EMF, internal resistance and terminal voltage

Pick up a brand-new cell stamped "1.5 V", touch a voltmeter across it while it powers nothing, and you read close to 1.5 V. Wire it into a torch bulb instead, and the reading sags — often noticeably. The cell has not changed; what has changed is that current is now flowing, and every cell resists its own current a little. That internal resistance is the missing piece between the label on a battery and what a circuit actually receives. This article separates three quantities that get blurred in careless reading: EMF, the cell's maximum push with no current flowing; terminal voltage, what you actually measure once current is drawn; and internal resistance, the culprit that eats the difference. One equation, $E = V + Ir$ , ties them together and drives every numerical in this topic — current-finding, terminal-voltage-finding, and internal-resistance-finding, all from the same relation rearranged three ways.

The core insight

A real cell's terminal voltage $V$ always falls short of its EMF $E$ by exactly $Ir$ , the voltage used up driving current through the cell's own internal resistance $r$ .

The core idea

Electromotive force (EMF), $E$ . This is the energy the cell gives to each coulomb of charge it pushes around the circuit, measured in volts, and it is defined for the open-circuit condition — no current flowing. It is the cell's maximum possible potential difference, fixed by its internal chemistry, and does not depend on what is connected outside.

Internal resistance, $r$ . Every real cell is built from an electrolyte and electrodes that themselves resist the flow of charge. This resistance sits inside the cell, in series with whatever external circuit you connect. It is a property of the cell alone — never of the external wires or resistors.

Terminal voltage, $V$ . This is the potential difference you actually measure across the cell's terminals once current $I$ is flowing. Because some of the EMF is spent driving current through $r$ before the charge even reaches the terminals, $V$ is always less than $E$ whenever $I > 0$ . The three quantities connect through one relation:

E = V + Ir

where: $E$ is EMF in volts, $V$ is terminal voltage in volts, $I$ is the current in amperes, and $r$ is internal resistance in ohms. Rearranged for terminal voltage, this becomes $V = E - Ir$ — terminal voltage equals EMF minus the "voltage lost" inside the cell. If the cell drives current through a single external resistor $R$ , then $V = IR$ as well (Ohm's law on the external part). Substituting into $E = V + Ir$ gives $E = IR + Ir = I(R + r)$ , so:

I = \frac{E}{R + r}

This is the master formula for a single-loop circuit with one cell: current is EMF divided by the total resistance of the loop, external plus internal — never external alone. A cell with very low $r$ behaves almost like an ideal source; a worn or heavily loaded cell has a larger $r$ , and its terminal voltage sags visibly under load — the reason a car's headlights dim while the starter motor cranks and draws a large current through the battery's internal resistance.

Figure

A single loop circuit with a cell of EMF E and internal resistance r drawn as a small resistor inside the cell symbol connected to an external resistor R showing current I flowing around the loop and the voltmeter reading V across the external terminals

Derivation

Derivation AGetting the current formula I = E/(R + r) from E = V + Ir

The goal is to show, starting only from the defining relation $E = V + Ir$ and Ohm's law applied to the external resistor, that the single-loop current is $I = E/(R + r)$ — the formula every numerical in this topic ultimately reduces to, and the one place internal resistance shows up as simply "one more resistance added in series" from the current's point of view.

Step 1. Start from the defining relation. By definition of internal resistance, $E = V + Ir$ , where $V$ is the terminal voltage that appears across the external circuit.

Step 2. Bring in Ohm's law for the external resistor. Since $V$ is also the potential difference driving current $I$ through the external resistance $R$ , Ohm's law gives $V = IR$ .

Step 3. Substitute $V = IR$ into $E = V + Ir$ . This gives $E = IR + Ir$ .

Step 4. Factor out the common current $I$ on the right-hand side: $E = I(R + r)$ .

Step 5. Divide both sides by $(R + r)$ to isolate the current, giving $I = \dfrac{E}{R + r}$ . Physically, this says the internal resistance simply adds to the external resistance in the denominator — a cell's own resistance always shrinks the current below the ideal value $E/R$ that a resistance-free cell would drive.

Worked example

A cell of EMF $2.0\ \text{V}$ and internal resistance $0.5\ \Omega$ is connected to an external resistor of $3.5\ \Omega$ . Find (a) the current in the circuit, (b) the terminal voltage, and (c) the internal resistance a different battery would need if a $6.0\ \text{V}$ cell driving $12\ \Omega$ of external resistance produced a current of $0.5\ \text{A}$ .

Step 1 — Find the current using $I = \dfrac{E}{R+r}$ . With $E = 2.0\ \text{V}$ , $R = 3.5\ \Omega$ and $r = 0.5\ \Omega$ : $I = \dfrac{2.0}{3.5 + 0.5} = \dfrac{2.0}{4.0} = 0.5\ \text{A}$ .

Step 2 — Find the terminal voltage using $V = E - Ir$ . $V = 2.0 - (0.5 \times 0.5) = 2.0 - 0.25 = 1.75\ \text{V}$ .

Step 3 — Cross-check with $V = IR$ . $V = 0.5 \times 3.5 = 1.75\ \text{V}$ , matching Step 2 exactly, confirming the arithmetic is consistent.

Step 4 — Part (c): a second battery, $E = 6.0\ \text{V}$ , $R = 12\ \Omega$ , $I = 0.5\ \text{A}$ . Rearrange $I = \dfrac{E}{R+r}$ to solve for $r$ : $0.5 = \dfrac{6.0}{12 + r}$ , so $12 + r = \dfrac{6.0}{0.5} = 12.0$ , giving $r = 0\ \Omega$ .

The current in the first circuit is $\mathbf{0.5\ A}$ , the terminal voltage is $\mathbf{1.75\ V}$ (less than the $2.0\ \text{V}$ EMF, as expected), and the second battery has negligible internal resistance ( $\mathbf{r \approx 0\ \Omega}$ ), behaving almost like an ideal source for that load.

Competitive stretch

Two identical cells, each of EMF $1.5\ \text{V}$ and internal resistance $0.5\ \Omega$ , are connected in series to a $5\ \Omega$ external resistor. Step 1. Series cells add their EMFs and their internal resistances: $E_{\text{total}} = 1.5 + 1.5 = 3.0\ \text{V}$ , $r_{\text{total}} = 0.5 + 0.5 = 1.0\ \Omega$ . Step 2. Apply $I = \dfrac{E}{R+r}$ : $I = \dfrac{3.0}{5 + 1.0} = \dfrac{3.0}{6.0} = 0.5\ \text{A}$ . Step 3. Terminal voltage of the combination: $V = E_{\text{total}} - Ir_{\text{total}} = 3.0 - (0.5 \times 1.0) = 2.5\ \text{V}$ . The combined current is $\mathbf{0.5\ A}$ and the terminal voltage of the series pair is $\mathbf{2.5\ V}$ .

Common traps

Treating EMF and terminal voltage as interchangeable — they coincide only when $I = 0$ (open circuit) or when $r = 0$ (ideal cell); otherwise $V < E$ whenever current is drawn.
Writing $I = E/R$ and forgetting internal resistance entirely — the correct denominator is always $R + r$ , the total resistance of the loop.
Assuming internal resistance belongs to the external circuit — it is a property of the cell/battery only, arising from its electrolyte and electrodes, never from the connected wires or resistors.
Sign confusion between $V = E - Ir$ (correct, terminal voltage drops under load) and mistakenly writing $V = E + Ir$ .
For cells in series, forgetting to add the internal resistances along with the EMFs — a common slip in multi-cell numericals like the stretch example above.

Ask the tutor: Stuck on a multi-cell or mixed series-parallel internal-resistance numerical — ask the AI tutor to walk through it with you step by step.

If EMF and terminal voltage are usually different, why does my textbook sometimes just write V for both?

Many introductory problems assume an "ideal" cell with negligible internal resistance, so $E \approx V$ and the distinction does not matter numerically. But conceptually they are always different quantities: EMF is defined at open circuit (no current), while terminal voltage is what you measure once current flows. The moment a numerical gives you a nonzero internal resistance $r$ , you must use $E = V + Ir$ and cannot treat them as the same number.

Does internal resistance behave just like an ordinary resistor you could remove from the circuit?

Not quite — you cannot physically separate it out and wire it elsewhere, because it is built into the cell itself, from the resistance of the electrolyte and the electrodes. Mathematically, though, it behaves exactly like an extra resistor placed in series with the external circuit: it adds directly into the denominator of $I = E/(R+r)$ , and it produces its own voltage drop $Ir$ just like any series resistor would.

How can internal resistance come out to exactly zero, like in the worked example

A calculated $r = 0\ \Omega$ from a numerical simply means that, for the given data, the current predicted assuming an ideal cell matches the current actually observed — so the cell's internal resistance is too small to detect with those numbers. In reality no cell has exactly zero internal resistance; a fresh, high-quality cell can have $r$ small enough (a fraction of an ohm) that it rounds to zero within the precision of a typical exam numerical.