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Potential difference and electromotive force (EMF) are two of the most important concepts in electricity. Though both are measured in volts, they describe fundamentally different things. Understanding the distinction is essential for circuit analysis.
Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.2 — Current/voltage characteristics, specifically the defining relationships V = W/Q and ε = W/Q, the role of EMF as the energy-supply quantity per coulomb in a source, and the connection between p.d., current and power through P = IV, P = I²R and P = V²/R. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- 3.2.1 — Particles and quantum phenomena: the electronvolt introduced at the end of this lesson is the natural energy unit used throughout particle physics (photoelectric work functions, accelerator energies, mass–energy of fundamental particles). The relationship 1 eV = eV is exactly the W = QV relation applied to one electronic charge.
- 3.4 — Mechanics, energy and momentum: the W = QV used here is the same work-energy theorem as F·s in mechanics — charge times potential difference is the electrical analogue of force times displacement, with V playing the role of a potential gradient integrated along the path.
- 3.5.1.6 — EMF and internal resistance: the distinction between EMF (energy supplied per coulomb) and terminal p.d. (energy delivered per coulomb to the external circuit) becomes operational in Lesson 7, where lost volts V_lost = Ir is the difference between the two. The careful definitional work done here pays off there.
Potential difference (p.d.) between two points is the energy transferred per unit charge as charge moves between those two points.
V=QW
where:
One volt equals one joule per coulomb (1 V = 1 J C⁻¹).
When charge flows through a component (such as a resistor or lamp), electrical energy is transferred to other forms (heat, light, etc.). The p.d. across the component tells us how much energy is transferred per coulomb of charge.
Key Definition: The potential difference across a component is the work done (energy transferred) per unit charge as charge passes through the component.
A 12 V battery drives a charge of 50 C through a circuit. Calculate the total energy transferred.
Solution:
V = W/Q, so W = QV
W = 50 × 12 = 600 J
A heater operates at 230 V and transfers 46 kJ of energy. How much charge flows through it?
Solution:
W = 46 kJ = 46 000 J
Q = W/V = 46 000 / 230 = 200 C
EMF (ε) is the energy transferred per unit charge by a source (battery, cell, generator, solar cell) in driving charge around a complete circuit.
ε=QW
where:
EMF is also measured in volts (V) because it has the same dimensions as potential difference (energy per unit charge).
| EMF (ε) | Potential Difference (V) | |
|---|---|---|
| What it measures | Energy supplied per unit charge by a source | Energy transferred per unit charge by a component |
| Where it applies | Sources (batteries, cells, generators) | Components (resistors, lamps, motors) |
| Energy conversion | Other forms → electrical energy | Electrical energy → other forms |
Common Misconception: EMF is not a "force" despite the name. It is an energy per unit charge (measured in volts), not a force (which would be measured in newtons). The name is historical and unfortunately misleading.
When current I flows through a component with resistance R, the potential difference across it is:
V=IR
This is one of the most frequently used equations in electricity. It can be rearranged to:
A 470 Ω resistor carries a current of 25 mA. Calculate the potential difference across it.
Solution:
I = 25 mA = 25 × 10⁻³ = 0.025 A
V = IR = 0.025 × 470 = 11.75 V
A 6.0 V battery is connected across a 240 Ω resistor (assume the battery has no internal resistance). Calculate the current.
Solution:
I = V/R = 6.0 / 240 = 0.025 A = 25 mA
Power is the rate of energy transfer:
P=tW
Since W = QV and Q = It:
P=tQV=tIt×V=IV
So: P = IV
Using V = IR, we can derive two more forms:
P=IV=I(IR)=I2R
P=IV=RV×V=RV2
| Power equation | Best used when you know |
|---|---|
| P = IV | Current and voltage |
| P = I²R | Current and resistance |
| P = V²/R | Voltage and resistance |
A 60 W filament lamp operates at 230 V. Calculate (a) the current through it and (b) its resistance when operating.
Solution:
(a) P = IV, so I = P/V = 60 / 230 = 0.261 A
(b) R = V/I = 230 / 0.261 = 882 Ω
Alternatively: R = V²/P = 230² / 60 = 52900 / 60 = 882 Ω ✓
A 100 Ω resistor carries a current of 0.50 A for 5 minutes. Calculate the energy dissipated as heat.
Solution:
P = I²R = (0.50)² × 100 = 0.25 × 100 = 25 W
t = 5 × 60 = 300 s
W = Pt = 25 × 300 = 7500 J = 7.5 kJ
Alternatively: W = I²Rt = (0.50)² × 100 × 300 = 7500 J ✓
Exam Tip: Learn all three forms of the power equation. The examiner may give you different combinations of known quantities. Using the correct form avoids unnecessary intermediate calculations and reduces rounding errors.
The electronvolt is a unit of energy commonly used in atomic and particle physics. It is the energy gained by an electron when it is accelerated through a potential difference of 1 volt.
1 eV=1.60×10−19 J
This comes from W = QV: if Q = e = 1.60 × 10⁻¹⁹ C and V = 1 V, then W = 1.60 × 10⁻¹⁹ J.
An electron is accelerated through a p.d. of 5000 V. Calculate its kinetic energy in (a) electronvolts and (b) joules.
Solution:
(a) By definition, an electron accelerated through 5000 V gains 5000 eV = 5.0 keV
(b) E = QV = 1.60 × 10⁻¹⁹ × 5000 = 8.0 × 10⁻¹⁶ J
Check: 5000 × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹⁶ J ✓
A 12 V car battery is used to power a 36 W bulb for 2.0 hours. Calculate (a) the current drawn, (b) the resistance of the bulb when operating, (c) the total charge passing through it in the 2.0 hours, and (d) the total energy delivered, expressed in both joules and kilowatt-hours.
Solution:
(a) Using P = IV, I = P/V = 36/12 = 3.0 A.
(b) Using V = IR, R = V/I = 12/3.0 = 4.0 Ω. Check using P = V²/R: 144/4.0 = 36 W ✓.
(c) t = 2.0 × 3600 = 7200 s. Q = It = 3.0 × 7200 = 2.16 × 10⁴ C (about 22 kC).
(d) Total energy W = Pt = 36 × 7200 = 2.59 × 10⁵ J = 259 kJ. In kilowatt-hours: W = 36 × 10⁻³ kW × 2.0 h = 0.072 kWh.
The kWh is the unit your electricity meter actually uses; converting between SI joules and "domestic" kWh is a common A-Level synoptic move. The conversion factor is 1 kWh = 3.6 × 10⁶ J.
A 12 V battery delivers 4,800 J of energy to a circuit while powering a small device. (a) Calculate the total charge that has flowed through the device. (b) Determine the number of electrons that this corresponds to. (c) Suppose the device is replaced with a load that draws the same charge in the same time, but the battery now has a non-zero internal resistance r = 0.50 Ω and the terminal p.d. measured across the load is only 11.4 V. By comparing the energy delivered to the external load with the energy dissipated inside the battery, evaluate the efficiency of energy transfer.
Solution — Part (a): Total charge.
The defining relationship between energy, charge and potential difference is W = QV, where V here is the e.m.f. of the cell (12 V) because all the energy reaches the external circuit when internal resistance is ignored.
Rearranging: Q = W / V = 4,800 / 12 = 400 C.
A charge of 400 coulombs is substantial — for context, a typical lithium-ion mobile-phone battery stores around 10,000 C of charge at 3.7 V. So our scenario corresponds to roughly a 1.2 Wh task, comparable to running a small LED torch for an hour.
Part (b): Number of electrons.
Each electron carries the elementary charge e = 1.60 × 10⁻¹⁹ C.
N = Q / e = 400 / (1.60 × 10⁻¹⁹) = 2.50 × 10²¹ electrons.
Sanity check: 2.5 × 10²¹ is around a millionth of Avogadro's number, which is a sensible figure for a moderate-current circuit running for a few minutes. If you ever calculate an N that comes out smaller than about 10¹⁰ or larger than about 10²⁵ for a typical lab scenario, suspect a unit error.
Part (c): Energy budget with internal resistance.
In this revised scenario, the e.m.f. ε = 12 V is still the energy each coulomb leaves the cell with, but only a terminal p.d. of V_terminal = 11.4 V actually appears across the external load. The remaining 0.6 V is "lost" to the internal resistance.
For the same charge transfer Q = 400 C:
Energy delivered to the external load: W_ext = Q × V_terminal = 400 × 11.4 = 4,560 J
Energy dissipated inside the battery: W_internal = Q × (ε − V_terminal) = 400 × 0.6 = 240 J
Total chemical energy used: W_total = Q × ε = 400 × 12 = 4,800 J (matches the original scenario, as it must).
Energy-transfer efficiency: η = W_ext / W_total = 4,560 / 4,800 = 0.95 = 95%.
Interpretation: Even with a modest internal resistance of half an ohm, the battery loses 5% of its stored chemical energy as heat inside itself. This is why batteries get warm during heavy use — and why high-current applications such as starter motors in cars demand batteries with very low internal resistance (typically 0.01 Ω or less). The remaining lesson on e.m.f. and internal resistance will formalise this with the equation ε = I(R + r), but the conceptual point is already on the table here: the e.m.f. is the "raw" energy per coulomb supplied by the chemistry of the cell, whereas the terminal p.d. is what the external circuit actually receives. The two coincide only in the limit of zero internal resistance, which is an idealisation that no real cell ever quite reaches.
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