Electromotive force and internal resistance

Section: D.C. Circuits  |  Syllabus: Cambridge AS Level Physics (9702)

Electromotive Force (e.m.f.)

Electromotive Force (e.m.f.) The energy transferred per unit charge by a source in driving charge around a complete circuit.

= (W/Q)

where = e.m.f. (V), W = work done/energy transferred (J), Q = charge (C) Important

Despite its name, electromotive force is NOT a force. It is measured in volts (V) and represents energy per unit charge, just like potential difference. The term is historical and predates our modern understanding of electricity.

Physical Meaning

The e.m.f. of a source tells us how much energy (in joules) is transferred to each coulomb of charge that passes through it. This energy comes from:

Worked Example: Calculating e.m.f.

Problem: A battery transfers 54 J of energy when 6 C of charge passes through it. What is the e.m.f. of the battery?

Solution:

= (W/Q) = (54/6) = 9 V

E.m.f. vs Potential Difference

Both e.m.f. and p.d. are measured in volts, but they describe different energy transfers:

Property Electromotive Force (e.m.f.) Potential Difference (p.d.)
Definition Energy transferred TO each coulomb by the source Energy transferred BY each coulomb in a component
Energy conversion Other forms → electrical energy Electrical energy → other forms
Where it applies Sources (cells, batteries, generators) Components (resistors, lamps, motors)
Symbol (epsilon) V
Energy Perspective

As charge moves around a circuit: it gains electrical potential energy when passing through the source (equal to the e.m.f. × charge), and loses this energy when passing through external components (equal to the p.d. × charge). By conservation of energy, the total energy gained equals the total energy lost.

Internal Resistance

Every real source of e.m.f. has some internal resistance, denoted r. This is caused by the resistance of the materials inside the source (wires, electrodes, electrolyte, etc.).

Internal Resistance The resistance within a source of e.m.f. that causes energy to be dissipated inside the source when current flows. FIG 10.15: Real Cell with Internal Resistance

Show a cell represented as two parts within a dashed box: an ideal e.m.f. source (ε) in series with an internal resistance (r). This is connected to an external resistance (R). Label the current I flowing through the circuit, the terminal p.d. (V) across the external resistance, and the "lost volts" (Ir) across the internal resistance.

Effect of Internal Resistance

When current flows through a source, some energy is dissipated as heat inside the source due to its internal resistance. This reduces the useful energy available to the external circuit.

The Potential Divider Model

A useful way to understand internal resistance is to treat the cell as a potential divider. The internal resistance r and external resistance R are in series, so the e.m.f. divides between them:

FIG 10.16: Cell as a Potential Divider

Show two circuit representations side by side: (a) A cell within a dashed box showing e.m.f. ε with internal resistance r inside, connected to external resistance R. (b) The equivalent potential divider: e.m.f. ε at the top, with r and R in series below it, showing that the p.d. across R (terminal p.d.) is a fraction of ε determined by the ratio R/(R+r).

Using the potential divider principle:

V_R = (R/R + r) ×

This shows that the terminal p.d. is always less than the e.m.f. (unless r = 0, which is impossible for real sources). The larger the internal resistance relative to external resistance, the greater the "lost" fraction of the e.m.f.

Terminal Potential Difference

The terminal p.d. is the p.d. measured across the terminals of a source when current is flowing. It equals the p.d. across the external circuit.

The Key Equation

Applying Kirchhoff's second law to a circuit with e.m.f. , internal resistance r, and external resistance R:

E.m.f. Equation = V + Ir

Or equivalently: = IR + Ir = I(R + r)

where:
= e.m.f. of source (V)
V = terminal p.d. (V)
I = current (A)
r = internal resistance (Ω)
R = external resistance (Ω)

Rearranging: Terminal p.d. = e.m.f. − lost volts

V = - Ir Lost Volts

The quantity Ir is called the lost volts or lost p.d. It represents the energy per coulomb dissipated inside the source. The larger the current, the greater the lost volts, and the lower the terminal p.d.

Special Cases

Worked Example: Internal Resistance

Problem: A battery has e.m.f. 12 V and internal resistance 0.5 Ω. It is connected to a 5.5 Ω resistor. Calculate (a) the current, (b) the terminal p.d., and (c) the lost volts.

Solution:

(a) Total resistance = R + r = 5.5 + 0.5 = 6\,Ω I = (/R + r) = (12/6) = 2 A (b) Terminal p.d. V = IR = 2 × 5.5 = 11 V Or: V = - Ir = 12 - (2 × 0.5) = 12 - 1 = 11 V (c) Lost volts = Ir = 2 × 0.5 = 1 V

Why High Internal Resistance is Problematic

Sources with high internal resistance are less useful because:

Example

Fruit and vegetable "batteries" (lemon cells, potato batteries) have very high internal resistance (hundreds of ohms). Although they have measurable e.m.f., they cannot deliver useful current because most of the energy is lost internally.

Measuring E.m.f. and Internal Resistance

The e.m.f. and internal resistance of a cell can be determined experimentally by measuring terminal p.d. and current for different external resistances.

FIG 10.17: Circuit for Measuring E.M.F. and Internal Resistance

Show a circuit with: a cell (e.m.f. ε, internal resistance r), a variable resistor, an ammeter in series, and a voltmeter in parallel across the cell terminals. Label all components.

Method

  1. Vary the external resistance using the variable resistor
  2. Record corresponding values of current I and terminal p.d. V
  3. Plot a graph of V against I

Graph Analysis

From V = - Ir, comparing with y = c + mx:

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