E.m.f. and internal resistanceEdexcel A-Level Physics: Revision notes
Section 1
E.m.f. and terminal p.d.
The electromotive force (e.m.f.), ε, of a source is the energy transferred to electrical energy per unit charge passing through the source: ε = E/Q. It is measured in volts.
The terminal p.d., V, is the energy per unit charge delivered to the external circuit by the charge, measured across the terminals.
They differ because the source has internal resistance. When there is no current, V = ε. When a current flows, some energy per coulomb is transferred as thermal energy inside the source, so V is smaller than ε.
E.m.f. is not a force. Do not say it is 'the p.d. across the cell'; the terminal p.d. is only equal to the e.m.f. when no current is drawn.
Section 2
The equations
The e.m.f. drives current through both the external resistance R and the internal resistance r:
ε = I(R + r) = V + Ir
so the terminal p.d. is V = ε − Ir. The quantity Ir is the 'lost volts'.
Worked example: ε = 1.5 V, r = 0.50 Ω and R = 2.5 Ω. Current I = 1.5/(2.5 + 0.50) = 0.50 A. Lost volts = 0.50 × 0.50 = 0.25 V, so V = 1.5 − 0.25 = 1.25 V (also IR = 0.50 × 2.5 = 1.25 V). The power dissipated inside the cell is I²r = 0.125 W.
Short circuit: if R is zero, the current is the maximum, I = ε/r, and V = 0.
A source with a small internal resistance gives a large short-circuit current, which is why a car battery can be dangerous while a high-voltage supply with a large internal resistance is limited to a tiny current.
Section 3
Why the terminal p.d. falls with current
As the current drawn increases, the lost volts Ir increases, so the terminal p.d. falls. This is why car headlamps dim when the starter motor draws a large current: with ε = 12 V, r = 0.020 Ω and I = 150 A, the lost volts are 3.0 V and V = 9.0 V.
The lamps' power, V²/R, falls with the lower p.d. Energy is also wasted inside the battery at the rate I²r (450 W in this example).
Section 4
Core practical: e.m.f. and internal resistance
Circuit: the cell in series with a variable resistor, an ammeter and a switch. A voltmeter is connected directly across the terminals of the cell.
Method: adjust the variable resistor to give a range of currents (at least five). Record V and I each time. Close the switch only while taking readings, so that the cell does not heat or run down.
Analysis: V = ε − Ir is of the form y = c + mx, so a graph of V against I is a straight line with y-intercept ε and gradient −r. The x-intercept is the short-circuit current ε/r.
A small protective resistor in series with the cell stops the current becoming too large for the cell.
Must Know
- ε = E/Q: energy transferred to electrical energy per coulomb
- ε = I(R + r) and V = ε − Ir
- V = ε only when no current flows
- Short-circuit current = ε/r
- V against I: intercept ε, gradient −r
- Energy wasted inside the source: I²r
That's the notes covered.
Carry on to the next subtopic.
Exam questions on E.m.f. and internal resistance
- A cell of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected in series with a 2.5 Ω resistor and a switch.Calculate the energy transferred as thermal energy in the internal resistance of the cell in 60 s.2 marks
- A student measures the terminal p.d. V of a battery for different currents I drawn from it, using a variable resistor. She plots V against I and draws a straight line of best fit. The line meets the V axis at 6.2 V and has a gradient of −0.80 V A⁻¹.Explain why the terminal p.d. of the battery is less than its e.m.f. when a current is drawn.2 marks
- A 12 V car battery has an internal resistance of 0.020 Ω. The headlamps are connected across the battery terminals. When the driver starts the engine, the starter motor draws a current of 150 A from the battery.Calculate the terminal p.d. of the battery while the starter motor is operating.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).