Electrical sources and internal resistance

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Concept 1

EMF, terminal p.d. & lost volts

On the relationship sheet — given to you in the exam Derived — rearrange the given one, or just remember it
On the relationship sheet
EMF
E = V + Ir
EMF = terminal p.d. + lost volts
Derived
Lost volts
lost volts = Ir = E − V
only there when a current flows
Where the lost-volts equation comes from Start with the one on the sheet:  E = V + Ir Take V off both sides:  Ir = E − V So the lost volts are just the EMF minus what is left at the terminals.
  • A real cell can be pictured as a perfect (ideal) source of EMF E joined in series with a small internal resistance r. You can't physically separate the two — r is built into the cell.
  • EMF (E) is the energy given to each coulomb of charge passing through the source (volt = joule per coulomb). An EMF of 1.5 V means 1.5 J is given to every coulomb.
  • Terminal p.d. (t.p.d., V) is the p.d. across the cell's terminals — this equals the p.d. across the external components (V = IR).
  • Lost volts (Ir) is the energy lost per coulomb driving current through the internal resistance: lost volts = E − V.
  • Ideal supply: r = 0, so there are no lost volts and the t.p.d. always equals the EMF.
  • Open circuit: no current (I = 0), so no lost volts → a voltmeter across the terminals reads the full EMF. (This is how you measure E.)
  • Short circuit: external resistance ≈ 0, so the t.p.d. ≈ 0 and the current is at its maximum, Ishort = E ÷ r.
A real cell: EMF E and internal resistance r, supplying an external resistor R cell E r R V
The voltmeter across the terminals reads the t.p.d. (E − Ir), which is also the p.d. across R.

Fill the gaps — say it like the SQA

Pick the precise word for each gap, then check. The marks here are for using the right terms.

The EMF is the energy supplied to each of charge passing through the source. When a current flows, energy is lost driving charge through the cell's ; this used-up p.d. is called the . What is left is measured across the terminals — the p.d. So terminal p.d. = EMF lost volts.

Match the term to its meaning

Tap a term on the left, then its meaning on the right. Correct pairs lock green.

Term

Meaning

Substitute & solve
A cell drives a current of 0.50 A round a circuit. A voltmeter across the cell's terminals reads 1.3 V, and the cell's internal resistance is 0.40 Ω. Calculate the EMF of the cell. 3 marks
Find the EMF from the terminal p.d., the current and the internal resistance cell E = ? r = 0.40 Ω A I = 0.50 A R V 1.3 V
The voltmeter reads the terminal p.d. (1.3 V); the EMF is this plus the lost volts (Ir).
E=V + Iron the sheet
E=1.3 + (0.50 × 0.40)numbers in first
E=1.3 + 0.20lost volts = 0.20 V
E=1.5 V
Rearrange — numbers in first
A battery of EMF 6.0 V has an internal resistance of 0.40 Ω. When connected to a lamp the current is 1.5 A. Calculate the terminal potential difference across the battery. 3 marks
Find the terminal p.d. of a battery driving a lamp battery E = 6.0 V r = 0.40 Ω A I = 1.5 A lamp V V = ?
The terminal p.d. is the EMF minus the lost volts driven through r.
E=V + Ir
6.0=V + (1.5 × 0.40)numbers in first
V=6.0 − 0.60lost volts = 0.60 V
V=5.4 V
Lost volts = Ir = 0.60 V, and t.p.d. = EMF − lost volts. The terminal p.d. is always less than the EMF whenever a current flows.
Practice 1

A car battery has an EMF of 12.8 V and an internal resistance of 0.10 Ω. The starter motor draws a current of 80 A. Show that the terminal p.d. falls to 4.8 V while the motor turns. 3 marks

Car battery driving a starter motor battery E = 12.8 V r = 0.10 Ω A I = 80 A M motor V V = ?
The huge 80 A current makes the lost volts (Ir) large, so the terminal p.d. drops sharply while the motor turns.
Answer
V = E − Ir (a "show" Q must start from a correct relationship) V = 12.8 − (80 × 0.10) V = 12.8 − 8.0 V = 4.8 V (target value, with unit)

Never start from 4.8 V and work back — that scores 0 in a "show that".

Concept 2

Solving circuits with internal resistance

Derived
Whole circuit
E = I(R + r)
EMF drives the current through R and r
Where this comes from — it is not on the sheet Start with the one on the sheet:  E = V + Ir The terminal p.d. is the p.d. across R, so  V = IR Put that in:  E = IR + Ir Take out the common factor I:  E = I(R + r) In words: the EMF pushes the same current through the outside resistor and the cell's own resistance.
  • Treat the internal resistance r as just another series resistor: total resistance = R + r.
  • The usual steps: add up the total resistance → find the current from E = I(R + r) → then work out whatever the question asks for: t.p.d. (V = IR), lost volts (Ir) or power.
  • Power: power to the external circuit = I2R; power wasted inside the cell = I2r; total power from the source = EI.
  • Cells in series add their EMFs and their internal resistances.

Check your calculation

A cell of EMF 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Work out the current, type your answer, then check — you'll get a hint if it's a common slip.

A cell of EMF 1.5 V and internal resistance 0.50 Ω connected to a 2.5 Ω resistor cell E = 1.5 V r = 0.50 Ω I = ? R = 2.5 Ω
Internal resistance r is in series with R, so the current sees a total of R + r.

Substitute & solve
A battery of EMF 9.0 V and internal resistance 1.0 Ω is connected to a 3.5 Ω resistor in series with a 0.5 Ω resistor. Calculate the current in the circuit. 3 marks
A battery driving two resistors in series battery E = 9.0 V r = 1.0 Ω R₁ = 3.5 Ω R₂ = 0.5 Ω A I = ?
The same current flows through R₁, R₂ and the internal resistance r — add all three to get the total.
E=I(R + r)
9.0=I(3.5 + 0.5 + 1.0)include r in the total
I=9.0 ÷ 5.0now rearrange
I=1.8 A
Several steps — find lost volts, then r
A cell of EMF 1.50 V is connected to a 4.0 Ω resistor. The current is 0.30 A. Calculate the internal resistance of the cell. 4 marks
Find the internal resistance from EMF, current and external resistance cell E = 1.50 V r = ? A I = 0.30 A R 4.0 Ω V
Find the terminal p.d. V = IR first, then use E = V + Ir to get r.
V=IRterminal p.d. first
V=0.30 × 4.0 = 1.2 V
E=V + Ir
1.50=1.2 + (0.30 × r)numbers in first
0.30 r=0.30now rearrange
r=1.0 Ω

Predict & justify

A battery of EMF 12.8 V and internal resistance 0.10 Ω supplies one lamp. A second identical lamp is now switched on in parallel. The terminal p.d. of the battery will…

Adding a second lamp in parallel across a battery battery E = 12.8 V r = 0.10 Ω lamp 1 lamp 2 switch
The second lamp adds a parallel path, so the total outside resistance falls — watch what that does to the current.
Practice 2

Two identical cells, each of EMF 1.5 V and internal resistance 0.20 Ω, are connected in series with a 2.6 Ω resistor. Calculate (a) the current and (b) the terminal p.d. of the battery. 4 marks

Two cells in series driving a 2.6 Ω resistor cell 1 cell 2 each: 1.5 V, 0.20 Ω R = 2.6 Ω A
Cells in series add up: the battery's EMF is 1.5 + 1.5 and its internal resistance is 0.20 + 0.20.
Answer
Series cells: E = 1.5 + 1.5 = 3.0 V ; r = 0.20 + 0.20 = 0.40 Ω (a) E = I(R + r) → 3.0 = I(2.6 + 0.40) → I = 3.0 ÷ 3.0 = 1.0 A (b) V = IR = 1.0 × 2.6 = 2.6 V (or V = E − Ir = 3.0 − 0.40 = 2.6 V)
Justify it

A battery of EMF 12.8 V and internal resistance 0.10 Ω supplies two identical lamps. A switch adds a third lamp in parallel. State and justify the effect on the terminal p.d. 3 marks

Three lamps in parallel across a battery, the third on a switch battery E = 12.8 V r = 0.10 Ω lamp 1 lamp 2 lamp 3 switch
Each lamp added in parallel lowers the total outside resistance, raising the current — and a bigger current means bigger lost volts.
Answer
Terminal p.d. decreases. (1) Adding a parallel lamp lowers the total external resistance, so the current increases. (1) Larger current → larger lost volts (Ir), so t.p.d. = E − Ir falls. (1)

Mark scheme: correct direction first; wrong physics in the justification caps the marks.

Concept 3

Measuring EMF & internal resistance — graphs & experiment

Derived
Straight line
V = E − rI
rearranged from E = V + Ir
Where the graph equation comes from Start with the one on the sheet:  E = V + Ir Make V the subject:  V = E − Ir Write Ir as rI so it lines up with a straight line:  V = −rI + E
V = −rI + E  is just  y = mx + c
Straight lineOur experimentWhat you read off the graph
yterminal p.d. Vplotted up the y-axis
xcurrent Iplotted along the x-axis
gradient m−rso r = − gradient (steepness)
y-intercept cEthe EMF (the V value when I = 0)
  • Written as V = −rI + E, plotting terminal p.d. V (y-axis) against current I (x-axis) gives a straight line:
    • y-intercept = EMF E (the t.p.d. when I = 0, i.e. open circuit);
    • gradient = −r (internal resistance = magnitude of the gradient);
    • x-intercept = short-circuit current, Ishort = E ÷ r (where V = 0).
  • Experiment: connect the cell to a variable resistor, with an ammeter in series and a voltmeter across the cell's terminals. Change the variable resistor to get a range of settings; for each one record the current I and terminal p.d. V.
  • Plot V against I and draw the best-fit straight line: the y-intercept is the EMF and the magnitude of the gradient is the internal resistance.
Experiment to measure EMF and internal resistance A R V
The ammeter reads the current I; the voltmeter reads the terminal p.d. V. Vary R and plot V against I.

V–I graph simulator

Change the EMF and internal resistance and watch the line tilt. Read off the y-intercept (EMF), the gradient (−r) and the x-intercept (short-circuit current).

Terminal p.d. against current terminal p.d. V (V) current I (A) E I(short)
—

Read the V–I graph

Use the graph above, then check both answers.

1. On a graph of terminal p.d. (y) against current (x), the y-intercept gives you…

2. The internal resistance is found from…

Read EMF, r and short-circuit current
The graph of terminal p.d. against current for a cell is a straight line. It cuts the p.d. axis at 6.0 V and has a gradient of −2.0 V A⁻¹. Determine (a) the EMF, (b) the internal resistance and (c) the short-circuit current. 4 marks
V–I graph with p.d.-axis intercept 6.0 V and gradient −2.0 V per amp terminal p.d. V (V) current I (A) 6.0 V gradient = −2.0 V A⁻¹ 3.0 A
The p.d.-axis intercept is the EMF; the current-axis intercept is the short-circuit current; the internal resistance is the size of the gradient.
(a) E=y-intercept = 6.0 V
(b) gradient=−r → r = 2.0 Ω
(c) E=Ishort rat V = 0
Ishort=6.0 ÷ 2.0 = 3.0 A
Two-point data — no graph drawn
When the current from a cell is 0.50 A the terminal p.d. is 1.40 V; when the current is 1.00 A the terminal p.d. is 1.20 V. Determine the EMF and the internal resistance of the cell. 4 marks
Two data points on a V–I graph terminal p.d. V (V) current I (A) 1.60 V (0.50, 1.40) (1.00, 1.20)
Read the EMF where the best-fit line meets the p.d. axis (1.60 V); the internal resistance is − gradient between the two points.
r=−gradient
r=−(1.20 − 1.40) ÷ (1.00 − 0.50)numbers in first
r=0.20 ÷ 0.50 = 0.40 Ω
E=V + Ir = 1.40 + (0.50 × 0.40)use either data point
E=1.60 V
Practice 3

In an experiment to find the EMF and internal resistance of a cell, a graph of terminal p.d. against current is plotted. The line meets the p.d. axis at 1.50 V and the current axis at 5.0 A. Determine (a) the EMF, (b) the short-circuit current and (c) the internal resistance. 4 marks

V–I graph with p.d.-axis intercept 1.50 V and current-axis intercept 5.0 A terminal p.d. V (V) current I (A) 1.50 V 5.0 A
EMF = p.d.-axis intercept; short-circuit current = current-axis intercept; then r = E ÷ Ishort.
Answer
(a) EMF = y-intercept = 1.50 V (b) short-circuit current = x-intercept = 5.0 A (c) I_short = E ÷ r → r = E ÷ I_short = 1.50 ÷ 5.0 = 0.30 Ω
Go further — the R against 1/I method

Some papers vary R and plot it against 1/I. Rearranging E = I(R + r) gives R = E(1/I) − r.

A graph of R (y) against 1/I (x) is a straight line: gradient = E y-intercept = −r (Same physics, different axes.)

Check yourself

Recap — fill the gaps

Pull the whole topic together: choose the right word for each gap, then check. Counts towards your badges.

A real cell behaves like a source of EMF in series with its internal . When a current flows, energy is lost inside the cell — the , equal to Ir. A voltmeter on the terminals then reads less than the EMF; this reading is the p.d. The relationship given on the sheet is E = V + . On a graph of terminal p.d. against current, the y-intercept is the and the size of the gradient is the .

Mixed multiple choice

Exam-style multiple choice, 1 mark each. Choose an answer for each, then mark them.

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