Internal resistance and EMF — Physics

Before this

This leads to

The potential divider circuit
Electrolysis

A battery that loses a volt as soon as you use it

Take a new 9 V battery and connect a voltmeter across its two terminals. The meter reads 9.0 V. Now connect a 4.0 Ω resistor across the same terminals, and leave the voltmeter where it is. The reading drops to 8.0 V.

Nothing else has changed. The battery is the same battery, a second later. So where did the missing volt go?

The last lecture gave you a rule that answers this. Round any closed loop, the e.m.f.s add up to the p.d.s across the resistors. It also promised to use that rule on something inside every real battery. Here it is. A real battery has a small resistance of its own, inside it. When a current flows, some of the battery's energy is used up pushing charge through its own insides. That is where the volt went.

This one idea explains several things you have probably seen. A torch dims when its batteries are old. A car's headlights dim for a moment when the engine is started. A phone can switch itself off in the cold while it still shows charge. By the end of this lecture, you will be able to measure the hidden resistance yourself.

Two different jobs for one battery

Start with the word you met in the last lecture. The electromotive force (e.m.f.) of a battery, written ε, is the energy the battery gives each coulomb of charge that passes through it. Its unit is the volt, because 1 V = 1 J/C. It is not a force. The name is older than our understanding of it.

The energy comes from chemical reactions inside the battery. But the charge must also travel through the battery, through its chemicals and plates, and they are not perfect conductors. They have a resistance. This is the internal resistance, written r.

So we model a real battery as two parts in series, inside one box: a perfect source of e.m.f. ε, and a small resistor r. You cannot take them apart, and you cannot reach the point between them. You can only reach the two outside terminals.

The dashed box is the battery. Inside it: a perfect e.m.f. ε and the internal resistance r. The voltmeter can only reach the outside wires, so it reads the terminal p.d. V, never ε directly while a current flows.

The p.d. you measure between the two terminals is the terminal p.d., written V. Some books call it the "terminal voltage".

Now apply the loop rule to the whole circuit. Walk round once, in the direction of the current. You gain ε inside the battery. You lose IR across the outside resistor. You also lose Ir across the internal resistance, because the same current flows through it.

ε = I R + I r

The p.d. across the outside resistor is the terminal p.d., V = IR. So the same equation can be written in two more ways.

E.m.f. and internal resistance

$ε = V + I r$ , so the terminal p.d. is $V = ε - I r$ .

With an outside resistance R: $ε = I (R + r)$ , so $I = \frac{ε}{R + r}$ .

The p.d. Ir across the internal resistance is often called the lost volts.

Read $ε = V + I r$ as an energy account for each coulomb. The battery gives each coulomb ε joules. Of these, V joules are delivered to the outside circuit. The other Ir joules are turned into thermal energy inside the battery. That is why a battery that delivers a large current gets warm.

Now the opening puzzle has numbers. The battery has ε = 9.0 V. With the 4.0 Ω resistor, the terminal p.d. is 8.0 V. So the current is $8.0 \div 4.0 = 2.0$ A, the lost volts are $9.0 - 8.0 = 1.0$ V, and the internal resistance is $1.0 \div 2.0 = 0.50$ Ω. The missing volt was used inside the battery.

Stop and think

Before the resistor was connected, the voltmeter read 9.0 V. Why did it show the full e.m.f. then, if the battery has an internal resistance all the time?

Show the answer

Because almost no current was flowing. The lost volts are Ir. With nothing connected, I is (almost) zero, so Ir is (almost) zero, and V = ε − Ir = ε. A good voltmeter has a very high resistance, so it draws only a tiny current. This is why the e.m.f. is often described as "the terminal p.d. when no current flows". That is a way to measure it. The meaning of e.m.f. is still the energy given to each coulomb.

More current, more lost volts

The equation V = ε − Ir says something you can test. The bigger the current you draw, the more volts are lost inside, and the lower the terminal p.d. becomes. The series and parallel lecture said that a real battery's p.d. "rises a little when less current is drawn". This is the reason.

It also changes a result from that lecture. There, each lamp in parallel got the full battery p.d., so adding a lamp did not change the others. That was true for a perfect battery. Test your feeling for a real one.

Predict — a second lamp in parallel

A battery with e.m.f. 6.0 V and internal resistance 1.0 Ω lights one lamp of resistance 6.0 Ω. You connect a second, identical lamp in parallel with the first. What happens to the first lamp?

This box shows what happens to the first lamp when a second, identical lamp is added in parallel. The battery has ε = 6.0 V and r = 1.0 Ω, and each lamp is 6.0 Ω.

Lamps in parallel	Outside resistance	Current from battery	Terminal p.d.	Power in each lamp
1	6.0 Ω	0.86 A	5.14 V	4.41 W
2	3.0 Ω	1.50 A	4.50 V	3.38 W

The first lamp gets dimmer. The second lamp makes the total current bigger, so more volts are lost inside the battery, and the p.d. across both lamps falls.

So "parallel branches are independent" is only as true as the battery is perfect. With a small internal resistance compared with the lamps, the change is small. With a large one, it is large. The same thing happens in a car. The starter motor draws a very large current from the battery, so for a moment the terminal p.d. drops, and the headlights, which share that battery, dim. Worked Example 3 puts typical numbers on this.

Measuring ε and r with a graph

You cannot open a battery to measure r. But you can find both ε and r from the outside. Connect the battery to a variable resistor, an ammeter in series, and a voltmeter across the terminals. Change the resistance, and record the current I and the terminal p.d. V each time.

Now rearrange the equation into the form of a straight line, y = c + mx:

V = ε - r I

So plot V up the side and I along the bottom. The points lie on a straight line. Its intercept on the V axis is ε, the terminal p.d. when no current flows. Its gradient is −r, so the internal resistance is the size of the gradient.

Measurements for one cell (invented, without the usual scatter). The intercept on the V axis gives ε = 1.50 V. The gradient is

- \frac{1.50}{3.0} = - 0.50

Ω, so r = 0.50 Ω. The dashed part is not measured: it is where the line would reach V = 0.

Notice the axes. In the Ohm's law lecture, I–V graphs put the current up the side. This graph puts the terminal p.d. up the side, so that the intercept is ε and the gradient is −r directly. Always check which way round a graph is drawn before you read a gradient.

The line also reaches the I axis. There, V = 0: all of the e.m.f. is lost inside the battery. This happens when the two terminals are joined by a thick wire with almost no resistance, a short circuit. The current is then as large as it can be, $I = ε \div r$ . For the cell in the graph, that is $1.50 \div 0.50 = 3.0$ A, and all of its power, $ε I = 4.5$ W, is turned into thermal energy inside the cell. This is why a short-circuited battery can get dangerously hot. In practice, the line is not quite straight at large currents, because r itself changes as the battery heats up. So nobody measures the short-circuit current directly.

Worked examples

Worked Example 1 — Following the energy

A battery of e.m.f. 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. Find the current, the terminal p.d., and the power delivered to the resistor and wasted in the battery.

Think first. The current goes through both resistances in series, so the total resistance is R + r.

I = \frac{ε}{R + r} = \frac{9.0}{4.0 + 0.50} = 2.0 A

Terminal p.d.: $V = ε - I r = 9.0 - 2.0 \times 0.50 = 8.0$ V. Check it another way: $V = I R = 2.0 \times 4.0 = 8.0$ V. ✓

Power from the e.m.f.: $ε I = 9.0 \times 2.0 = 18$ W. Power to the resistor: $I^{2} R = 2. 0^{2} \times 4.0 = 16$ W. Power wasted inside: $I^{2} r = 2. 0^{2} \times 0.50 = 2.0$ W.

Check. $16 + 2.0 = 18$ W, so every joule is accounted for. ✓ The battery delivers $16 \div 18 \approx 89%$ of its energy to the outside.

Worked Example 2 — Two readings give ε and r

A cell is connected to a 10 Ω resistor, and the terminal p.d. is 5.5 V. With a 2.0 Ω resistor instead, the terminal p.d. is 3.0 V. Find ε and r.

Think first. There are two unknowns, so you need two equations. Each reading gives one. First find each current from $I = V \div R$ , because V is the p.d. across the outside resistor.

Reading 1: $I_{1} = 5.5 \div 10 = 0.55$ A, so $ε = 5.5 + 0.55 r$ .

Reading 2: $I_{2} = 3.0 \div 2.0 = 1.5$ A, so $ε = 3.0 + 1.5 r$ .

Both equal ε, so they equal each other: $5.5 + 0.55 r = 3.0 + 1.5 r$ . That gives $2.5 = 0.95 r$ , so $r = \frac{50}{19} \approx 2.63$ Ω.

Keep the exact value for the next step: $ε = 5.5 + 0.55 \times \frac{50}{19} \approx 6.95$ V.

Check. Put r into the second equation: $3.0 + 1.5 \times \frac{50}{19} \approx 6.95$ V. ✓ The larger current gave the lower terminal p.d., as it must. ✓ Two readings are the least you need. A graph of many readings is better, because one bad reading cannot spoil the answer.

Worked Example 3 — Starting a car

These are typical values, not measurements of one car. A fully charged car battery reads about 12.6 V when nothing is drawing current. While the starter motor turns the engine, it draws about 150 A, and the terminal p.d. falls to about 10.0 V. Find the internal resistance, and the power wasted inside the battery while the engine is being started.

Think first. With no current, the reading is ε. With current, the reading is V. The difference is the lost volts.

Lost volts: $12.6 - 10.0 = 2.6$ V. So $r = 2.6 \div 150 \approx 0.017$ Ω.

Power wasted inside: $I^{2} r = 15 0^{2} \times \frac{2.6}{150} = 150 \times 2.6 = 390$ W.

Check. The internal resistance is tiny, much less than 1 Ω, but the current is huge, so the lost volts are large. ✓ For those few seconds, the headlights on the same battery get about 10 V instead of 12.6 V, so they dim. As a battery ages, r grows, and the terminal p.d. while starting falls further. This is why a battery can read a healthy 12.6 V with nothing connected and still fail to start the engine.

Now explore how the outside resistance changes where the energy goes.

Explore — where does the battery's energy go?

A battery has ε = 12 V and r = 2.0 Ω. Change the outside resistance R. Watch the power delivered to R (the curve), and how the battery's total power is split between R and its own insides (the bar).

Try:

Outside resistance R: 4.0 Ω

This table shows a battery with ε = 12 V and r = 2.0 Ω connected to different outside resistances R.

R (Ω)	I (A)	V (V)	Power to R (W)	Power wasted in r (W)	Share to R
1.0	4.0	4.0	16	32	33%
2.0	3.0	6.0	18	18	50%
4.0	2.0	8.0	16	8.0	67%
8.0	1.2	9.6	11.5	2.9	80%

The power delivered to R is largest, 18 W, when R equals r. But then only half of the battery's power reaches R. A larger R wastes a smaller share, but receives less power.

You should have seen two things that pull in opposite directions. When R is small, the current is large, and most of the energy is wasted inside the battery. When R is large, almost all the energy reaches R, but the current is small, so the power is small. The power delivered to R is greatest when R equals r, and then exactly half of the energy is wasted.

Three traps

Trap 1 — Calling the e.m.f. "the p.d. of the battery"

The e.m.f. is the energy given to each coulomb. The terminal p.d. is what reaches the outside circuit. They are equal only when no current flows. With a current, $V = ε - I r$ is always less than ε.

Trap 2 — Forgetting r when you find the current

$I = ε \div R$ is wrong for a real battery, because the current also passes through r. Use $I = ε \div (R + r)$ . Only the terminal p.d. divided by R gives the current directly: $I = V \div R$ .

Trap 3 — Reading the wrong thing from the graph

On a graph of V against I, the gradient is −r. So r is the size of the gradient, not one over it. If a question plots I up the side instead, the gradient is −1/r. Check the axes first. Also read ε from the V axis, not from the I axis: the I intercept is the short-circuit current.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A battery of e.m.f. 6.0 V and internal resistance 0.40 Ω is connected to a 2.6 Ω resistor. Find the current, the terminal p.d. and the lost volts.

Hint 1
The current passes through R and r in series.

Hint 2
$I = \frac{6.0}{2.6 + 0.40}$ . Then $V = ε - I r$ .

Full answer
$I = 6.0 \div 3.0 = 2.0$ A. Lost volts $= I r = 2.0 \times 0.40 = 0.80$ V. Terminal p.d. $V = 6.0 - 0.80 = 5.2$ V. Check: $V = I R = 2.0 \times 2.6 = 5.2$ V. ✓ And $5.2 + 0.80 = 6.0$ V. ✓
A high-resistance voltmeter across a cell reads 1.58 V when nothing else is connected. When a resistor is connected, the voltmeter reads 1.42 V and the current is 0.32 A. Find the e.m.f., the internal resistance and the resistance of the resistor.

Hint 1
With almost no current, the reading is the e.m.f.

Hint 2
The lost volts are $1.58 - 1.42$ V, and they equal Ir.

Full answer
$ε \approx 1.58$ V. Lost volts $= 1.58 - 1.42 = 0.16$ V, so $r = 0.16 \div 0.32 = 0.50$ Ω. The resistor: $R = V \div I = 1.42 \div 0.32 \approx 4.44$ Ω. Check: $I = 1.58 \div (4.4375 + 0.50) = 0.32$ A. ✓
A student measures a cell: at 0.20 A the terminal p.d. is 1.40 V, and at 0.80 A it is 1.10 V. The points lie on a straight line. Find the internal resistance, the e.m.f., and the current if the cell were short-circuited (assuming the line stays straight).

Hint 1
The gradient of V against I is −r.

Hint 2
Gradient $= \frac{1.10 - 1.40}{0.80 - 0.20}$ . Then use $ε = V + I r$ with either point.

Full answer
Gradient $= - 0.30 \div 0.60 = - 0.50$ Ω, so $r = 0.50$ Ω. Then $ε = 1.40 + 0.20 \times 0.50 = 1.50$ V. Check with the other point: $1.10 + 0.80 \times 0.50 = 1.50$ V. ✓ Short-circuit current $= ε \div r = 1.50 \div 0.50 = 3.0$ A. In practice the line bends at large currents, so this is an estimate.
A stretch. You have two identical cells, each with e.m.f. 1.5 V and internal resistance 0.50 Ω, and a lamp of resistance 1.0 Ω. Which gives the lamp more power: the cells in series, or the cells in parallel? (For two identical cells in parallel, the e.m.f. stays 1.5 V and the internal resistance halves.)

Hint 1
In series, the e.m.f.s add and the internal resistances add.

Hint 2
Series: ε = 3.0 V, r = 1.0 Ω. Parallel: ε = 1.5 V, r = 0.25 Ω. In each case, find I, then $P = I^{2} R$ .

Full answer
Series: $I = 3.0 \div (1.0 + 1.0) = 1.5$ A, so $P = 1. 5^{2} \times 1.0 = 2.25$ W. Parallel: $I = 1.5 \div (1.0 + 0.25) = 1.2$ A, so $P = 1. 2^{2} \times 1.0 = 1.44$ W. Series gives more power here. Why the parallel rule works: by symmetry each cell carries half the current, so the loop rule through one cell gives $1.5 = V + \frac{I}{2} \times 0.50$ , the same as one cell of 0.25 Ω. Check: in parallel the lost volts are only $1.2 \times 0.25 = 0.30$ V, but the e.m.f. is half as big, and that matters more for this lamp. ✓ Parallel cells are useful when the outside resistance is very small, or to make the battery last longer.

The bigger idea

The explorer showed that a battery delivers the most power when the outside resistance equals its internal resistance. This result is usually credited to Moritz von Jacobi, around 1840, and is sometimes called Jacobi's law. It was also misread. According to the usual account, some engineers concluded that an electric motor could never be more than 50% efficient. Their reason was that half the energy is lost at maximum power. Around 1880, Thomas Edison's team realised the mistake. Maximum power is not the same as maximum efficiency. Make the source's internal resistance very small compared with the load, and almost all of the energy reaches the load. (Source: "Maximum power transfer theorem", Wikipedia, citing Harrison, Physics Education, 2013.)

That is still how the world's electricity supply is designed: the resistance inside generators and cables is kept very small compared with what is connected to them. Matching is used where the aim is the most power from a weak source, for example in some radio receivers.

Internal resistance also explains a winter puzzle. The chemistry inside a lithium-ion cell slows down in the cold, so its internal resistance rises. When the phone suddenly draws a larger current, the lost volts can push the terminal p.d. below the level at which the phone protects itself by switching off. The energy is still in the cell. Warm it up, r falls, and the phone works again.

Key concepts in this lesson

The e.m.f. ε of a battery is the energy it gives each coulomb of charge, and it is measured in volts.
A real battery behaves like a perfect e.m.f. in series with a small internal resistance r.
The terminal p.d. is $V = ε - I r$ , so it falls as the current rises, and it equals ε only when no current flows.
The current in a circuit with outside resistance R is $I = ε \div (R + r)$ .
A graph of terminal p.d. against current is a straight line with intercept ε and gradient −r.
The power delivered to the outside resistance is greatest when R equals r, but then half of the energy is wasted inside the battery.