_{\text{source}}}{q}
\]

where:

• \(\varepsilon\) is the EMF in volts (V).
• \(W_{\text{source}}\) is the energy supplied by the source in joules (J).
• \(q\) is the charge in coulombs (C).

For example, a cell with an EMF of 1.50 V supplies 1.50 J of energy for every coulomb of charge passing through it, under the assumed operating conditions.

Despite its name, EMF is not a mechanical force. It is an energy-per-charge quantity.

Potential difference (PD)

Potential difference describes the change in electrical potential energy per unit charge between two points.

It tells us how much energy is transferred when charge moves between those points.

For a potential difference \(V\),

\[
V=\frac{\Delta E}{q}
\]

where:

• \(V\) is the potential difference in volts (V).
• \(\Delta E\) is the energy transferred in joules (J).
• \(q\) is the charge in coulombs (C).

For example, a lamp with a potential difference of 1.20 V transfers 1.20 J of energy for every coulomb passing through it.

This is why the voltage across a lamp tells us about the energy transferred to the lamp, rather than the total energy originally supplied by the battery.

Quick comparison

A useful way to remember the distinction is:

• EMF: How much energy does the source give each coulomb?
• Potential difference: How much energy is transferred per coulomb between these two points?

The word voltage is often used informally for either quantity. Always check what the question is actually asking.

02Why EMF and terminal potential difference can be different

Consider a battery connected to a lamp.

You might initially predict that the entire EMF of the battery appears across the lamp. This would be correct for an ideal battery.

Real batteries, however, have internal resistance.

Introducing internal resistance

A real battery can be modelled as an ideal EMF source connected in series with an internal resistance, represented by \(r\).

This internal resistance represents the electrical losses inside the battery. Some of the energy supplied by the source is transferred into thermal energy within the battery itself.

Think of a water pump pushing water through a pipe. The pump supplies energy, but friction within the pipe dissipates some of it before the water reaches its destination.

The comparison is useful for understanding energy losses, although electric charge is not literally flowing water.

When the battery supplies current, some energy is transferred inside the battery itself.

Consequently, the terminal potential difference is smaller than the EMF for a discharging real cell.

Using conservation of energy:

\[
\boxed{\varepsilon=V_t+Ir}
\]

Rearranging gives:

\[
\boxed{V_t=\varepsilon-Ir}
\]

where:

• \(\varepsilon\) is the EMF (V).
• \(V_t\) is the terminal potential difference (V).
• \(I\) is the current leaving the positive terminal (A).
• \(r\) is the internal resistance (\(\Omega\)).
• \(Ir\) is the potential difference across the internal resistance (V).

The equation tells us that the terminal voltage is reduced by the voltage drop inside the battery.

What happens when the current increases?

The equation \(V_t=\varepsilon-Ir\) gives us an important prediction.

If the EMF and internal resistance remain constant, increasing the current increases the internal voltage drop.

This means the terminal potential difference decreases.

For example, consider a cell with:

• EMF = 1.50 V
• Internal resistance = 0.50 Ω

Notice that the EMF remains 1.50 V throughout this example.

The terminal potential difference changes because the internal voltage drop changes.

This is the central reason why EMF and terminal PD must not be treated as interchangeable quantities.

For an ideal battery, \(r=0\). Therefore,

\[
V_t=\varepsilon
\]

An ideal battery has no internal voltage drop, so its terminal voltage equals its EMF even when current flows.

03How an experiment distinguishes EMF from potential difference

How could we experimentally determine whether a battery’s EMF is different from its terminal potential difference?

The important measurement problem is that connecting a battery to a circuit changes the conditions under which its voltage is measured.

We can investigate this by measuring the battery under different loads.

First measurement: the open-circuit voltage

Start with a cell connected directly to a voltmeter, with no external load.

The voltmeter measures the potential difference between the positive and negative terminals.

Since there is almost no current flowing, the internal voltage drop is almost zero.

Using

\[
V_t=\varepsilon-Ir
\]

when \(I\) is approximately zero,

\[
V_t\approx\varepsilon
\]

Therefore, the voltmeter reading gives a good estimate of the cell’s EMF.

Suppose the voltmeter reads 1.50 V.

We estimate the cell’s EMF to be 1.50 V.

Second measurement: connect an external load

Now connect a resistor across the battery terminals.

The battery supplies current through the resistor. This current also passes through the battery’s internal resistance.

What should happen to the voltmeter reading?

If the battery were ideal, the reading would remain at 1.50 V.

For a real battery, however, the reading decreases because some energy per coulomb is transferred inside the battery.

The voltmeter now measures the terminal potential difference under load, not the open-circuit EMF.

Revising the experiment

A single open-circuit measurement cannot tell us the battery’s internal resistance.

To investigate this, we can vary the external resistance and measure both current and terminal voltage.

The apparatus is arranged as follows:

1. Connect the cell, switch, ammeter, and variable resistor in one series circuit.
2. Connect a voltmeter in parallel across the cell’s external terminals.
3. Close the switch and adjust the variable resistor to change the current.
4. Record the current and corresponding terminal voltage for each setting.

The ammeter measures the current supplied by the cell. The voltmeter measures the terminal potential difference.

Illustrative experimental results

Suppose the following measurements are obtained.

These are illustrative data, not measurements from a real experiment.

There is a clear pattern: as the current increases, the terminal potential difference decreases.

This supports the internal resistance model.

Using a voltage-current graph

We can investigate the relationship more clearly by plotting terminal voltage against current.

Starting with

\[
V_t=\varepsilon-Ir
\]

compare this with the straight-line equation:

\[
y=c+mx
\]

The two equations have the same structure.

Therefore:

• The vertical intercept represents the EMF, \(\varepsilon\).
• The gradient represents negative internal resistance, \(-r\).

We can calculate the internal resistance using two measurements:

\[
r=\frac{1.50-1.20}{0.60-0}
\]

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

\[
\boxed{r=0.50\ \Omega}
\]

The graph therefore gives:

\[
\boxed{\varepsilon=1.50\text{ V}}
\]

and

\[
\boxed{r=0.50\ \Omega}
\]

The experiment has allowed us to distinguish the source’s EMF from its terminal potential difference.

It has also allowed us to estimate the internal resistance.

Experimental limitation: This model assumes that the cell’s EMF and internal resistance remain approximately constant during the measurements. Real cells can change with temperature, state of charge, and current history. The simple linear relationship may therefore only be an approximation.

04A decision process for HSC questions

When a question mentions EMF, potential difference, or terminal voltage, use the following process.

Step 1: Identify what is being measured.

Is the question asking about energy supplied by the source, or energy transferred between two circuit points?

• Energy supplied per coulomb by a source: EMF.
• Energy transferred per coulomb between two points: potential difference.

Step 2: If measuring across a battery, check whether current flows.

For a cell with negligible current:

\[
V_t\approx\varepsilon
\]

For a cell supplying current:

\[
V_t=\varepsilon-Ir
\]

Step 3: If measuring across a resistor, use the resistor’s potential difference.

For an ohmic resistor:

\[
V=IR
\]

Do not automatically substitute the battery’s EMF for the resistor’s potential difference.

Step 4: Check the energy balance.

For a discharging cell:

\[
\varepsilon=V_t+Ir
\]

The energy supplied per coulomb must account for both the energy transferred externally and the energy dissipated internally.

A common misconception: terminal voltage is always less than EMF

This statement is not always correct.

It is true for a real cell supplying current, but consider what happens when a rechargeable cell is being charged.

During charging, an external power supply forces current into the positive terminal of the cell.

The supply must overcome the cell’s EMF as well as its internal resistance.

If \(I_c\) is the magnitude of the charging current, then:

\[
\boxed{V_t=\varepsilon+I_c r}
\]

The terminal potential difference is now greater than the EMF.

This does not violate energy conservation. The external charger supplies the extra energy needed to drive current against the cell’s EMF and overcome internal resistance.

The direction of current matters when deciding which equation to use.

05Practical extension: real voltmeters and ammeters

In standard circuit questions, meters are usually treated as ideal unless the question supplies additional information.

Real meters, however, have resistance.

This matters when measuring the EMF of a cell, particularly when the cell has appreciable internal resistance.

A real voltmeter

An ideal voltmeter has infinite resistance. It draws no current and does not affect the circuit.

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

When connected across a cell with no other external load, it draws a small current.

The current is:

\[
I_V=\frac{\varepsilon}{r+R_V}
\]

The voltmeter reading is:

\[
V_V=I_VR_V
\]

Combining the equations:

\[
\boxed{V_V=\frac{\varepsilon R_V}{r+R_V}}
\]

The reading is slightly smaller than the actual EMF because the voltmeter draws a small current through the internal resistance.

For example, suppose:

• \(\varepsilon=1.50\text{ V}\)
• \(r=0.50\ \Omega\)
• \(R_V=100\ \Omega\)

Then:

\[
V_V=\frac{1.50\times100}{100+0.50}
\]

\[
\boxed{V_V\approx1.49\text{ V}}
\]

The voltmeter reads approximately 1.49 V rather than exactly 1.50 V.

A voltmeter with a higher input resistance draws less current, making its reading closer to the true EMF.

Importantly, a high-resistance voltmeter does not eliminate the voltage drop caused by an external load. If a lamp is still connected, the meter measures the terminal potential difference under that loaded condition.

A real ammeter

An ideal ammeter has zero resistance.

A real ammeter has a small resistance, \(R_A\), which adds to the resistance of the circuit.

This slightly reduces the current compared with the ideal model.

For most standard HSC calculations, ignore these effects unless meter resistance is explicitly supplied or the question asks you to consider measurement limitations.

06Worked examples

07Questions and solutions

Question 1

A cell has an EMF of 12.0 V and an internal resistance of 0.50 Ω. It supplies a current of 3.0 A to a lamp.

Calculate:

(a) The terminal potential difference of the cell.

(b) The potential difference across the lamp, assuming negligible connecting-wire resistance and no other external components.

(c) The energy dissipated inside the cell per coulomb of charge.

Solution 1

The terminal potential difference is 10.5 V, and the energy dissipated inside the cell is 1.5 J per coulomb.

Part (a): Terminal potential difference

Use:

\[
V_t=\varepsilon-Ir
\]

Substitute:

\[
V_t=12.0-(3.0\times0.50)
\]

\[
V_t=12.0-1.5
\]

\[
\boxed{V_t=10.5\text{ V}}
\]

Part (b): Potential difference across the lamp

The lamp is the only external component.

Therefore, the entire terminal potential difference appears across the lamp.

\[
\boxed{V_{\text{lamp}}=10.5\text{ V}}
\]

The lamp does not receive the full 12.0 V because some energy is dissipated inside the cell.

Part (c): Energy dissipated inside the cell

The internal potential difference is:

\[
V_r=Ir
\]

\[
V_r=3.0\times0.50
\]

\[
V_r=1.5\text{ V}
\]

Since 1 V is equivalent to 1 J/C:

\[
\boxed{1.5\text{ J per coulomb}}
\]

The cell supplies 12.0 J per coulomb, of which 1.5 J per coulomb is dissipated internally and 10.5 J per coulomb is transferred to the lamp.

Question 2

During a controlled discharge experiment, a cell produces the following measurements:

Assume the EMF and internal resistance remain constant.

(a) Determine the internal resistance of the cell.

(b) Determine its EMF.

(c) Calculate the terminal potential difference when the cell is connected to an 8.0 Ω resistor.

(d) Explain why the original measurements of 5.70 V and 5.10 V are not the cell’s EMF.

Solution 2

The cell has an EMF of 6.0 V and an internal resistance of 1.0 Ω. Its terminal potential difference with the 8.0 Ω resistor connected is approximately 5.33 V.

Part (a): Internal resistance

Use the relationship:

\[
V_t=\varepsilon-Ir
\]

Rearranging for the internal resistance:

\[
r=\frac{V_1-V_2}{I_2-I_1}
\]

Substitute:

\[
r=\frac{5.70-5.10}{0.90-0.30}
\]

\[
r=\frac{0.60}{0.60}
\]

\[
\boxed{r=1.0\ \Omega}
\]

Part (b): EMF

Rearrange:

\[
\varepsilon=V_t+Ir
\]

Using the first measurement:

\[
\varepsilon=5.70+(0.30\times1.0)
\]

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

Part (c): Terminal potential difference with an 8.0 Ω resistor

First calculate the total circuit resistance:

\[
R_{\text{total}}=R+r
\]

\[
R_{\text{total}}=8.0+1.0=9.0\ \Omega
\]

Calculate the current:

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

\[
I=\frac{6.0}{9.0}
\]

\[
I=0.667\text{ A}
\]

Now calculate the terminal potential difference:

\[
V_t=IR
\]

\[
V_t=0.667\times8.0
\]

\[
\boxed{V_t=5.33\text{ V}}
\]

Part (d): Why the original readings are not the EMF

Both original measurements were taken while current was flowing through the cell.

Therefore, some energy was dissipated across its internal resistance.

The voltmeter measured the terminal potential difference under each load condition, rather than the full EMF.

The EMF is 6.0 V, even though the terminal readings were 5.70 V and 5.10 V.

Question 3

Two identical cells are connected in series, with their EMFs acting in the same direction.

Each cell has:

• EMF = 1.50 V
• Internal resistance = 0.20 Ω

The cells are connected to a 2.60 Ω external resistor. Assume negligible connecting-wire resistance.

Calculate:

(a) The total EMF.

(b) The total internal resistance.

(c) The current in the circuit.

(d) The terminal potential difference across the two cells combined and across each individual cell.

Explain why the sum of the individual terminal potential differences is smaller than the total EMF.

Solution 3

The combined EMF is 3.00 V, the total internal resistance is 0.40 Ω, and the current is 1.00 A. The terminal PD is 2.60 V across the pair, or 1.30 V across each cell.

Part (a): Total EMF

The cells are connected in series aiding, so their EMFs add.

\[
\varepsilon_{\text{total}}=1.50+1.50
\]

\[
\boxed{\varepsilon_{\text{total}}=3.00\text{ V}}
\]

Part (b): Total internal resistance

The internal resistances are also in series.

\[
r_{\text{total}}=0.20+0.20
\]

\[
\boxed{r_{\text{total}}=0.40\ \Omega}
\]

Part (c): Circuit current

The total resistance is:

\[
R_{\text{total}}=2.60+0.40
\]

\[
R_{\text{total}}=3.00\ \Omega
\]

Using Ohm’s law:

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

\[
I=\frac{3.00}{3.00}
\]

\[
\boxed{I=1.00\text{ A}}
\]

Part (d): Terminal potential differences

The combined terminal potential difference is the voltage across the external resistor:

\[
V_t=IR
\]

\[
V_t=1.00\times2.60
\]

\[
\boxed{V_t=2.60\text{ V}}
\]

Since the two cells are identical and connected in series, the current through each is 1.00 A.

The terminal PD across each cell is:

\[
V_{\text{cell}}=\varepsilon-Ir
\]

\[
V_{\text{cell}}=1.50-(1.00\times0.20)
\]

\[
\boxed{V_{\text{cell}}=1.30\text{ V}}
\]

Adding the two individual terminal voltages:

\[
1.30+1.30=2.60\text{ V}
\]

This equals the terminal PD across the external resistor.

The total EMF is still 3.00 V. The difference of 0.40 V is the combined voltage drop across the two internal resistances.

This demonstrates that the sum of the external potential differences is equal to the total EMF only when internal voltage drops are also accounted for.

Question 4

A rechargeable cell has an EMF of 6.0 V and an internal resistance of 0.75 Ω.

It is connected to an ideal 9.0 V charger with the correct polarity. Assume negligible resistance in the connecting wires.

(a) Calculate the charging current.

(b) Calculate the terminal potential difference across the cell.

(c) Explain why the terminal potential difference is greater than the EMF during charging.

Solution 4

The charging current is 4.0 A, and the terminal potential difference across the cell is 9.0 V.

Part (a): Charging current

During charging, the external supply must overcome the cell’s EMF and the internal resistance.

Therefore:

\[
V_{\text{charger}}=\varepsilon+I_cr
\]

Rearranging:

\[
I_c=\frac{V_{\text{charger}}-\varepsilon}{r}
\]

Substitute:

\[
I_c=\frac{9.0-6.0}{0.75}
\]

\[
\boxed{I_c=4.0\text{ A}}
\]

Part (b): Terminal potential difference

Use:

\[
V_t=\varepsilon+I_cr
\]

\[
V_t=6.0+(4.0\times0.75)
\]

\[
\boxed{V_t=9.0\text{ V}}
\]

Part (c): Explanation

During charging, current enters the positive terminal of the cell.

The charger must supply enough voltage to overcome the cell’s EMF and the voltage drop across its internal resistance.

For every coulomb passing through the cell, 6.0 J is transferred into chemical energy, while 3.0 J is dissipated internally.

Therefore, the charger must provide 9.0 J per coulomb.

This is why the terminal potential difference is greater than the EMF during charging.

The statement that terminal voltage is always smaller than EMF is incorrect. The direction of current must be considered.

Question 5

An actual cell has an EMF of 12.0 V and an internal resistance of 1.0 Ω.

Initially, the cell powers a single 4.0 Ω lamp. An ideal voltmeter is connected across the cell’s terminals.

A switch then connects a second identical 4.0 Ω lamp in parallel with the first lamp.

Assume that:

• The cell’s EMF and internal resistance remain constant.
• Both lamps behave as ohmic resistors.
• Connecting-wire resistance is negligible.
• The voltmeter is ideal.

(a) Calculate the total current supplied by the cell before and after the second lamp is connected.

(b) Calculate the terminal potential difference in both situations.

(c) Determine the current through the original lamp and its electrical power before and after the switch is closed.

(d) Calculate the internal voltage drop in both situations.

(e) Once both lamps are disconnected, what does the voltmeter read?

(f) A student argues that the current through the original lamp cannot change because it remains connected in parallel with the battery. Explain the precise error in this reasoning. Would the student’s prediction be correct if the battery were an ideal source with zero internal resistance?

Solution 5

Adding the second lamp increases the total current from 2.4 A to 4.0 A. However, the terminal potential difference decreases from 9.6 V to 8.0 V, causing the current through the original lamp to fall from 2.4 A to 2.0 A.

The cell’s EMF remains 12.0 V throughout.

Part (a): Total current before and after connecting the second lamp

Initially, there is only one external resistor.

The total circuit resistance is:

\[
R_{\text{total}}=R+r
\]

\[
R_{\text{total}}=4.0+1.0
\]

\[
R_{\text{total}}=5.0\ \Omega
\]

Therefore:

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

\[
I=\frac{12.0}{5.0}
\]

\[
\boxed{I_{\text{initial}}=2.4\text{ A}}
\]

When the second lamp is connected, the two 4.0 Ω resistors are in parallel.

Their equivalent resistance is:

\[
\frac{1}{R_{\text{eq}}}
=
\frac{1}{4.0}+\frac{1}{4.0}
\]

\[
R_{\text{eq}}=2.0\ \Omega
\]

The new total circuit resistance is:

\[
R_{\text{total}}=2.0+1.0
\]

\[
R_{\text{total}}=3.0\ \Omega
\]

Therefore:

\[
I=\frac{12.0}{3.0}
\]

\[
\boxed{I_{\text{final}}=4.0\text{ A}}
\]

The total current increases because the equivalent external resistance decreases.

Part (b): Terminal potential difference

Initially:

\[
V_t=\varepsilon-Ir
\]

\[
V_t=12.0-(2.4\times1.0)
\]

\[
\boxed{V_{t,\text{initial}}=9.6\text{ V}}
\]

After the second lamp is connected:

\[
V_t=12.0-(4.0\times1.0)
\]

\[
\boxed{V_{t,\text{final}}=8.0\text{ V}}
\]

The terminal potential difference decreases because the increased current produces a greater voltage drop across the internal resistance.

Part (c): Current and power through the original lamp

Initially, the original lamp is the only external resistor.

Its current is therefore:

\[
\boxed{I_{\text{lamp, initial}}=2.4\text{ A}}
\]

Its electrical power is:

\[
P=I^2R
\]

\[
P=(2.4)^2\times4.0
\]

\[
\boxed{P_{\text{initial}}=23.04\text{ W}}
\]

After the second lamp is connected, both lamps have the same potential difference because they are in parallel.

The potential difference across the original lamp is 8.0 V.

Using Ohm’s law:

\[
I_{\text{lamp, final}}=\frac{V}{R}
\]

\[
I_{\text{lamp, final}}=\frac{8.0}{4.0}
\]

\[
\boxed{I_{\text{lamp, final}}=2.0\text{ A}}
\]

Its new power is:

\[
P=I^2R
\]

\[
P=(2.0)^2\times4.0
\]

\[
\boxed{P_{\text{final}}=16.0\text{ W}}
\]

The original lamp’s power decreases from 23.04 W to 16.0 W. Assuming the same lamp behaviour, it becomes dimmer.

Part (d): Internal voltage drop

Before the second lamp is connected:

\[
V_r=Ir
\]

\[
V_r=2.4\times1.0
\]

\[
\boxed{V_{r,\text{initial}}=2.4\text{ V}}
\]

After the second lamp is connected:

\[
V_r=4.0\times1.0
\]

\[
\boxed{V_{r,\text{final}}=4.0\text{ V}}
\]

The internal voltage drop increases because the total current increases.

Part (e): Both lamps disconnected

When both lamps are disconnected, there is no current flowing through the cell in the ideal-voltmeter model.

Therefore:

\[
I=0
\]

Using:

\[
V_t=\varepsilon-Ir
\]

\[
V_t=12.0-(0\times1.0)
\]

\[
\boxed{V_t=12.0\text{ V}}
\]

The voltmeter now reads the cell’s EMF.

Part (f): Why the original lamp’s current changes

The student’s argument confuses two different ideas.

It is correct that parallel components have the same potential difference across them. However, this does not mean that the potential difference must remain constant when the circuit changes.

The voltage across both lamps is determined by the battery’s terminal potential difference.

When the second lamp is connected, the equivalent external resistance decreases. The total current increases, which increases the internal voltage drop.

Consequently, the terminal potential difference decreases from 9.6 V to 8.0 V.

The voltage across the original lamp decreases as well, so its current decreases.

The student’s prediction would be correct for an ideal battery with zero internal resistance.

In that case:

\[
r=0
\]

and:

\[
V_t=\varepsilon=12.0\text{ V}
\]

The original lamp would continue to receive 12.0 V, giving:

\[
I_{\text{lamp}}=\frac{12.0}{4.0}
\]

\[
\boxed{I_{\text{lamp}}=3.0\text{ A}}
\]

Its current would remain unchanged when the second lamp was connected.

The crucial difference is the internal resistance of the real cell. Parallel branches share the same voltage, but that voltage is not necessarily fixed.

This is why you must distinguish an ideal voltage source from a real battery when analysing a circuit.

Once you have identified whether a question involves EMF or potential difference, the next step is understanding how much energy a charge transfers through a component. The relationship \(W=qV\) is covered in Potential Difference and Electrical Work: Using W = qV. For questions involving potential difference in a uniform electric field, see Electric Field and Potential Difference Using E = V/d.