Edexcel IAL Physics
The full notes page in 12 sections · live simulations · practice questions
Press F for full screen · → or click to move on
ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only.
Pearson Edexcel and Cambridge are trademarks of their owners; ExaminerPrep is not affiliated with any exam board.
E.m.f. as the energy transferred per unit charge by a source, and how it differs from potential difference
Internal resistance, terminal p.d. and lost volts; ε= I(R + r) and V = ε- Ir
Finding ε and r from meter readings
The maximum (short-circuit) current a source can give
Power delivered to a load and power dissipated inside the source
Cells connected in series, including a cell connected the wrong way round
Core Practical 8: determining the e.m.f. and internal resistance of a cell
A real source behaves like a perfect e.m.f. ε in series with an internal resistance r, so its terminal p.d. falls as the current it supplies rises.
E.m.f. = energy transferred to electrical energy per unit charge by the source; p.d. = energy transferred from electrical energy per unit charge. Both are in volts (J C⁻¹).
ε= I(R + r), so the terminal p.d. is V = ε- Ir. The Ir term is the lost volts. With no current, V = ε.
Short circuit (R = 0): maximum current ε/r, with all the power dissipated inside the source.
Power to the load is greatest when R = r, but then only half the energy reaches the load.
Plot V against I: the V-intercept is ε and the gradient is −r.
Until now every battery has been perfect: its voltage never changed. Real sources are not like that. A torch dims as the battery works harder, and a car's lights dim when the engine starts. One extra resistor, hidden inside the source, explains all of this. First, the two energy definitions need to be sharp.
E.m.f. ε of a source: the energy transferred to electrical energy per unit charge passing through the source.
Potential difference V across a component: the energy transferred from electrical energy per unit charge passing through the component.
As an equation, ε= W/Q, where W is the energy transferred to electrical energy by the source (J) and Q the charge passing through it (C). Not in the formulae list: learn it, noting that it has the same form as V = W/Q, which is. 1 V = 1 J C⁻¹, so a 1.5 V cell gives each coulomb that passes through it 1.5 J of electrical energy.
| E.m.f. ε | Potential difference V | |
|---|---|---|
| Where | a source: cell, battery, generator, solar cell | a component: resistor, lamp, motor, or across the terminals of a source |
| Energy transfer | into electrical energy (from chemical, kinetic, light…) | out of electrical energy (to thermal, light, kinetic…) |
| Unit | volt = J C⁻¹ | volt = J C⁻¹ |
| Depends on current? | no: a property of the source | yes, for the terminal p.d. of a real source |
Despite the name, "electromotive force" is not a force; it is an energy per unit charge. Around a complete circuit energy is conserved, so the e.m.f. equals the sum of all the p.d.s, including the one across the internal resistance.
WORKED EXAMPLE · QUESTION
A cell of e.m.f. 1.50 V drives a current through a torch bulb until 540 C of charge has passed. The p.d. across the cell's terminals is 1.35 V throughout. Calculate (a) the energy transferred by the cell, (b) the energy transferred to the bulb and (c) the energy dissipated inside the cell.
SOLUTION
(a) The e.m.f. gives the energy supplied per coulomb: W = εQ = 1.50 ×540 = 810 J.
(b) The terminal p.d. gives the energy delivered to the external circuit per coulomb: W = VQ = 1.35 ×540 = 729 J.
(c) The difference stays inside the cell: 810 - 729 = 81 J (0.15 J for every coulomb). The 0.15 V "missing" from the terminals is the energy per coulomb dissipated by the cell's internal resistance.
Internal resistance r: the resistance of the materials inside a source, which dissipates energy whenever current flows through the source.
Charge moving through the chemicals and electrodes of a cell meets resistance, just as it does in a wire. The model is a perfect e.m.f. ε in series with a resistor r; you cannot separate them or connect to the point between them. The terminal p.d. V is the p.d. across the terminals of the source, which is what a voltmeter across it reads. The lost volts Ir is the p.d. across the internal resistance: the energy per coulomb dissipated as thermal energy inside the source.
Apply V = IR to the whole circuit, whose total resistance is R + r:
ε= I(R + r) ⇒ ε= IR + Ir = V + Ir ⇒ V = ε- Ir
Here R is the total external resistance. ε= I(R + r) is in the formulae list; V = ε- Ir comes from expanding the brackets and using V = IR for the external circuit. In words: e.m.f. = terminal p.d. + lost volts, which is energy conservation for each coulomb going once round the circuit.
WORKED EXAMPLE · QUESTION
A cell of e.m.f. 1.50 V and internal resistance 0.80 Ω is connected to a 4.2 Ω resistor. Calculate (a) the current, (b) the terminal p.d. and (c) the lost volts.
SOLUTION
(a) The current depends on the total resistance: I = ε/(R + r) = 1.50/(4.2 + 0.80) = 0.300 A.
(b) The terminal p.d. is the p.d. across the external resistor: V = IR = 0.300 ×4.2 = 1.26 V.
(c) Ir = 0.300 ×0.80 = 0.24 V. Check: 1.26 + 0.24 = 1.50 V = ε.
How the terminal p.d. depends on the current
V = ε- Ir has the form y = c + mx, so a graph of V against I is a straight line with intercept ε and gradient −r. Reducing R increases I, increases the lost volts and so reduces the terminal p.d. With no current (open circuit) there are no lost volts, so V = ε.
Common error
Using I = ε/R and forgetting the internal resistance. Here that would give 1.50/4.2 = 0.357 A instead of 0.300 A. The e.m.f. drives current through the whole circuit, and r is part of it.
The e.m.f.: connect a high-resistance voltmeter (for example a digital meter) across the source with nothing else connected. The current is negligible, so the lost volts are negligible and the reading is very close to ε.
The internal resistance: connect a known load and measure the terminal p.d. V and the current I. Then lost volts = ε- V and r = (ε- V)/I.
If there is no ammeter, find I from the load: I = V/R.
WORKED EXAMPLE · QUESTION
A high-resistance voltmeter connected across a battery reads 6.20 V. A 5.0 Ω resistor is then connected across the battery, and the voltmeter reading falls to 5.40 V. Determine the internal resistance of the battery.
SOLUTION
With no load, the reading is the e.m.f.: ε= 6.20 V. Current through the resistor: I = V/R = 5.40/5.0 = 1.08 A. Lost volts: Ir = 6.20 - 5.40 = 0.80 V, so r = 0.80/1.08 = 0.74 Ω.
Exam tip
When you meet "a high-resistance voltmeter is connected across the cell" with no other load, read it as "this is the e.m.f." and say why: the current is (almost) zero, so Ir is (almost) zero.
The current is largest when the external resistance is zero, a short circuit. Then ε= Ir, so I_(max) = ε/r, the intercept on the current axis of the V–I graph. In a short circuit the terminal p.d. is zero and all the power, I²r, is dissipated inside the source, which can heat it dangerously.
Low r wanted: car batteries and rechargeable packs must deliver large currents without a large drop in terminal p.d.
High r wanted: high-voltage laboratory supplies are built with a very large internal resistance so that, even if the terminals are touched or shorted, the current stays small and safe.
WORKED EXAMPLE · QUESTION
A car battery has an e.m.f. of 12.6 V and an internal resistance of 0.020 Ω. While the starter motor turns the engine, the current from the battery is 180 A. (a) Calculate the terminal p.d. while the starter motor runs. (b) Calculate the rate at which energy is dissipated inside the battery. (c) Calculate the current if the terminals were short-circuited.
SOLUTION
(a) Lost volts Ir = 180 ×0.020 = 3.6 V, so V = 12.6 - 3.6 = 9.0 V.
(b) P = I²r = 180² ×0.020 = 650 W (648 W).
(c) R = 0: I_(max) = ε/r = 12.6/0.020 = 630 A.
While the starter runs, the headlamps get 9.0 V instead of about 12.6 V and visibly dim.
Common error
Thinking a battery delivers a fixed current. A battery has a (nearly) fixed e.m.f.; the current depends on the total resistance it is connected to, and the terminal p.d. depends on that current.
Multiply ε= V + Ir by the current:
εI = VI + I²r
Power supplied by the source = power delivered to the load + power dissipated inside. The power delivered to the load is P = VI = I²R = V²/R. The fraction of the energy that reaches the load (the efficiency) is VI/εI = V/ε= R/(R + r). For a fixed source, the power in the load is greatest when R = r, but then only half the energy reaches the load.
WORKED EXAMPLE · QUESTION
A supply of e.m.f. 6.0 V and internal resistance 2.0 Ω is connected in turn to loads of 1.0 Ω, 2.0 Ω and 4.0 Ω. For each load, calculate the current, the power delivered to the load and the fraction of the supplied energy that reaches the load. Comment on the results.
SOLUTION
Use I = ε/(R + r), then P = I²R and fraction = R/(R + r):
| R / Ω | I / A | power in load / W | power inside / W | fraction to load |
|---|---|---|---|---|
| 1.0 | 2.0 | 4.0 | 8.0 | 33% |
| 2.0 | 1.5 | 4.5 | 4.5 | 50% |
| 4.0 | 1.0 | 4.0 | 2.0 | 67% |
Greatest power, 4.5 W, when R = r = 2.0 Ω, but only 50% of the energy then reaches the load. A larger load receives less power but wastes less: 67% of the energy reaches the 4.0 Ω load. Sources such as mains generators are designed for efficiency, not maximum power.
Exam tip
Always say which power you have calculated. εI is the power supplied by the source, VI or I²R the power delivered to the load, and I²r the power wasted inside.
Cells in series, all the same way round (+ to −): the e.m.f.s add and the internal resistances add. A cell connected the wrong way round subtracts its e.m.f., so the total drops by 2ε compared with all cells correct, but its internal resistance still adds. The internal resistances add because they are in series with each other and with the external circuit, exactly like resistors in series.
WORKED EXAMPLE · QUESTION
A torch holds four cells in series, each of e.m.f. 1.5 V and internal resistance 0.30 Ω. The bulb has a resistance of 6.0 Ω. (a) Calculate the current in the bulb and the p.d. across it. (b) One cell is put in the wrong way round. Calculate the new current, and the factor by which the power in the bulb falls.
SOLUTION
(a) Totals: ε= 4 ×1.5 = 6.0 V and r = 4 ×0.30 = 1.20 Ω. I = 6.0/(6.0 + 1.2) = 0.83 A (0.833 A) and V = 0.833 ×6.0 = 5.0 V.
(b) One cell reversed: ε= 1.5 + 1.5 - 1.5 + 1.5 = 3.0 V, r still 1.20 Ω. I = 3.0/7.2 = 0.42 A (0.417 A). Power ∝ I² for the same bulb resistance, so P falls from 4.17 W to 1.04 W, a factor of 4.
This treats the bulb's resistance as constant; in reality a dimmer filament is cooler and has a lower resistance (page 2.7).
V = ε- Ir is a straight line on a graph of V (y-axis) against I (x-axis), with V-intercept ε and gradient −r.
Circuit: the cell in series with an ammeter, a variable resistor and a switch; a voltmeter connected directly across the cell's terminals.
Method: close the switch, set the variable resistor, read I and V quickly, then open the switch. Repeat for at least six settings spread over as wide a range of current as possible.
Graph: plot V against I, draw the line of best fit, read ε from the intercept at I = 0 and find r = −gradient from a large triangle.
Practical issues
The cell heats up. Current in the internal resistance dissipates energy inside the cell, and its temperature, and with it r, changes. Open the switch between readings and keep the currents moderate.
The cell runs down. A cell left delivering current loses e.m.f. and its internal resistance rises. Take the readings promptly so ε and r stay constant.
Protective resistor: a few ohms in series stops the current becoming very large if the variable resistor is turned to zero. It does not change the graph, because the voltmeter is across the cell only.
Meters: a high-resistance (digital) voltmeter takes negligible current. Choose ranges that give readings to 0.01 V and 0.01 A.
Safety: never short-circuit the cell; rechargeable cells have very low r and can deliver currents large enough to make wires dangerously hot.
Uncertainty: a reading on a meter of resolution 0.01 V has an uncertainty of ±0.005 V, about 0.45% of a 1.11 V reading.
WORKED EXAMPLE · QUESTION
Use the specimen graph above to determine the e.m.f. and the internal resistance of the cell.
SOLUTION
Extend the line to I = 0 and read the intercept: ε= 1.58 V. Gradient from a large triangle on the line, from (0.05 A, 1.55 V) to (0.75 A, 1.08 V): (1.08 - 1.55)/(0.75 - 0.05) = -0.47/0.70 = -0.67 V A⁻¹. So r = 0.67 Ω.
The gradient has units of V A⁻¹ = Ω, a quick check that it really is a resistance.
Exam tip
Only read ε from the intercept if the I-axis starts at zero. A false origin on the V-axis is fine and makes the graph larger. If the I-axis does not start at zero, calculate ε from one point on the line instead: ε= V + Ir.
Exam tip 1 — start from ε = I(R + r)
Write the formulae-list equation ε= I(R + r) first, then decide what R is: the total external resistance, worked out with the series and parallel rules before you substitute.
Exam tip 2 — the definitions differ in one direction word
E.m.f.: energy transferred to electrical energy per unit charge, by a source. P.d.: energy transferred from electrical energy per unit charge, in a component. "Per unit charge" is needed in both; "the voltage of a battery" is not a definition.
Exam tip 3 — the lost volts are ε − V
If you know the e.m.f. and the terminal p.d., you do not need r to find the lost volts: subtract. Then r = lost volts ÷ current.
Exam tip 4 — explain dimming with a chain of reasons
For "why does the lamp dim when another device is switched on?": total current increases → lost volts Ir increase → terminal p.d. decreases → the lamp's p.d., current and power decrease. Each arrow is a separate step.
Exam tip 5 — the graph gives two answers
For the core practical, state both results with units: ε from the intercept (V) and r from minus the gradient (Ω). Show the gradient triangle on the line, spanning more than half of it.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| E.m.f. is a force that pushes charge round the circuit. | Its unit is the volt, J C⁻¹, not the newton. | E.m.f. is energy transferred to electrical energy per unit charge. |
| A voltmeter across a battery always reads its e.m.f. | When current flows, part of the e.m.f. is used across the internal resistance. | It reads the terminal p.d., ε- Ir; this equals ε only when the current is zero. |
| A battery supplies a fixed current. | The current depends on the resistance it is connected to. | I = ε/(R + r); the e.m.f. is (nearly) fixed, not the current. |
| The lost volts are energy that disappears. | Energy is conserved. | The lost volts are the energy per coulomb dissipated as thermal energy inside the source. |
| A cell connected the wrong way round just stops contributing. | It actively opposes the others, and current still passes through its resistance. | Its e.m.f. subtracts and its internal resistance still adds. |
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Maximum power to the load means the source is working most efficiently. | At R = r as much power is wasted inside as is delivered. | Maximum power transfer comes with only 50% efficiency; efficiency = R/(R + r). |
6 Current, p.d. and resistance
E.m.f. is defined the same way as p.d., as energy per unit charge, and the power in the internal resistance uses the familiar P = I²R.
7 Series and parallel circuits; 2.8 The potential divider
The internal resistance is one more resistor in series with the external circuit; switching on more devices in parallel lowers R, raises I and lowers the terminal p.d. r and R share the e.m.f. like a potential divider: V = εR/(R + r).
Practical skills (Unit 3)
Rearranging an equation into y = mx + c and reading two constants from the intercept and the gradient is the pattern behind many practical questions.
| Equation | Use | In the formulae list? |
|---|---|---|
| ε= I(R + r) | current from a real source (ε in V, I in A, R and r in Ω) | Yes |
| V = ε- Ir | terminal p.d.; lost volts = Ir | No — derive it from ε= I(R + r) |
| ε= W/Q | definition of e.m.f. (W in J, Q in C) | No — learn it (same form as V = W/Q) |
| I_(max) = ε/r | short-circuit current | No — derive it: R = 0 |
| P = VI = I²R = V²/R | power delivered to a component (P in W) | Yes |
| εI = VI + I²r | power supplied = power to load + power wasted | No — derive it: multiply ε= V + Ir by I |
| fraction to load = V/ε= R/(R + r) | efficiency of energy delivery | No — derive it |
| εₜₒₜₐₗ = Σε (with signs), rₜₒₜₐₗ = Σr | cells in series | No — learn it |
1. Why does a high-resistance voltmeter connected on its own across a battery read (almost exactly) the e.m.f.?
2. A battery of e.m.f. 9.0 V and internal resistance 1.0 Ω delivers a current of 0.50 A. Calculate the terminal p.d.
3. A cell of e.m.f. 2.0 V drives a current of 0.40 A through a 4.5 Ω resistor. Calculate its internal resistance.
4. Three cells, each of e.m.f. 1.2 V and internal resistance 0.20 Ω, are connected in series. Calculate the maximum current they can deliver.
5. In the core practical a student's line of best fit has a V-intercept of 1.46 V and a gradient of −0.85 V A⁻¹. State the e.m.f. and the internal resistance, and give one reason for opening the switch between readings.
6. A 12 V battery of internal resistance 0.50 Ω supplies a 5.5 Ω heater. Calculate the power delivered to the heater and the power wasted in the battery.
1. The voltmeter takes almost no current, so the lost volts Ir are almost zero and the terminal p.d. ε- Ir is almost equal to ε.
2. V = 9.0 - 0.50 ×1.0 = 8.5 V.
3. Total resistance = 2.0/0.40 = 5.0 Ω, so r = 5.0 - 4.5 = 0.50 Ω.
4. ε= 3.6 V and r = 0.60 Ω, so I_(max) = 3.6/0.60 = 6.0 A.
5. ε= 1.46 V and r = 0.85 Ω. Opening the switch stops the cell heating up and running down, so that ε and r stay constant during the experiment.
6. I = 12/(5.5 + 0.50) = 2.0 A. Heater: I²R = 2.0² ×5.5 = 22 W. Battery: I²r = 2.0² ×0.50 = 2.0 W. Check: εI = 12 ×2.0 = 24 W = 22 + 2.0.
Check understanding
The answer and each step appear one click at a time.
A cell has an e.m.f. of 3.0 V and an internal resistance of 0.50 Ω. It delivers a current of 0.80 A.
What is the potential difference across its terminals?
Mark scheme click to reveal each step
Answer: B
Lost volts = Ir = 0.80 × 0.50 = 0.40 V. Terminal p.d. V = ε − Ir = 3.0 − 0.40 = 2.6 V.
A power supply of e.m.f. 12 V and internal resistance 1.5 Ω is connected to a resistor of resistance 6.5 Ω.
What is the current in the resistor?
Figure: power supply (dashed box: e.m.f. and internal resistance) connected to the resistor
Mark scheme click to reveal each step
Answer: A
ε = I(R + r), so I = 12 ÷ (6.5 + 1.5) = 12 ÷ 8.0 = 1.5 A.
A battery of e.m.f. 9.0 V and internal resistance 3.0 Ω is connected to a 6.0 Ω resistor.
At what rate is energy dissipated inside the battery?
Figure: battery (dashed box: e.m.f. and internal resistance) connected to the resistor
Mark scheme click to reveal each step
Answer: A
I = 9.0 ÷ (6.0 + 3.0) = 1.0 A. Power dissipated inside = I²r = 1.0² × 3.0 = 3.0 W.
A torch bulb is lit by a single cell. While the bulb is lit its resistance is 3.5 Ω. The cell has an e.m.f. of 1.60 V and its internal resistance is 0.50 Ω.
Figure: torch bulb connected to the cell (dashed box: e.m.f. and internal resistance)
(a) Calculate the current in the bulb. [2]
(b) Calculate the potential difference across the terminals of the cell. [2]
(c) State the value of the lost volts. [1]
Mark scheme click to reveal each step
I = ε/(R + r) = 1.60 ÷ (3.5 + 0.50)
= 0.40 A
V = IR = 0.40 × 3.5
= 1.4 V (allow error carried forward)
1.60 − 1.4 = 0.20 V / 0.40 × 0.50 = 0.20 V
A high-voltage supply used in a school laboratory has an e.m.f. of 5.0 kV. For safety it is designed with an internal resistance of 50 MΩ.
(a) Calculate the maximum current the supply can deliver. [1]
(b) Explain why a large internal resistance makes the supply safer to use. [2]
(c) The supply is connected to a resistor of resistance 10 MΩ. Calculate the potential difference across the resistor. [2]
Mark scheme click to reveal each step
Iₘₐₓ = ε/r = 5000 ÷ (50 × 10⁶) = 1.0 × 10⁻⁴ A (0.10 mA)
the internal resistance is in series with anything connected to the terminals, including a person / a short circuit
so the current can never exceed ε/r, which is small / about 0.1 mA
I = 5000 ÷ ((10 + 50) × 10⁶) = 8.3 × 10⁻⁵ A
V = IR = 8.33 × 10⁻⁵ × 10 × 10⁶ = 830 V (833 V)
Keep practising
Full notes, every simulation and hundreds of questions with mark schemes, free:
ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only.
Pearson Edexcel and Cambridge are trademarks of their owners; ExaminerPrep is not affiliated with any exam board.