EMF, Internal Resistance, and Terminal Voltage
The electromotive force (emf, ε) of a source is the energy transferred to each coulomb of charge that passes through it, and a real source has an internal resistance r, so when a current I flows the terminal voltage is V = ε − Ir, which is less than the emf. In brief: the emf is the open-circuit voltage (no current). The “lost volts” Ir are dissipated inside the source. With an external resistance R, the current is I = ε ÷ (R + r). The power delivered to the external circuit is greatest when R = r (maximum power transfer), but the efficiency is then only 50 percent.
The headlights of a car dim when the starter motor is switched on, and a battery that is nearly flat may show 1.5 V on a voltmeter but fail to power a torch. Both are due to internal resistance. This guide defines emf and terminal voltage, derives the relations, explains how to measure ε and r experimentally, compares emf and potential difference in a table, works through numerical examples and corrects common mistakes.
What is emf?
The emf of a source (battery, cell, generator, solar cell) is the work done by the source per unit charge in moving charge around the complete circuit: ε = W ÷ Q. It is measured in volts. Despite the name, it is not a force. The emf is a property of the chemical reactions in a battery (about 1.5 V for an alkaline cell, 2.0 V for a lead-acid cell, 3.7 V for a lithium-ion cell). A voltmeter of very high resistance connected across the terminals of a source in an open circuit reads the emf, as almost no current flows.
What is internal resistance?
The materials inside a source (electrolyte, electrodes, wires) resist the flow of charge, which gives the source an internal resistance r. When a current I flows, a potential drop Ir occurs inside the source, so the voltage at the terminals is smaller than the emf:
V = ε − Ir, and with an external load R, V = IR, so ε = I(R + r).
The larger the current, the more the terminal voltage falls. Fresh cells have a small r (a fraction of an ohm for a car battery, a few tens of milliohms), and old, cold or tiny batteries have a larger r. The energy per coulomb ε is divided between the load (IR) and the inside of the source (Ir). The power produced by the source, εI, is divided between the useful power I2R in the load and the wasted power I2r. The efficiency of the source is R ÷ (R + r) = V ÷ ε.
How are emf and internal resistance measured?
Connect the cell to a variable resistor (rheostat) and an ammeter in series, with a voltmeter across the terminals of the cell. Vary the resistance to obtain pairs of values of V and I, and plot V against I. From V = ε − Ir, the graph is a straight line: the intercept on the V axis (I = 0) is the emf ε, and the gradient is −r. The line meets the I axis at the short-circuit current Isc = ε ÷ r. To keep the cell from running down, the current should be small and switched on only briefly.
When is the power maximum?
The power in the load is P = I2R = ε2R ÷ (R + r)2. This has a maximum when R = r, giving Pmax = ε2 ÷ 4r. At that point, half of the power is wasted in the source, so the efficiency is 50 percent. Maximum power transfer is important for signal sources (audio, radio), where the matching of impedance matters, while power systems operate with R much larger than r for high efficiency.
EMF and terminal voltage compared
| Feature | EMF (ε) | Terminal p.d. (V) |
|---|---|---|
| Meaning | Energy per coulomb given by the source | Energy per coulomb delivered to the external circuit |
| Value | Constant for a given source | ε − Ir (depends on the current) |
| Equal when | No current flows (open circuit) or r = 0 | |
Worked example: battery calculations
- Terminal voltage. A battery of ε = 12 V and r = 0.50 Ω is connected to a resistor of R = 5.5 Ω. I = ε ÷ (R + r) = 12 ÷ 6.0 = 2.0 A. The terminal voltage V = IR = 2.0 × 5.5 = 11 V (also ε − Ir = 12 − 1.0 = 11 V). The lost volts are 1.0 V.
- Power and efficiency. The total power produced is εI = 24 W. The power in the load = VI = 22 W. The power wasted in the battery = I2r = 4.0 × 0.50 = 2.0 W. Efficiency = 22 ÷ 24 = 92 percent.
- Car battery and starter. A car battery has ε = 12.6 V and r = 0.020 Ω. The starter motor has an effective resistance of 0.10 Ω. I = 12.6 ÷ (0.10 + 0.020) = 105 A. The terminal voltage falls to 12.6 − 105 × 0.020 = 10.5 V, which dims the lights.
- Finding ε and r from data. With I = 1.0 A the terminal voltage is 1.40 V, and with I = 3.0 A it is 1.20 V. The gradient: r = (1.40 − 1.20) ÷ (3.0 − 1.0) = 0.10 Ω. The emf: ε = V + Ir = 1.40 + 1.0 × 0.10 = 1.50 V.
- Maximum power. For a source with ε = 6.0 V and r = 2.0 Ω, the maximum power is delivered to a load with R = 2.0 Ω: I = 6.0 ÷ 4.0 = 1.5 A, P = I2R = 2.25 × 2.0 = 4.5 W (= ε2 ÷ 4r = 36 ÷ 8 = 4.5 W).
- Short circuit. If the terminals of the 12 V, 0.50 Ω battery are connected by a wire of negligible resistance, I = 12 ÷ 0.50 = 24 A, and the 288 W (I2r = 576 × 0.5) heats the battery.
How were these ideas developed?
Alessandro Volta’s electric pile (1800) was the first battery. Georg Ohm’s work on the circuit as a whole (1827) included the effect of the internal resistance of the source. Kirchhoff’s laws handle internal resistance as an extra resistor in series with the emf. Moritz von Jacobi formulated the maximum power transfer theorem in 1840. Lithium-ion batteries, developed from the 1970s to the 1990s, have a low internal resistance and a high energy density.
Which mistakes are common?
Mistake 1: Treating the terminal voltage as always equal to the emf
It is equal only when no current flows (or when r is zero).
Mistake 2: Forgetting r in the total resistance of the circuit
The current is ε ÷ (R + r), not ε ÷ R.
Mistake 3: Using a low-resistance voltmeter to measure the emf
It draws a current and shows the terminal voltage. A voltmeter with very high resistance is required.
Mistake 4: Thinking that maximum power means maximum efficiency
At R = r, the power is maximum but the efficiency is only 50 percent.