3.5.1.4ElectricityCore

EMF and Internal Resistance

A brand-new battery labelled "1.5 V" never quite gives you 1.5 V once it’s actually driving a current — connect a voltmeter across its terminals under load and you’ll read a little less. The missing energy hasn’t vanished: every real source of e.m.f. has its own $internal resistance$ , and it dissipates some of its own energy pushing charge through itself before that charge ever reaches the rest of the circuit.

35 min Video by Science Shorts 3.5.2 DC circuits

EMF & Internal Resistance — A-level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Define e.m.f. as the energy transferred per unit charge by a source, including energy dissipated internally
Explain internal resistance and "lost volts"
Use the equation $ε = I (R + r)$
Calculate terminal potential difference for a source with internal resistance
Describe how to determine e.m.f. and internal resistance experimentally from a graph

EMF and internal resistance

The $electromotive force (e.m.f.)$ , $ε$ , of a source is the total energy transferred to each unit of charge that passes through it — not just the energy delivered to the external circuit, but $all$ of it, including whatever is dissipated inside the source itself. Every real cell or battery has some $internal resistance$ , $r$ , arising from the resistance of the chemicals and materials the current must pass through inside the source.

Because current must flow through this internal resistance too, some of the e.m.f. is inevitably "used up" driving current through the source itself, before the charge ever reaches the external circuit. This lost energy per unit charge is called the $lost volts$ , and what’s left over — the p.d. actually available across the external circuit — is the $terminal potential difference$ .

ε = I (R + r) = I R + I r

ε = terminal p.d. (IR) + lost volts (Ir).

1Lost volts:

I r = 2.0 \times 0.50 = 1.0

2Terminal p.d.:

V = ε - I r = 6.0 - 1.0

AnswerTerminal p.d. = 5.0 V; lost volts = 1.0 V.

Tip — E.m.f. is measured with (effectively) no current flowing — that’s when there are no lost volts and the terminal p.d. equals the full e.m.f.

Short-circuit current

If the external resistance is reduced to zero (a direct short circuit across the terminals), the only resistance left in the circuit is the source’s own internal resistance. This gives the maximum possible current the source can ever deliver — and is exactly why short-circuiting a battery is dangerous: a large current flows, dissipating a lot of power inside the source itself, which can generate significant heat.

I_{short-circuit} = \frac{ε}{r}

Maximum current when the external resistance is zero.

1Use

I = \frac{ε}{r}

with

R = 0

I = \frac{9.0}{1.5}

Answer

I = 6.0

Measuring EMF and internal resistance

Rearranging $ε = I R + I r$ in terms of the terminal p.d. $V = I R$ gives $V = ε - I r$ — a straight-line equation if you plot terminal p.d. $V$ against current $I$ (varying the external resistance, for instance with a rheostat, to get several data points). The line has $y$ -intercept $ε$ (the terminal p.d. you’d extrapolate to at zero current) and gradient $- r$ (the internal resistance, as a negative slope).

This graphical method is a standard AQA required-practical technique: it finds both e.m.f. and internal resistance from a single set of readings, without ever needing to measure either directly, and it lets you assess the uncertainty in each from how well the data fits the line.

V = ε - I r

Terminal p.d. against current: straight line, intercept ε, gradient −r.

1Comparing

V = ε - I r

with

y = c + m x

: the intercept

c

ε

, and the gradient

m

- r

2Intercept

= 4.5

V, so

ε = 4.5

3Gradient

= - 0.30 Ω

, so

r = 0.30 Ω

Answerε = 4.5 V, r = 0.30 Ω

Tip — Don’t forget the minus sign: the gradient of the V–I graph is −r, so a gradient of −0.30 Ω means the internal resistance itself is +0.30 Ω.

Equation recap

ε = I (R + r)

E.m.f. in terms of external resistance R and internal resistance r.

V = ε - I r

Terminal p.d. equation — used for the ε/r graphical method.

I_{short-circuit} = \frac{ε}{r}

Maximum current, when external resistance is zero.

Common mistakes to avoid

Treating e.m.f. and terminal p.d. as always equal.

They are only equal when no current flows; whenever current is drawn, terminal p.d. = ε − Ir, which is less than ε.

Forgetting internal resistance is itself part of the total circuit resistance when finding current.

Use ε = I(R+r) — internal resistance adds to the external resistance when calculating current, exactly like another series resistor.

Reading the internal resistance directly off the gradient of a V–I graph without the sign.

The gradient of V against I is −r, so internal resistance r is the positive value of (minus) the gradient.

Key takeaways

E.m.f. is the total energy per unit charge a source provides, including energy dissipated internally.
Every real source has internal resistance r, causing "lost volts" Ir and reducing terminal p.d. below ε.
ε = I(R + r); short-circuit current (R = 0) is ε/r, the maximum current a source can deliver.
Plotting terminal p.d. V against current I gives a straight line: y-intercept = ε, gradient = −r.

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EMF and Internal Resistance

internal resistance

, and it dissipates some of its own energy pushing charge through itself before that charge ever reaches the rest of the circuit.

35 min Video by Science Shorts 3.5.2 DC circuits