Electromotive force, internal resistance, terminal voltage and lost volts
AQA A-Level Physics 5
⚡Define EMF as energy transferred per unit charge: ε = E/Q
🔋Explain the concept of internal resistance in a real cell
📐Apply ε = I(R + r) and V = ε − Ir
📉Distinguish between EMF and terminal potential difference
📊Interpret V–I graphs to find ε and r
🔬Carry out the required practical: measuring ε and r
Electromotive Force (EMF)
Electromotive force (EMF, ε): The energy transferred to electrical energy per unit charge by a source. Measured in volts (V). It is NOT a force despite its name.
ε = E / Q
ε = EMF (V)
E = energy transferred by source (J)
Q = charge (C)
EMF is the "push" given to each coulomb of charge by the source (battery, solar cell, generator, etc.). A 12 V battery gives 12 J of electrical energy to every coulomb of charge that passes through it.
EMF can only be measured directly when no current flows (open circuit). As soon as a current flows, some energy is wasted inside the source due to internal resistance, and the measurable terminal voltage is less than ε.
Internal Resistance
Internal resistance (r): The resistance inside the source itself (e.g. chemical resistance inside a battery). It causes energy to be dissipated as heat inside the battery when current flows.
ε = I(R + r) = IR + Ir
ε = EMF of the cell (V)
I = current in the circuit (A)
R = external resistance (Ω)
r = internal resistance (Ω)
Terminal p.d. = V = IR = ε − Ir
Lost volts = Ir
"Lost volts" (Ir) is the voltage drop across the internal resistance. It is energy wasted per unit charge inside the battery. As current increases, lost volts increase and terminal p.d. decreases.
When no current flows (open circuit): V_terminal = ε (no lost volts). When a large current flows (short circuit): V_terminal ≈ 0 (all voltage lost internally, I = ε/r). This is why batteries get hot when short-circuited.
V–I Graph for a Real Cell
By varying the external resistance R (using a rheostat) and measuring I and V, a V–I graph can be plotted:
V = ε − Ir = −rI + ε
Comparing with y = mx + c:
y = V, x = I, gradient = −r, y-intercept = ε
Graph feature
Physical meaning
y-intercept (I = 0)
EMF ε (open-circuit voltage)
Gradient (magnitude)
Internal resistance r
x-intercept (V = 0)
Short-circuit current I_sc = ε/r
The graph has a negative gradient (−r): as current increases, terminal voltage decreases due to greater "lost volts." A steeper gradient indicates higher internal resistance.
Use when higher voltage is needed (e.g. 3 × 1.5 V cells in series = 4.5 V battery).
Identical cells in parallel: EMF stays the same; internal resistance halves (for 2 cells), reducing energy loss.
ε_total = ε; r_total = r/n (for n identical cells)
Cells in parallel are used when a larger current is needed without the terminal voltage dropping too much — each cell shares the total current, so less current flows through each internal resistance and there are fewer lost volts.
A battery of EMF 6.0 V and internal resistance 0.5 Ω is connected to an external resistor of 5.5 Ω. Calculate the current, terminal p.d., and lost volts.
3Terminal p.d. = ε − Ir = 6.0 − 0.5 = 5.5 V (or IR = 1.0 × 5.5 = 5.5 V ✓)
I = 1.0 A; Terminal p.d. = 5.5 V; Lost volts = 0.5 V
A student plots a V–I graph for a battery. The y-intercept is 4.5 V and the graph has a gradient of −0.8 V A⁻¹. Find the EMF and internal resistance. What would the terminal voltage be at I = 2.0 A?
1From V = ε − Ir: y-intercept = ε = 4.5 V
2Gradient = −r → r = 0.8 Ω
3At I = 2.0 A: V = 4.5 − (0.8 × 2.0) = 4.5 − 1.6 = 2.9 V
ε = 4.5 V; r = 0.8 Ω; V at 2.0 A = 2.9 V
A battery of EMF 9.0 V and internal resistance 2.0 Ω powers a device. The terminal voltage is measured as 7.5 V. Calculate the current and the resistance of the device.
1Lost volts = ε − V = 9.0 − 7.5 = 1.5 V
2I = lost volts / r = 1.5/2.0 = 0.75 A
3R = V/I = 7.5/0.75 = 10 Ω
I = 0.75 A; R = 10 Ω
Three identical cells, each EMF 1.5 V and internal resistance 1.2 Ω, are connected in series and then to a 5.4 Ω external resistance. Find the current and the terminal p.d. across the external resistor.
1. A cell has EMF 1.5 V and internal resistance 0.3 Ω. It drives a current of 2 A. What is the terminal p.d.?
V = ε − Ir = 1.5 − (2 × 0.3) = 1.5 − 0.6 = 0.9 V.
2. On a V–I graph for a cell, as the current through the cell increases, the terminal p.d.:
V = ε − Ir. As I increases, Ir (lost volts) increases, so V decreases. The relationship is linear (V = −rI + ε), giving a straight line with negative gradient.
3. The open-circuit voltage of a battery is 6.0 V. When a 2 Ω resistor is connected, the terminal voltage falls to 5.0 V. What is the internal resistance?
I = V/R = 5.0/2 = 2.5 A. Lost volts = 6.0 − 5.0 = 1.0 V. r = lost volts/I = 1.0/2.5 = 0.4 Ω.
4. Which statement correctly defines EMF?
EMF (ε = E/Q) is the energy given to each coulomb of charge by the source. It is measured in volts but is not a force. The terminal voltage (across the external circuit) is always less than ε when current flows.
5. A battery has EMF 12 V and internal resistance 1.5 Ω. What is the short-circuit current (when R = 0)?
1. A student investigates a battery using a rheostat. She measures the following data: at I = 0.20 A, V = 2.86 V; at I = 0.60 A, V = 2.58 V; at I = 1.00 A, V = 2.30 V; at I = 1.40 A, V = 2.02 V. Find the EMF and internal resistance by plotting or calculating the best-fit values.
Using V = ε − Ir. Calculate gradient from two points: gradient = ΔV/ΔI = (2.02 − 2.86)/(1.40 − 0.20) = −0.84/1.20 = −0.70 V A⁻¹ → r = 0.70 Ω. To find ε, use any point: ε = V + Ir = 2.86 + (0.20 × 0.70) = 2.86 + 0.14 = 3.00 V. Check with another point: ε = 2.30 + (1.00 × 0.70) = 3.00 V ✓. EMF = 3.00 V, internal resistance = 0.70 Ω.
2. A laptop battery has EMF 11.1 V and internal resistance 0.12 Ω. It supplies 45 W to the laptop. Calculate: (a) the current drawn; (b) the terminal voltage; (c) the power wasted in the internal resistance; (d) the efficiency of energy transfer to the laptop.
(a) P = IV → I = P/V. We need to be careful — P = 45 W is the power to the external load, P = IV_terminal. Let V = terminal voltage. V = ε − Ir, and P = IV → I = P/V. Substituting: P = (ε − Ir)I = εI − I²r → εI − I²r − 45 = 0 → 0.12I² − 11.1I + 45 = 0. Using quadratic: I = (11.1 ± √(11.1² − 4×0.12×45))/(2×0.12) = (11.1 ± √(123.21 − 21.6))/0.24 = (11.1 ± √101.61)/0.24 = (11.1 ± 10.08)/0.24. I = (11.1 − 10.08)/0.24 = 4.25 A (taking the physically reasonable solution). (b) V = ε − Ir = 11.1 − 4.25 × 0.12 = 11.1 − 0.51 = 10.59 V ≈ 10.6 V. Check: P = IV = 4.25 × 10.59 = 45.0 W ✓. (c) P_internal = I²r = 4.25² × 0.12 = 18.06 × 0.12 = 2.17 W. (d) P_total = εI = 11.1 × 4.25 = 47.2 W. Efficiency = 45/47.2 = 95.3%.
3. Explain why the terminal voltage of a car battery drops significantly when the starter motor is operated, but recovers when the engine starts. What damage can result from repeatedly operating a starter motor with a weak battery?
The starter motor draws a very large current (typically 100–200 A) because it has low resistance and the car battery must provide enough torque to turn the cold engine. With I very large, lost volts = Ir become significant (e.g. at 150 A and r = 0.02 Ω: Ir = 3 V). Terminal voltage drops from ~12.6 V to ~9–10 V, which is why dashboard lights dim and radio cuts out during cranking. Once the engine starts, the starter motor is disconnected (current drops to near zero) and the alternator begins charging the battery, restoring terminal voltage to 13.8–14.4 V (the charging voltage). With a weak battery (higher r due to sulfation or age), lost volts are greater at the same current: terminal voltage drops too low to sustain cranking. Repeatedly attempting to start with a weak battery causes: (1) excessive heating inside the battery (power = I²r), accelerating plate damage; (2) deep discharge, which permanently reduces capacity through sulfation; (3) potential warping or cracking of plates from thermal stress. Eventually the battery fails completely.
Required Practical Investigating EMF and Internal Resistance
Objective: Determine the EMF and internal resistance of a cell by measuring terminal voltage at different currents.
Equipment
Cell or battery under investigation
Variable resistor (rheostat, e.g. 0–20 Ω)
Ammeter (in series) and voltmeter (across the cell terminals)
Switch (to break circuit between readings)
Connecting leads
Method
Connect the cell in series with the ammeter, switch and rheostat. Connect the voltmeter directly across the cell terminals (not across the rheostat).
Set the rheostat to its maximum resistance. Close the switch. Read I and V. Open the switch between readings to prevent the battery running down.
Decrease the resistance in steps (e.g. 8 values). Record I and V at each step.
Plot V (y-axis) vs I (x-axis). Draw the best-fit straight line.
Determine ε from the y-intercept and r from the magnitude of the gradient.
Safety and Precautions
Keep the switch open between readings — continuous current drains the battery and changes r, making results inconsistent. Never short-circuit the cell directly. Do not use a mains power supply for this experiment — it does not behave like a cell with internal resistance.