What you'll learn
This topic examines how real batteries and cells behave in electrical circuits, focusing on electromotive force (EMF) and internal resistance. You'll learn why terminal voltage differs from EMF, how to calculate circuit quantities when internal resistance is present, and how to interpret experimental results. This topic appears regularly on CXC CSEC Physics Paper 02 and Paper 03, typically worth 6-12 marks per examination.
Key terms and definitions
Electromotive Force (EMF, ε) — the total energy supplied by a cell or battery per coulomb of charge that passes through it, measured in volts (V). EMF represents the maximum potential difference available when no current flows.
Internal resistance (r) — the opposition to current flow within a cell or battery itself, caused by the resistance of the electrolyte and electrodes, measured in ohms (Ω).
Terminal voltage (V or p.d.) — the potential difference across the terminals of a cell when current flows through an external circuit. Terminal voltage is always less than EMF when current flows due to voltage drop across internal resistance.
Lost volts — the voltage drop across the internal resistance of a cell, calculated as Ir, where I is the current flowing through the circuit.
Load resistance (R) — the external resistance connected across a cell's terminals, which could be a single resistor or combination of resistors.
Short circuit current — the maximum current that flows when a cell's terminals are connected directly together (when external resistance is zero), calculated as ε/r.
Core concepts
Understanding electromotive force
EMF is not actually a force despite its name — it's an energy concept. When a cell converts chemical energy (or other forms) into electrical energy, the EMF quantifies how much energy each coulomb of charge receives.
For a typical AA battery used in torches throughout Trinidad and Tobago, the EMF might be 1.5 V. This means each coulomb passing through the battery gains 1.5 joules of electrical energy.
Key characteristics of EMF:
- Measured in volts (V)
- Represents energy per unit charge (J/C)
- Remains constant for a healthy cell regardless of current drawn
- Can only be measured directly when no current flows (open circuit)
- Symbol: ε (Greek letter epsilon) or E
The relationship between EMF and energy is: ε = W/Q
Where W is work done or energy transferred (J) and Q is charge (C).
Internal resistance in real cells
No cell is perfect. Every battery and cell possesses internal resistance due to:
- Resistance of the electrolyte solution or paste
- Resistance at electrode surfaces
- Resistance of connecting strips inside the cell
Internal resistance causes two important effects:
- Voltage drop inside the cell — when current flows, some voltage is "lost" across the internal resistance, reducing the voltage available at the terminals
- Energy dissipation as heat — power is wasted inside the cell (P = I²r), which is why batteries get warm during use
The internal resistance of a new D-cell battery might be 0.1-0.5 Ω, while a car battery used in vehicles across Jamaica and other Caribbean islands typically has very low internal resistance (around 0.01-0.05 Ω) to deliver high currents for starting engines.
As batteries age or discharge:
- Internal resistance increases
- Terminal voltage decreases more rapidly under load
- Battery becomes less effective at delivering power
The relationship between EMF, terminal voltage, and internal resistance
When a cell is connected in a complete circuit and current flows, the terminal voltage is always less than the EMF. The fundamental equation is:
ε = V + Ir
Or rearranged: V = ε - Ir
Where:
- ε = EMF of the cell (V)
- V = terminal voltage (V)
- I = current in the circuit (A)
- r = internal resistance (Ω)
The term Ir represents the "lost volts" — voltage dropped across the internal resistance.
This equation can also be written using Ohm's law for the external circuit (V = IR):
ε = IR + Ir ε = I(R + r)
This form shows that the total EMF equals the current multiplied by the total resistance (external plus internal).
Circuit calculations with internal resistance
When solving problems involving internal resistance, treat the circuit as having two resistances in series:
- External load resistance R
- Internal resistance r
The total resistance in the circuit is (R + r), so the current flowing is:
I = ε/(R + r)
Once you know the current, you can find:
- Terminal voltage: V = IR or V = ε - Ir
- Lost volts: Ir
- Power delivered to load: P = I²R or P = VI
- Power wasted internally: P = I²r
Experimental determination of EMF and internal resistance
CXC CSEC Physics Paper 03 often tests the practical investigation of EMF and internal resistance. The standard method involves:
Procedure:
- Connect a cell in series with a variable resistor and ammeter
- Connect a voltmeter across the cell terminals
- Vary the resistance and record pairs of current (I) and terminal voltage (V) values
- Plot a graph of V (y-axis) against I (x-axis)
Analysis of results:
The equation V = ε - Ir is in the form y = mx + c, where:
- y-intercept = ε (EMF)
- Gradient (slope) = -r (negative of internal resistance)
The graph produces a straight line with:
- Y-intercept: where I = 0, V = ε (the EMF)
- Negative slope: the magnitude equals internal resistance
- X-intercept: where V = 0, I = ε/r (short circuit current)
Safety considerations for the practical:
- Avoid short-circuiting the cell (causes overheating and possible leakage)
- Use appropriate range on meters
- Don't allow large currents to flow for extended periods
- Ensure connections are secure to avoid heating at contacts
Factors affecting internal resistance
Several factors influence a cell's internal resistance:
Temperature: Higher temperature generally decreases internal resistance because electrolyte conductivity increases. This is why car batteries in hot Caribbean climates may perform differently than in cold regions.
Age and usage: As cells age, chemical degradation increases internal resistance, explaining why old batteries deliver less current.
State of charge: Discharged batteries have higher internal resistance than fully charged ones.
Cell construction: Larger cells typically have lower internal resistance because they have more electrode surface area and more electrolyte.
Current demand: Very high current draws can temporarily increase internal resistance due to depletion of ions near electrodes.
Worked examples
Example 1: Basic EMF and internal resistance calculation
A cell has an EMF of 1.5 V and internal resistance of 0.5 Ω. It is connected to a 4.5 Ω resistor.
(a) Calculate the current in the circuit. [2 marks]
Solution: Total resistance = R + r = 4.5 + 0.5 = 5.0 Ω
Using I = ε/(R + r): I = 1.5/5.0 = 0.30 A
(b) Calculate the terminal voltage. [2 marks]
Solution: Method 1: V = IR = 0.30 × 4.5 = 1.35 V
Method 2: V = ε - Ir = 1.5 - (0.30 × 0.5) = 1.5 - 0.15 = 1.35 V
(c) Calculate the lost volts. [1 mark]
Solution: Lost volts = Ir = 0.30 × 0.5 = 0.15 V
Example 2: Determining internal resistance from measurements
A student at a school in Barbados tests a battery and obtains these measurements:
| Current (A) | Terminal voltage (V) |
|---|---|
| 0.0 | 6.0 |
| 1.0 | 5.5 |
| 2.0 | 5.0 |
| 3.0 | 4.5 |
(a) State the EMF of the battery. [1 mark]
Solution: EMF = 6.0 V (terminal voltage when I = 0)
(b) Determine the internal resistance of the battery. [3 marks]
Solution: Using any two data points with V = ε - Ir:
When I = 2.0 A, V = 5.0 V: 5.0 = 6.0 - 2.0r 2.0r = 1.0 r = 0.5 Ω
Or calculate gradient: change in V/change in I = (4.5 - 6.0)/(3.0 - 0.0) = -1.5/3.0 = -0.5 Internal resistance = 0.5 Ω
(c) Calculate the power dissipated internally when the current is 3.0 A. [2 marks]
Solution: Power = I²r = (3.0)² × 0.5 = 9.0 × 0.5 = 4.5 W
Example 3: Application problem
A car battery in Jamaica has an EMF of 12 V and internal resistance of 0.02 Ω. Calculate the terminal voltage when the starter motor draws 200 A.
Solution: [3 marks] V = ε - Ir V = 12 - (200 × 0.02) V = 12 - 4 V = 8 V
This example shows why car batteries need very low internal resistance — even 0.02 Ω causes a significant 4 V drop at high currents.
Common mistakes and how to avoid them
Mistake: Confusing EMF with terminal voltage and using them interchangeably in calculations. Correction: Remember that terminal voltage V is what you measure across the terminals when current flows, while EMF ε is the maximum voltage available (measured with no current). Always use V = ε - Ir to relate them.
Mistake: Forgetting to add internal resistance to external resistance when calculating total circuit resistance. Correction: Total resistance is always R + r. The internal resistance is in series with external components. Use I = ε/(R + r) for current calculations.
Mistake: Calculating lost volts as εr instead of Ir. Correction: Lost volts depend on current flow. The voltage drop across internal resistance follows Ohm's law: V = IR, so lost volts = I × r (not ε × r).
Mistake: Stating that internal resistance causes energy loss from the circuit. Correction: Energy is not lost but converted to thermal energy inside the cell. The total energy from the EMF is conserved; some is delivered to the external load (useful) and some heats the cell (wasted).
Mistake: Believing that EMF changes when current changes. Correction: EMF remains constant for a healthy cell regardless of current (until the cell is significantly discharged). What changes is the terminal voltage, which decreases as current increases according to V = ε - Ir.
Mistake: Incorrectly identifying the y-intercept or gradient when analyzing V-I graphs in practicals. Correction: On a V vs I graph, the y-intercept (where the line crosses the V-axis) is the EMF, and the gradient (slope) is -r. The gradient is negative because terminal voltage decreases as current increases.
Exam technique for Electromotive Force and Internal Resistance
Command word awareness: Questions using "calculate" require numerical answers with units and working shown (typically 2-3 marks). "Explain" questions need physical reasoning about energy transfers or voltage drops (2-3 marks). "Determine from the graph" means reading values or calculating gradient/intercept (1-2 marks per value).
Show your working: Even if your final answer is incorrect, you can earn method marks by showing clear steps. Always write the equation first (e.g., V = ε - Ir), substitute values, then calculate. For Paper 02, 3-mark calculations typically award 1 mark for correct formula, 1 for substitution, 1 for answer with unit.
Graph skills for practicals: When plotting V against I in Paper 03, ensure your graph fills at least half the page, use sensible scales, label axes with quantities and units, and draw the best-fit line with a ruler. To find internal resistance, calculate gradient using points far apart on your line (not original data points), showing the triangle clearly. State the y-intercept as the EMF.
Unit awareness: EMF and voltage are in volts (V), current in amperes (A), resistance in ohms (Ω), and power in watts (W). Lost volts must have the same unit as EMF (V). Always include correct units in final answers for full marks.
Quick revision summary
EMF (ε) is the energy per coulomb supplied by a cell, measured in volts. Internal resistance (r) causes voltage drop inside the cell when current flows. Terminal voltage V = ε - Ir, where Ir represents lost volts. Total circuit resistance is R + r, giving current I = ε/(R + r). In practicals, plotting V against I gives a straight line with y-intercept = EMF and gradient = -r. As batteries age, internal resistance increases and performance decreases. Always distinguish between EMF (maximum voltage with no current) and terminal voltage (actual voltage when current flows). Power dissipated internally is I²r.