What causes internal resistance?

The cell’s own electrolyte and electrodes oppose the flow of charge through them, dissipating energy as heat inside the cell rather than delivering it to the circuit.

Why does a voltmeter on an unconnected cell read the e.m.f.?

A voltmeter has very high resistance, so almost no current flows. With I 0 the lost volts Ir are negligible and the terminal p.d. equals the e.m.f.

Why is a short-circuited battery dangerous?

With R 0 the current is /r, which can be very large, and all the power I is dissipated inside the cell. The heating can rupture or ignite it.

Is low internal resistance always better?

For efficiency, yes. But when the goal is maximum power transfer to a load, the optimum is R = r — a deliberate impedance match used in some electronics.

Back to course

45–46ElectricityStretch

E.m.f. and Internal Resistance

A real cell is not an ideal source. Its own materials resist the current passing through them, so some of the energy each coulomb receives is dissipated inside the cell before it ever reaches the circuit. That is $\textbf{internal resistance}$ , and it is why a battery’s terminal voltage falls as you draw more current.

35 min Topic 3 · Electric Circuits

What you'll be able to do

Define e.m.f. and internal resistance and explain terminal potential difference
Use $\varepsilon = I(R + r)$ and $V = \varepsilon - Ir$
Determine $\varepsilon$ and $r$ from a graph of $V$ against $I$
Explain why terminal p.d. falls as current increases

Where the lost volts go

The $\textbf{e.m.f.}$ $\varepsilon$ is the total energy the cell gives each coulomb. But the charge must travel through the cell itself, whose internal resistance $r$ opposes it, so some energy is dissipated there as heat.

What reaches the external circuit is the $\textbf{terminal potential difference}$ $V$ , the energy per coulomb actually delivered. The difference, $Ir$ , is called the $\textbf{lost volts}$ .

Applying conservation of energy round the loop: the e.m.f. equals the p.d. across the external resistance plus the p.d. across the internal resistance.

\varepsilon = I(R + r)

E.m.f. drives current through both resistances.

V = \varepsilon - Ir

Terminal p.d. is e.m.f. minus lost volts.

V = IR

And equals the p.d. across the external circuit.

Example 1A cell of e.m.f. 1.5 V and internal resistance

0.50\ \Omega

is connected to a

4.0\ \Omega

resistor. Find the current and the terminal p.d.

\varepsilon = I(R+r)

, so

1.5 = I(4.0 + 0.50) = 4.5I

I = \dfrac{1.5}{4.5} = 0.333

3Terminal p.d.:

V = IR = 0.333 \times 4.0

Answer

I = 0.33\ \mathrm{A}

;

V = 1.33\ \mathrm{V}

— the lost volts are

0.17

Tip — Check that $V + Ir = \varepsilon$ . Here $1.33 + 0.167 = 1.5$ ✓, which catches most arithmetic slips in one line.

Why terminal p.d. falls with current

The lost volts $Ir$ grow in direct proportion to the current, so drawing more current means more energy dissipated inside the cell and less available outside.

At the extremes the behaviour is clear. With no current drawn — an open circuit, or a voltmeter of very high resistance — there are no lost volts and the terminal p.d. equals the e.m.f. exactly. That is why a voltmeter across an unconnected cell reads its e.m.f.

At the other extreme, a short circuit makes $R \approx 0$ , so the current is limited only by the internal resistance: $I_{\text{max}} = \dfrac{\varepsilon}{r}$ . All the energy is then dissipated inside the cell, which is why a shorted battery heats up dangerously.

V = \varepsilon \text{ when } I = 0

Open circuit reads the e.m.f.

I_{\text{max}} = \dfrac{\varepsilon}{r}

Short-circuit current.

Example 1Explain why a car’s headlights dim when the starter motor is engaged.

1The starter motor draws a very large current from the battery.

2The lost volts

Ir

therefore become substantial.

3The terminal p.d.

V = \varepsilon - Ir

falls.

4The headlights receive a lower p.d. and so dissipate less power.

AnswerThe large starter current increases the lost volts, reducing the terminal p.d. available to the lights.

Finding e.m.f. and internal resistance graphically

Rearranging $V = \varepsilon - Ir$ into the form $y = mx + c$ shows that a graph of terminal p.d. against current is a straight line.

The $\textbf{intercept}$ on the $V$ -axis is $\varepsilon$ , since that is the terminal p.d. when no current flows. The $\textbf{gradient}$ is $-r$ , so the internal resistance is the magnitude of the (negative) gradient.

This is the standard experimental method: vary the external resistance, record $V$ and $I$ at each setting, plot $V$ against $I$ , and read both quantities off one graph.

V = -rI + \varepsilon

Straight line: gradient

-r

, intercept

\varepsilon

Example 1A graph of

V

against

I

has intercept 6.0 V and passes through

\left(2.0\ \mathrm{A},\ 4.8\ \mathrm{V}\right)

. Find

\varepsilon

and

r

1Intercept gives

\varepsilon = 6.0

2Gradient:

\dfrac{4.8 - 6.0}{2.0 - 0} = \dfrac{-1.2}{2.0} = -0.60

3Internal resistance is the magnitude.

Answer

\varepsilon = 6.0\ \mathrm{V}

;

r = 0.60\ \Omega

Tip — The gradient is negative and $r$ is its magnitude. Quoting a negative internal resistance is a sign slip, not a physical result.

Power and efficiency

The cell supplies total power $\varepsilon I$ . Of that, $I^{2}R$ is delivered usefully to the external circuit and $I^{2}r$ is wasted as heat inside the cell.

The efficiency is therefore $\dfrac{R}{R+r}$ , which approaches 1 when the external resistance greatly exceeds the internal resistance. Low internal resistance means an efficient source.

A subtlety worth knowing: maximum $\textbf{power transfer}$ to the external circuit occurs when $R = r$ , but the efficiency is then only 50%. Maximum power and maximum efficiency are different objectives, and which you want depends on the application.

P_{\text{total}} = \varepsilon I

Power supplied by the cell.

P_{\text{useful}} = I^{2}R, \quad P_{\text{wasted}} = I^{2}r

Split between outside and inside.

\eta = \dfrac{R}{R+r}

Efficiency.

Example 1A cell of e.m.f. 12 V and internal resistance

1.0\ \Omega

drives a

5.0\ \Omega

load. Find the power wasted internally and the efficiency.

1Current:

I = \dfrac{12}{6.0} = 2.0

2Power wasted:

I^{2}r = 4.0 \times 1.0 = 4.0

3Efficiency:

\dfrac{5.0}{6.0}

Answer

4.0\ \mathrm{W}

wasted; efficiency

83\%

Equation recap

\varepsilon = I(R+r)

The circuit equation.

V = \varepsilon - Ir

Terminal p.d.

\text{gradient} = -r

From a

V

–

I

graph.

I_{\max} = \tfrac{\varepsilon}{r}

Short-circuit current.

\eta = \tfrac{R}{R+r}

Efficiency.

Common mistakes to avoid

Using

\varepsilon = IR

and ignoring the internal resistance.

The current flows through

r

too:

\varepsilon = I(R+r)

Quoting a negative internal resistance from a graph.

The gradient is

-r

; the internal resistance is its magnitude.

Saying terminal p.d. equals e.m.f. under load.

They are equal only when no current flows. Otherwise

V = \varepsilon - Ir

Assuming maximum power transfer means maximum efficiency.

Maximum power occurs at

R = r

, where efficiency is only 50%.

Key takeaways

A real cell dissipates energy internally, so $V = \varepsilon - Ir$ .
The lost volts $Ir$ grow with current, so terminal p.d. falls under load.
On open circuit $V = \varepsilon$ ; on short circuit $I = \varepsilon/r$ .
A graph of $V$ against $I$ has intercept $\varepsilon$ and gradient $-r$ .
Efficiency is $R/(R+r)$ ; maximum power transfer at $R = r$ gives only 50%.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 6 lessons