Cells, emf and internal resistance

Emf and internal resistance

Narrated video. Transcript:

A steady current needs something to keep pushing charges round the circuit. That is the job of an electrolytic cell. Look inside one: two electrodes, P and N, dipped in an electrolyte.

The electrolyte gives the positive electrode P a potential V plus above it, and the negative electrode N a potential V minus below it. So V minus is a negative number.

With no current flowing, the potential difference between P and N is V plus minus V minus. This is called the electromotive force, or emf, of the cell, written ε.

Despite its name, emf is not a force. It is a potential difference, measured in volts: the open-circuit voltage between the terminals.

Now connect a resistor R across the cell. Current flows through R from P to N, and inside the cell, through the electrolyte, from N to P.

The electrolyte itself has a resistance, r, called the internal resistance of the cell. Draw it as a small resistor inside the cell.

Some of the emf is used up driving current through r. The terminal voltage is V equals ε minus I r. And since V equals I R, the current is I equals ε over R plus r.

Worked example. A battery of emf 12 volt and internal resistance 3 ohm drives a current of 0.5 ampere through a resistor. Then 12 equals 0.5 times, R plus 3, so R plus 3 is 24, and R is 21 ohm.

Terminal voltage: V equals I R, 0.5 times 21, which is 10.5 volt. The other 1.5 volt, I r, is lost inside the battery.

Short the terminals with R equal to zero, and the current is as large as it can be: ε over r. A car battery has a very small r, so it can give a huge current to start the engine.

Cells in series and in parallel

Narrated video. Transcript:

Two cells can be joined in series: the negative terminal of the first to the positive terminal of the second. Cell 1 has emf ε₁ and internal resistance r₁; cell 2 has ε₂ and r₂.

Walk from A to C through both cells with current I. The potential differences add: V AC equals ε₁ plus ε₂, minus I times r₁ plus r₂.

That is exactly one cell with ε eq equals ε₁ plus ε₂ and r eq equals r₁ plus r₂. For n identical cells in series, the emf is n ε and the internal resistance n r.

Turn cell 2 the other way round, negative to negative. Now its emf opposes the first: ε eq becomes ε₁ minus ε₂. But the internal resistances still add.

Now join the two cells in parallel: both positive terminals to one point, both negative terminals to another. Currents I₁ and I₂ leave the cells and join to give I equals I₁ plus I₂.

Both cells have the same terminal voltage V. So V equals ε₁ minus I₁ r₁, and also ε₂ minus I₂ r₂. Solve for I₁ and I₂ and add them.

The pair behaves like one cell with r eq equals r₁ r₂ over r₁ plus r₂, and ε eq equals ε₁ r₂ plus ε₂ r₁, over r₁ plus r₂.

Example: two cells of 1.5 volt and 0.5 ohm each, across a 2 ohm resistor. In series: ε eq is 3 volt, r eq is 1 ohm, and the current is 3 over 3, which is 1 ampere.

In parallel: ε eq stays 1.5 volt, r eq is 0.25 ohm, and the current is 1.5 over 2.25, about 0.67 ampere. Each cell supplies only half of it.

So joining cells in series raises the emf, while joining identical cells in parallel keeps the emf the same but lowers the internal resistance.

Try it: load a cell

Interactive simulation. Things to try:

• Tick open circuit: no current flows and the voltmeter reads the emf ε.
• Lower R. The current rises, the lost volts Ir grow, and the terminal voltage falls along the line V = ε − Ir.
• Set R = 0: a short circuit. The current is ε/r, and the terminal voltage is zero.
• Switch to the power graph. The power given to the load is largest when R equals r.

Try it: two cells together

Interactive simulation. Things to try:

• In series, the emfs add and the internal resistances add: ε_eq = ε₁ + ε₂, r_eq = r₁ + r₂.
• Reverse cell 2 in series: the emfs now subtract.
• Switch to parallel with two identical cells: ε_eq stays the same and r_eq halves.
• In parallel, make ε₂ much smaller than ε₁. Watch I₂ go negative: cell 2 is being charged by cell 1.

Terminal voltage

A cell of emf 2 V and internal resistance 0.5 Ω is connected to a 4 Ω resistor. What does a voltmeter across the cell’s terminals read?

1. 2.00 V
2. 1.78 V
3. 2.22 V
4. 0.22 V

Unequal cells in parallel

A cell of emf 2 V and internal resistance 1 Ω is connected in parallel with a cell of emf 1 V and internal resistance 1 Ω (positive to positive). What single cell are they equivalent to?

1. ε_eq = 3 V, r_eq = 2 Ω
2. ε_eq = 1.5 V, r_eq = 0.5 Ω
3. ε_eq = 2 V, r_eq = 0.5 Ω
4. ε_eq = 1.5 V, r_eq = 2 Ω

Summary

• Current and terminal voltage: I = ε/(R + r), V = ε − Ir
• Cells in series: ε_eq = ε₁ + ε₂, r_eq = r₁ + r₂
• Cells in parallel: ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂), r_eq = r₁r₂/(r₁ + r₂)

• The emf ε of a cell is the potential difference between its terminals when no current flows.
• The electrolyte has an internal resistance r, so the terminal voltage falls to ε − Ir when a current I is drawn. The largest possible current is ε/r.
• In series, emfs add (subtract if a cell is reversed) and internal resistances add. n identical cells give nε and nr.
• In parallel, the combination behaves like a single cell with the ε_eq and r_eq above; identical cells keep ε and divide r.