CP8 – EMF and Internal Resistance | Edexcel A Level Physics

Theory — EMF and Internal Resistance

Understanding why a real cell has a lower terminal voltage than its EMF

Objectives

Understand the difference between EMF (ε) and terminal voltage (V)
Use the equation ε = V + Ir to determine EMF and internal resistance of a cell
Plot a V vs I graph — gradient = −r, y-intercept = ε
Compare ε and r for different cell types (alkaline, lead-acid, lithium)

EMF and Terminal Voltage

The electromotive force (EMF) ε is the energy supplied per unit charge by the cell — it is the open-circuit voltage (when no current flows). The terminal voltage V is the potential difference across the cell terminals when current flows. It is always less than ε because some energy is lost driving current through the cell's own internal resistance r.

ε = V + Ir Rearranged: V = ε − Ir Also: ε = I(R + r)

ε = EMF (V) · V = terminal voltage (V) · I = current (A) · r = internal resistance (Ω) · R = external resistance (Ω)

The V vs I Graph

From V = ε − Ir, this is a straight line in the form y = c + mx where:

y-intercept = ε (terminal voltage when I = 0, i.e. open circuit) gradient = −r (negative because V decreases as I increases)

The x-intercept occurs at I = ε/r — the short-circuit current (when R = 0). In practice never reach this — it would damage the cell.

A steeper (more negative) gradient means a higher internal resistance. A battery with high r loses more voltage under load — it is less useful for high-current applications.

Cell Types Compared

Cell	ε / V	r / Ω	Notes
AA Alkaline	1.50	0.80	Common household cell. r increases significantly as it discharges.
Lead-acid cell	2.00	0.05	Very low r — can deliver large currents (car starter motors).
Lithium cell	3.70	0.20	High ε, low r. Good for portable electronics requiring steady voltage.

Lost Volts

The voltage drop across the internal resistance is called the lost volts:

Lost volts = Ir = ε − V

When I = 0 (open circuit): lost volts = 0, terminal voltage = ε.
When I is large: lost volts = Ir is large, terminal voltage drops significantly.

Procedure

Two methods to vary external resistance — use both for the most reliable results.

Equipment

Electrical cell (AA alkaline, lead-acid or lithium) · Variable resistor (rheostat) · Resistance substitution box · Ammeter · Voltmeter · Switch · Connecting leads

Important: Always include a switch — open it between readings to prevent the cell discharging when not measuring. Never short-circuit the cell (R = 0).

1

Set up the circuit

Connect: cell → switch → ammeter → variable resistor → back to cell. Connect the voltmeter directly across the cell terminals. The ammeter measures total circuit current I; the voltmeter measures terminal voltage V.

💡 The voltmeter must connect directly to the cell terminals — not via the ammeter — to measure the true terminal voltage including the internal resistance voltage drop.

2

Open-circuit reading

With the switch open (no current flowing), read the voltmeter. This gives the open-circuit voltage ≈ ε. Record this as your first data point: I = 0, V = ε.

💡 The open-circuit voltage is a very good approximation of ε as long as the voltmeter draws negligible current (high voltmeter resistance). Record it at the start before the cell has discharged at all.

3

Method A — Rheostat sweep

Close the switch. Set the rheostat to maximum resistance (minimum current). Record V and I. Slowly decrease the resistance to increase the current. Take readings at roughly equal current intervals across the full range.

💡 Take readings quickly and open the switch between readings to prevent the cell from discharging. Aim for 8–10 readings spread from minimum to maximum current.

4

Method B — Resistance substitution box

Replace the rheostat with a substitution box. Select discrete resistance values (e.g. 20, 10, 5, 3, 2, 1 Ω). For each R, close the switch briefly, read V and I, then open the switch. Calculate R = V/I to verify each setting.

💡 The substitution box gives precise, reproducible resistance values — useful when you need to repeat readings or compare results between different cells at the same R.

5

Plot V vs I and find ε and r

Plot terminal voltage V (y-axis) against current I (x-axis). Draw a best-fit straight line. The y-intercept = ε. The gradient = −r, so r = |gradient|.

💡 The line should be straight — a curved line indicates the internal resistance is changing (common in cells as they discharge or heat up). Use the gradient of the line rather than individual readings to reduce the effect of random errors.

Cell Type

External Resistance

Rheostat 10.0 Ω

1 Ω (high I)20 Ω (low I)

Substitution Box

Click to connect a fixed resistance. Rheostat adds in series.

Switch

Terminal Voltage V

— V

open switch to start

Voltage V

—

Current I

—

R_ext

—

Lost volts

—

Readings: 0/8 minimum

Results Data Table

V = ε − Ir. Plot V (y-axis) against I (x-axis). Gradient = −r, y-intercept = ε.

#	Cell	R_ext / Ω	V / V	I / A	R = V/I / Ω	Lost volts = ε − V / V	Ir check / V
No data yet.

Graph & Analysis — V vs I

Terminal voltage V against current I. Straight line: gradient = −r, y-intercept = ε.

Results from graph

—V — EMF ε (y-intercept)

—Ω — internal resistance r

Graph parameters

Gradient (= −r)—

y-intercept (= ε)—

R²—

Expected ε—

Expected r—

% error ε—

% error r—

Interpretation

Record at least 3 readings to see analysis.

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

Explain why the terminal voltage of a cell is always less than its EMF when current is flowing, and what happens to the terminal voltage as the external resistance decreases.

When current I flows, there is a voltage drop Ir across the internal resistance r of the cell. By Kirchhoff's voltage law, the terminal voltage V = ε − Ir. Since Ir > 0 whenever I > 0, the terminal voltage is always less than the EMF. As external resistance R decreases, the current I = ε/(R+r) increases, so the lost volts Ir increases and the terminal voltage V = IR decreases. At R = 0 (short circuit), V = 0 and all the EMF drives current through the internal resistance. In practice this would cause dangerous overheating and is never done deliberately.

Q2

Explain why the y-intercept of the V vs I graph gives the EMF, and why the gradient gives −r.

From V = ε − Ir, this is of the form y = c + mx where y = V, x = I, c = ε and m = −r. When I = 0 (extrapolating to the y-axis), V = ε — the y-intercept equals the EMF. This corresponds to the open-circuit condition where no current flows and there is no voltage drop across r. The gradient m = ΔV/ΔI = −r — negative because terminal voltage decreases as current increases. Taking the magnitude: r = |gradient| = |ΔV/ΔI|.

Q3

A lead-acid cell has much lower internal resistance than an AA alkaline cell. Explain why this makes lead-acid cells more suitable for starting a car engine.

A car starter motor requires a very large current (100–200 A) for a short time. From I = ε/(R+r), the current is determined by total resistance R+r. The starter motor has a very low resistance R (it is essentially a short circuit during starting). If r is large, most of the voltage is dropped across the internal resistance (lost volts = Ir) leaving very little terminal voltage to drive the motor, and the current I = ε/(R+r) is limited. A lead-acid cell with r ≈ 0.01–0.05 Ω allows very large currents with minimal voltage drop. An alkaline cell with r ≈ 0.5–2 Ω would lose most of its voltage internally at such currents and could not deliver enough power to the motor.

Q4

A student finds that their V vs I graph is curved rather than a straight line, curving downward at high currents. Suggest a reason for this and explain how it affects the determination of ε and r.

The internal resistance r increases as the current increases. This can happen because: (1) the cell heats up due to power dissipation P = I²r — for most cells, resistance increases with temperature; (2) chemical polarisation effects in the cell — reactants near the electrodes become depleted at high currents, increasing effective resistance. The result is that V decreases faster than expected at high currents. If a straight line is forced through curved data, the gradient overestimates r (too steep) and the y-intercept may underestimate ε. To minimise this effect, take readings quickly, open the switch between readings to allow the cell to recover, and restrict the current range to avoid high currents where the effect is most pronounced.

Q5

Why is it important to include a switch in the circuit and open it between readings? What would happen if the switch were left closed throughout?

If the switch is left closed, current flows continuously through the cell, causing it to discharge. As the cell discharges: its EMF ε decreases (fewer chemical reactants remain) and its internal resistance r increases (the cell's condition deteriorates). This means the values of ε and r are not constant throughout the experiment — early readings are taken from a fresh cell while later readings come from a partially discharged cell. The V vs I graph would not be a straight line, and any ε and r determined from it would not represent the initial state of the cell. Opening the switch between readings prevents this by allowing the cell to rest and minimising the total charge removed. The readings at different R values are then comparable as they are all taken from approximately the same cell state.