Sit in a parked car at night with the headlights on and turn the key. For a second or two, while the starter motor turns the engine, the headlights dim. Then the engine catches and they brighten again. Nothing was switched off. The same battery simply gave the lamps a smaller voltage for a moment.
What a cell does to a charge
In the external circuit, positive charge flows from the high-potential terminal to the low one, and the field does work on it. Inside the cell the charge must go the other way, from the negative terminal up to the positive one. The electric field inside pushes against that. Chemical reactions supply the push and do work on the charge.
The emf of a cell is that work per unit charge: \(\mathcal E = W/q\). Its unit is the joule per coulomb, the volt. A cell of emf \(1.5\,\text{V}\) does \(1.5\,\text{J}\) of chemical work on every coulomb it pushes through itself. The name "electromotive force" is older than the idea. The emf is a work per charge, not a force.
The emf \(\mathcal E\) of a cell is the work done by its non-electrostatic forces per unit charge. The charge is carried through the cell from the negative terminal to the positive one. It equals the potential difference across the terminals when no current flows.
The cell's own resistance
The charge crosses the electrolyte and the electrodes, and these conduct imperfectly. So every real cell has an internal resistance \(r\) in series with its emf. We draw a real cell as an ideal emf and a resistor \(r\) inside one dashed outline. No terminal lies between the two, so we can never measure them separately with a wire.
Now let a current \(I\) flow out of the positive terminal, as in a torch. Walk through the cell from the negative terminal B to the positive terminal A. Crossing the emf raises the potential by \(\mathcal E\). Crossing \(r\) in the direction of the current lowers it by \(Ir\). So the potential difference across the terminals, the terminal voltage, is \(V = \mathcal E - Ir\).
Now force a current into the positive terminal, as a charger does. The current inside now flows from A to B. Walking from B to A we cross \(r\) against the current, so the potential rises by \(Ir\). The terminal voltage becomes \(V = \mathcal E + Ir\), larger than the emf. A charger must push harder than the emf it works against.
In the figure above the emf always lifts the potential by \(12\,\text{V}\). At \(2\,\text{A}\) of discharge the internal resistance takes back \(0.5 \times 2 = 1\,\text{V}\), and A sits at \(11\,\text{V}\). At \(4\,\text{A}\) of charging it adds \(2\,\text{V}\), and A sits at \(14\,\text{V}\). At zero current the line is flat after the rise, and \(V = \mathcal E\).
A cell of emf \(12\,\text{V}\) and internal resistance \(1\,\Omega\) is joined to a \(5\,\Omega\) resistor. Find the current, the terminal voltage and the potential drop inside the cell.
The loop holds \(5 + 1 = 6\,\Omega\) in all, so \(I = \mathcal E/(R + r) = 12/6 = 2\,\text{A}\).
The terminal voltage is the drop across the resistor, \(V = IR = 2 \times 5 = 10\,\text{V}\). The drop inside is \(Ir = 2 \times 1 = 2\,\text{V}\).
Check: \(V = \mathcal E - Ir = 12 - 2 = 10\,\text{V}\).
Values used in this lesson: the cells and loads of the workbook's first sheet. The car battery has \(12\,\text{V}\) and \(r = 0.02\,\Omega\), and its starter draws \(100\,\text{A}\); these are typical round figures.
A charger pushes \(4\,\text{A}\) into the positive terminal of a battery of emf \(12\,\text{V}\) and internal resistance \(0.5\,\Omega\). Find the terminal voltage, the power stored as chemical energy and the power lost as heat.
The current enters the positive terminal, so the battery is being charged.
\(V = \mathcal E + Ir\). Stored power \(= \mathcal EI\); heat \(= I^2r\).
\(V = 12 + 4 \times 0.5 = 14\,\text{V}\). Stored: \(12 \times 4 = 48\,\text{W}\). Heat: \(16 \times 0.5 = 8\,\text{W}\).
The terminals stand at \(14\,\text{V}\). Of the \(56\,\text{W}\) supplied, \(48\,\text{W}\) is stored and \(8\,\text{W}\) warms the battery.
When is the terminal voltage of a cell greater than its emf?
Two readings give \(\varepsilon\) and \(r\)
An ideal voltmeter across a cell with nothing else joined reads the emf, because no current flows. Join a load and the reading falls by \(Ir\). One reading at zero current and one at a known current therefore give both \(\mathcal E\) and \(r\).
An ideal voltmeter across a cell reads \(6\,\text{V}\) with the switch open. The switch is closed and a current flows through the outer resistor. What happens to the voltmeter reading?
With the switch open the voltmeter reads \(\mathcal E = 6\,\text{V}\). With it closed the ammeter reads \(2\,\text{A}\) and the voltmeter \(5\,\text{V}\). So \(5 = 6 - 2r\), which gives \(r = 0.5\,\Omega\), and the outer resistance is \(R = V/I = 5/2 = 2.5\,\Omega\). A cell that reads \(2\,\text{V}\) idle and \(1.5\,\text{V}\) while it drives \(1\,\text{A}\) has \(r = 0.5\,\Omega\) and drives \(R = 1.5\,\Omega\).
Plot the terminal voltage against the current drawn. Since \(V = \mathcal E - Ir\), the graph is a straight line. It meets the \(V\) axis at the emf and falls with slope \(-r\). It meets the \(I\) axis where \(V = 0\), at the short-circuit current \(\mathcal E/r\). A cell of \(6\,\text{V}\) and \(2\,\Omega\) gives a line from \(6\,\text{V}\) down to \(3\,\text{A}\).
The short-circuit current is the largest current a cell can give, and all of its power then heats the cell itself. A thick wire across a car battery can carry hundreds of amperes, melt and burn. Never join the terminals of a cell directly.
A cell gives a terminal voltage of \(5.5\,\text{V}\) while it supplies \(1\,\text{A}\) and \(5\,\text{V}\) while it supplies \(2\,\text{A}\). Find its internal resistance and its emf.
Write \(V = \mathcal E - Ir\) twice: \(5.5 = \mathcal E - r\) and \(5 = \mathcal E - 2r\).
Subtract: \(0.5 = r\), so \(r = 0.5\,\Omega\). Then \(\mathcal E = 5.5 + 0.5 = 6\,\text{V}\).
One more ampere lowered the terminal voltage by \(0.5\,\text{V}\); that fall per ampere is \(r\).
A battery shows \(9\,\text{V}\) across its terminals while it supplies \(1\,\text{A}\) and \(8\,\text{V}\) while it supplies \(3\,\text{A}\). Find its emf, in volts.
Where the cell's power goes
The chemistry does work \(\mathcal E\) on each coulomb, and \(I\) coulombs pass each second, so it supplies the power \(\mathcal EI\). The load outside receives \(VI\). The internal resistance turns \(I^2r\) into heat inside the cell. Energy is conserved, so the three must balance.
\[ \mathcal E I = VI + I^2 r \]
Why. the power the chemistry supplies equals the power delivered outside plus the heat inside the cell.
\[ \mathcal E = V + Ir \]
Why. divide every term by the current \(I\), which is not zero.
\[ V = \mathcal E - Ir \]
Read it. rearrange; the terminal voltage is the emf less the drop across the internal resistance.
The efficiency of the cell is the share of its power that reaches the load: \(VI/\mathcal EI = V/\mathcal E = R/(R + r)\). A cell of \(6\,\text{V}\) and \(2\,\Omega\) on a \(4\,\Omega\) load drives \(1\,\text{A}\). The load gets \(4\,\text{W}\) at \(4\,\text{V}\), the cell wastes \(2\,\text{W}\), and the efficiency is \(4/6 = 2/3\).
Now the car of the evening. Suppose the battery has \(\mathcal E = 12\,\text{V}\) and \(r = 0.02\,\Omega\), round values for a car battery. The headlights draw only a few amperes, so the drop is tiny. The starter motor draws about \(100\,\text{A}\). The drop inside becomes \(100 \times 0.02 = 2\,\text{V}\), and the terminals fall to about \(10\,\text{V}\).
The headlights sit across those same terminals. With the filament's resistance fixed, their power goes as \(V^2\): \((10/12)^2 \approx 0.69\). For those seconds they give about two-thirds of their usual power, and they dim. When the engine runs and the starter stops, \(I\) falls, \(Ir\) vanishes and the lamps recover.
Explain why the headlights of a car dim for a moment while the starter motor turns the engine.
The starter draws a very large current from the battery. That current crosses the battery's internal resistance, so the drop \(Ir\) becomes large. The terminal voltage \(V = \varepsilon - Ir\) falls. The headlights are joined across the terminals, so they receive a smaller voltage and a smaller power, and they dim. When the starter stops, the current and the drop \(Ir\) fall, and the voltage recovers.
The load that takes the most power
A cell drives a load \(R\) with \(I = \mathcal E/(R + r)\), so the load receives \(P = I^2R = \mathcal E^2R/(R + r)^2\). A tiny load takes a large current but has almost no voltage across it. A huge load has the full voltage but almost no current. Both extremes deliver almost nothing, so the power must peak somewhere between.
In the figure above the power climbs, peaks at \(4.5\,\text{W}\) when \(R = 2\,\Omega\), and falls again. The peak sits where the load matches the internal resistance. There \(V = 3\,\text{V}\), half the emf, and the efficiency is only \(50\,\%\): the cell wastes as much as it delivers.
\[ P = \dfrac{\mathcal E^2 R}{(R + r)^2} \]
Why. the current \(\mathcal E/(R + r)\) squared, times the load \(R\).
\[ (R + r)^2 = (R - r)^2 + 4Rr \]
Why. expand both sides; each equals \(R^2 + 2Rr + r^2\).
\[ P = \dfrac{\mathcal E^2}{(R - r)^2/R + 4r} \]
Why. substitute step 2 and divide the top and the bottom by \(R\).
\[ P_{\max} = \dfrac{\mathcal E^2}{4r} \text{ at } R = r \]
Read it. the term \((R - r)^2/R\) is never negative and is zero only at \(R = r\), so the denominator is least there.
Calculus reaches the same peak. Differentiate \(P = \mathcal E^2R/(R + r)^2\) with respect to \(R\). The derivative is \(\mathcal E^2(r - R)/(R + r)^3\), which is zero at \(R = r\), positive below it and negative above it.
The graph can be read backwards. If a cell's power curve peaks at \(4.5\,\text{W}\) at \(2\,\Omega\), then \(r = 2\,\Omega\). And \(\mathcal E^2/4r = 4.5\) gives \(\mathcal E^2 = 36\), so \(\mathcal E = 6\,\text{V}\). At \(R = 6\,\Omega\) the same cell delivers \(36 \times 6/64 = 3.375\,\text{W}\).
The curve is not symmetric in \(R\), but it pairs loads. The loads \(r/2\) and \(2r\) both receive \(\tfrac{2}{9}\,\mathcal E^2/r\), while \(R = r\) receives \(\tfrac14\,\mathcal E^2/r\). In general \(R\) and \(r^2/R\) receive the same power. So maximum power and maximum efficiency are different aims: a torch wants \(R \gg r\), a loudspeaker matched to an amplifier wants \(R = r\).
The power a cell delivers to a load peaks at \(4.5\,\text{W}\) when the load is \(2\,\Omega\). Find the internal resistance, the emf, and the power delivered to a \(6\,\Omega\) load.
The peak of the power curve is at \(R = r\).
\(r = 2\,\Omega\) and \(P_{\max} = \mathcal E^2/4r\); then \(P = \mathcal E^2R/(R + r)^2\).
\(4.5 = \mathcal E^2/8\), so \(\mathcal E = 6\,\text{V}\). At \(6\,\Omega\): \(P = 36 \times 6/8^2 = 3.375\,\text{W}\).
The cell has \(r = 2\,\Omega\) and \(\mathcal E = 6\,\text{V}\); a \(6\,\Omega\) load receives \(3.375\,\text{W}\).
A battery has an emf of \(12\,\text{V}\) and an internal resistance of \(3\,\Omega\). What is the largest power it can deliver to an outside load, in watts?
A cell delivers the largest possible power to a load. What is its efficiency then?
- Treating the emf as the voltage across the terminals at every current.The terminals show \(\mathcal E\) only when no current flows. Under load subtract \(Ir\); while charging add it.
- Writing \(V = \mathcal E - Ir\) for a cell that is being charged.Look at the direction of the current inside the cell. From + to − inside means charging, and then \(V = \mathcal E + Ir\).
- Using \(\mathcal EI\) as the power delivered to the load.\(\mathcal EI\) is what the chemistry supplies. The load receives \(VI\); the difference \(I^2r\) heats the cell.
- Thinking that the largest power means the best efficiency.At \(R = r\) the efficiency is only \(50\,\%\). High efficiency needs \(R \gg r\).