EG1216 · Physics 2 — Electricity & Magnetism
Theme 3 · Current & DC circuits

EMF & internal resistance

A real battery has its own small resistance inside, so the voltage you actually get sags as you draw more current.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §28.1.

Before you start

What you need first

  • Topic 17 — Ohm's law and resistance.
  • Topic 9 — voltage as energy per charge.
  • Topic 18 — power \(P = I^2R\).

What you'll be able to do

  • Say what EMF \(\varepsilon\) is.
  • Account for a battery's internal resistance \(r\).
  • Find the terminal voltage, \(V = \varepsilon - Ir\).
  • Explain why the voltage drops under load.

The battery's push

EMF — the full push

EMF \(\varepsilon\) (electromotive "force," though it's a voltage) is the push a battery gives when nothing is connected — its full voltage. It is measured in volts; a "9 V battery" has \(\varepsilon = 9\) V.

Water-pump picture: a battery is a pump that lifts charge up to a higher voltage so it can flow back down through the circuit.

Real batteries: internal resistance

A real battery also resists the current a little, inside itself — its internal resistance \(r\). So the voltage you actually get at the terminals (the terminal voltage) is a bit less than the EMF:

$$ V = \varepsilon - Ir $$
r ε real battery R I
Inside the dashed "real battery": the EMF \(\varepsilon\) in series with internal resistance \(r\), driving the load \(R\).
where:
SymbolMeaningSI unit
Vterminal voltage (what the circuit actually gets)V
εEMF (the full push)V
Icurrent drawnA
rinternal resistanceΩ
More current → bigger "lost volts" \(Ir\) → less voltage left for the circuit. With nothing connected (\(I=0\)) the terminals read the full \(\varepsilon\).
📐 Worked example 1

Terminal voltage

A battery has \(\varepsilon = 12\) V and internal resistance \(r = 0.50\ \Omega\). It supplies \(I = 2.0\) A. Find the terminal voltage.

1Use \(V = \varepsilon - Ir\):
$$ V = 12 - (2.0)(0.50) $$
2Work it out:
$$ V = 11\ \text{V} $$
The internal resistance "ate" 1 V; the circuit gets the other 11 V.
📐 Worked example 2

Battery driving a load

A battery (\(\varepsilon = 9.0\) V, \(r = 0.30\ \Omega\)) is connected to an external resistor \(R = 2.7\ \Omega\). Find (a) the current and (b) the terminal voltage.

1The EMF drives current through \(r\) and \(R\) in series: \(I = \dfrac{\varepsilon}{R+r}\):
$$ I = \frac{9.0}{2.7 + 0.30} = \frac{9.0}{3.0} = 3.0\ \text{A} $$
2Terminal voltage \(V = \varepsilon - Ir\) (= the voltage across \(R\)):
$$ V = 9.0 - (3.0)(0.30) = 8.1\ \text{V} $$
Check: \(V = IR = (3.0)(2.7) = 8.1\) V. ✓

✏️ Try it yourself

A car battery has \(\varepsilon = 12.6\) V and internal resistance \(r = 0.020\ \Omega\). When starting the engine it delivers \(I = 150\) A.

(a) Find the terminal voltage while starting.
(b) Find the power lost inside the battery.

(a) \(V = \varepsilon - Ir = 12.6 - (150)(0.020) = 12.6 - 3.0 = 9.6\ \text{V}\) (the lights dim!). (b) \(P = I^2 r = (150)^2(0.020) = 450\ \text{W}\) wasted as heat inside the battery.

Common mistakes

MistakeFix
Thinking the terminal voltage always equals the EMF.Only when \(I=0\). Under load, \(V = \varepsilon - Ir\) is less.
Leaving \(r\) out of the total resistance.The current sees \(R + r\): \(I = \varepsilon/(R+r)\).
Forgetting the battery itself heats up.It dissipates \(I^2 r\) internally.
Confusing EMF with "force."EMF is a voltage (in volts), despite the name.

Recap — the whole topic on one screen

IdeaWhat you own now
EMF\(\varepsilon\) = full push (terminal voltage at zero current).
Terminal voltage\(V = \varepsilon - Ir\); drops as current rises.
Current\(I = \dfrac{\varepsilon}{R+r}\) for a single load \(R\).
Internal loss\(P = I^2 r\) heats the battery.

Next topic

Resistors in series & parallel

Most circuits have several resistors. Next we reduce them to one: series ones add, parallel ones combine "upside-down" — the mirror image of the capacitor rules.

→ Topic 20 · Resistors in series & parallel