Cells, emf and internal resistance

What are emf and internal resistance?

In plain words: a cell is like a pump that lifts charge through a fixed height, its emf. But some of the lift is used up pushing the current through the cell’s own insides. The more current you draw, the more is lost inside, and the less is left at the terminals.

Emf of a cell

• The source here is a galvanic (electrochemical) cell; an electrolytic cell strictly uses external electricity to drive a reaction. This source cell has two electrodes, positive (P) and negative (N), immersed in an electrolytic solution. Dipped in the solution, the electrodes exchange charges with the electrolyte (NCERT pages 93–94, Figure 3.12).
• The positive electrode develops a potential difference V₊ (> 0) relative to the electrolyte next to it (point A); the negative electrode develops −V₋ (V₋ ≥ 0) relative to the electrolyte next to it (point B) (page 94).
• With no current, the electrolyte is at one potential throughout, so the potential difference between P and N is V₊ + V₋, the emf (NCERT Eq. (3.36), page 94):

• So the emf is the potential difference between the positive and negative electrodes in an open circuit, when no current flows through the cell (NCERT Eq. (3.37), page 94).

Internal resistance

The constant-emf, constant-r model is an approximation over the conditions considered. Real internal voltage losses include effects beyond the electrolyte alone. Short-circuit and maximum-power formulae here assume r > 0; cell chemistry and state of charge can change both parameters.

• Connect a resistor R across the cell. A steady current I flows through R from P to N, and through the electrolyte from N to P (NCERT page 94).
• The electrolyte through which this current flows has a finite resistance r, the internal resistance of the cell (page 94).
• Internal resistances vary from cell to cell; the internal resistance of dry cells is much higher than that of common wet cells. It may be neglected in calculations when ε ≫ Ir (page 94).

Terminal potential difference

• With a current I, the potential difference between P and N is (NCERT Eq. (3.38), page 94):

• The minus sign appears because inside the electrolyte the current flows from B to A, so the potential drops by Ir across the internal resistance (page 94).
• Since V is also the potential difference across R, V = IR by Ohm’s law (NCERT Eq. (3.39)). Combining, IR = ε − Ir, so (NCERT Eq. (3.40), page 95):

• The largest current is drawn when R = 0, a short circuit: I_max = ε/r, and the terminal potential difference is then zero. In most cells the safe current is much lower, to prevent permanent damage (page 95).
• While a cell is being charged, the current is forced through it from P to N, and then V = ε + Ir: the terminal potential difference is more than the emf (NCERT Eq. (3.60), page 97, which gives this when the current through the cell runs from P to N; see labelling cells in Kirchhoff’s rules).

Solved examples

Common mistakes to avoid

Quick revision summary

• A cell’s emf ε is the potential difference between its electrodes in an open circuit, ε = V_₊ + V_₋; it is a potential difference, not a force.
• The electrolyte’s resistance r is the internal resistance; dry cells have much higher r than common wet cells.
• Supplying a current I: V = ε − Ir and I = ε/(R + r); being charged: V = ε + Ir.
• The largest current, with R = 0, is ε/r; cells are usually limited well below this.
• r can be neglected when ε ≫ Ir.

Questions students ask

What is the difference between emf and terminal voltage?

The emf of a cell is the work its source does per unit charge in moving positive charge internally from the negative to the positive terminal, equal to the potential difference between its terminals when no current flows, while the terminal voltage is the potential difference across the terminals, both with current and in an open circuit (NCERT Summary point 2, page 102; Eqs. (3.37)–(3.38), page 94). The emf is set by the cell itself, whatever the external circuit; the terminal voltage depends on the current: V = ε − Ir when the cell supplies current, ε + Ir when it is charged (NCERT Eq. (3.60), page 97). So a 9.0 V battery of internal resistance 1.5 Ω supplying 0.40 A has 9.0 − 0.40 × 1.5 = 8.4 V across its terminals.

Is the internal resistance of a cell constant, and what does it depend on?

The internal resistance of a cell is not a fixed constant, though NCERT’s calculations treat it as one: it is the resistance of the electrolyte between the electrodes (NCERT page 94), so, like a column of solution between two electrodes, R = ρl/A, it grows with the distance between the electrodes and falls with the area of electrode in the electrolyte (reasoned from NCERT Chemistry Part I (Class 12), Unit 2, pages 43–44). It also depends on the nature and concentration of the electrolyte, and it falls as the cell warms, because the conductivity of an electrolytic solution rises with temperature (page 43).

Does the emf of a cell depend on its size?

No, the emf of a cell does not depend on its size or on the amount of electrolyte: it is the sum of the potential differences at its two electrodes, ε = V₊ + V₋ (NCERT Eq. (3.36), page 94), set by the materials, and NCERT Chemistry calls a cell’s emf an intensive parameter (NCERT Chemistry Part I (Class 12), Unit 2, page 40). It does depend on the ion concentrations, which change as the cell is used, and on temperature (NCERT Chemistry, Eq. (2.11), pages 38–39). Beyond the NCERT text, a bigger cell of the same kind has the same emf but more chemicals, so it delivers more charge, and its larger electrodes usually lower its internal resistance.

How do you find the emf and internal resistance of a cell from a graph?

A graph of a cell’s terminal voltage V against its current I gives its emf and internal resistance: since V = ε − Ir (NCERT Eq. (3.38), page 94), points taken with several external resistances lie on a straight line whose intercept on the V-axis is the emf ε and whose slope is −r, and the line would meet the I-axis at the short-circuit current ε/r. For example, a line from 3.0 V at I = 0 down to 2.4 V at 1.5 A gives ε = 3.0 V and r = (3.0 − 2.4)/1.5 = 0.40 Ω, so ε/r = 7.5 A. Plotted against the external resistance R instead, V = εR/(R + r) is a curve that rises from zero towards ε (NCERT Summary point 11, page 103).

Does a voltmeter connected across a cell read its emf?

A voltmeter across a cell reads a little less than its emf, because it draws a small current through the internal resistance: a voltmeter of resistance R_V reads εR_V/(R_V + r) (NCERT Summary point 11, page 103, with the voltmeter as the external resistance). The reading is close to the emf only when R_V ≫ r: a 1000 Ω voltmeter across a cell of emf 2.0 V and internal resistance 0.80 Ω reads 2.0 × 1000/1000.8 ≈ 1.998 V. Across a cell that is also supplying a load, the same voltmeter reads the terminal voltage, V = ε − Ir, not the emf (NCERT Eq. (3.38), page 94).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A cell of emf 1.5 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. Find the current and the terminal voltage.

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   I = ε/(R + r) = 1.5/3.0 = 0.50 A; V = IR = 0.50 × 2.5 = 1.25 V (NCERT Eqs. (3.38)–(3.40), pages 94–95).

2. Q2. When is the terminal potential difference of a cell equal to its emf?

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   When no current flows through it, in an open circuit (NCERT page 94).

3. Q3. Which usually has the higher internal resistance, a dry cell or a common electrolytic cell?

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   A dry cell (NCERT page 94).