EMF and internal resistance

Switch on a torch with a fresh battery and, for a fraction of a second, the beam is slightly dimmer than the battery's rating promises. Nothing is wrong with the cell — every real cell has some internal resistance, and the moment current flows, a little of the cell's own driving voltage is spent pushing charge through its own electrolyte before any of it reaches the bulb. This gap between what a cell can deliver (its EMF) and what it actually delivers under load (its terminal voltage) is one of the most tested ideas in the ISC Class 12 Current Electricity chapter. It sits directly after Ohm's law and feeds forward into cell combinations, Kirchhoff's laws, and the potentiometer. This article builds the relation $\varepsilon = I(R+r)$ from a single series circuit, works two standard numerical patterns — direct substitution, and the two-reading method for finding an unknown $\varepsilon$ and $r$ — and clears up the sign confusion between charging and discharging.

The core insight

A real cell's terminal voltage always falls short of its EMF by exactly $Ir$ , the voltage spent driving current through the cell's own internal resistance, so $\varepsilon = I(R+r)$ links the ideal driving voltage to what a load actually receives.

The core idea

EMF, $\varepsilon$ , is the work done per unit charge by the cell's internal energy source (chemical, in a battery) in driving charge around the complete circuit, including through the cell itself. It is measured in volts, and despite the name it is not a force — it is an energy-per-charge quantity. Crucially, EMF is the maximum potential difference the cell can offer, realised only when no current flows (open circuit).

Internal resistance, $r$ , is the resistance of the electrolyte and electrodes inside the cell to the charge passing through them. The instant current flows, some of the cell's energy is dissipated inside the cell itself, so the terminal potential difference $V$ — what is actually available across the cell's terminals to drive the external circuit — is less than $\varepsilon$ . Picture a cell of EMF $\varepsilon$ and internal resistance $r$ wired in series with an external load $R$ . Applying Ohm's law to the whole loop, the current is

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

where: $I$ is the circuit current (A), $\varepsilon$ is the EMF (V), $R$ is the external (load) resistance ( $\Omega$ ), and $r$ is the cell's internal resistance ( $\Omega$ ). Rearranging gives the key relation used in almost every problem on this topic:

\varepsilon = I(R+r) = IR + Ir

Here $IR$ is the terminal voltage $V$ (the p.d. across the load, and hence across the cell's terminals), and $Ir$ is the voltage "lost" pushing current through the cell's own resistance. So

V = \varepsilon - Ir

On discharge (the cell supplying current to a circuit), $V = \varepsilon - Ir$ , so $V < \varepsilon$ whenever $I > 0$ , and $V \to \varepsilon$ only as $I \to 0$ — this is why an ideal voltmeter, which draws negligible current, reads very close to the true EMF across an idle cell. On charge (an external source forcing current backward through the cell), the sign flips: $V = \varepsilon + Ir$ , since the charging source must now overcome both the EMF and the internal resistance. If the external resistance is shorted to $R \to 0$ , the current is capped, not infinite: $I_{sc} = \varepsilon/r$ , set entirely by the cell's own internal resistance.

Figure

A series circuit with a cell of EMF epsilon and internal resistance r connected to an external resistor R showing the terminal voltage across R equal to epsilon minus I r

Derivation ATerminal voltage from EMF and internal resistance

The goal is to derive $V = \varepsilon - Ir$ from first principles for a cell of EMF $\varepsilon$ and internal resistance $r$ driving current $I$ through an external load $R$ in a single series loop, showing exactly where the terminal voltage comes from and why it must always be less than the EMF whenever current flows.

Step 1. Treat the loop as one series circuit. The cell (EMF $\varepsilon$ , internal resistance $r$ ) and the external resistor $R$ carry the same current $I$ , since they are in series with no branching. By Ohm's law applied to the total resistance of the loop, $I = \dfrac{\varepsilon}{R + r}$ .

Step 2. Multiply both sides by $(R + r)$ to recover the EMF as a sum of two voltage drops: $\varepsilon = I(R + r) = IR + Ir$ . Physically, the cell's total driving voltage is used up in two places as the current goes around the loop — some across the external resistor, some across the cell's own internal resistance.

Step 3. Identify each term. $IR$ is the voltage across the external resistor $R$ ; since $R$ is directly connected to the cell's terminals, this is also the terminal potential difference, $V = IR$ . $Ir$ is the voltage drop developed across the internal resistance itself, inside the cell, which never appears at the external terminals — it is the "lost volts."

Step 4. Substitute $V = IR$ back into $\varepsilon = IR + Ir$ to isolate $V$ : $\varepsilon = V + Ir$ , so $V = \varepsilon - Ir$ . Because $I$ and $r$ are both non-negative for a discharging cell, $Ir \geq 0$ , hence $V \leq \varepsilon$ always, with equality only when $I = 0$ (open circuit) — confirming that EMF is the upper bound on terminal voltage.

Worked example

A cell of EMF $\varepsilon = 2.0\ \text{V}$ and internal resistance $r = 0.5\ \Omega$ is connected to an external resistor $R = 3.5\ \Omega$ . Find the circuit current and the terminal potential difference; then use a second reading — the same cell connected instead to $R_2 = 9\ \Omega$ giving current $I_2 = 0.25\ \text{A}$ — to check the internal resistance independently.

Step 1. Find the current for the first circuit using $I = \dfrac{\varepsilon}{R+r}$ : $I = \dfrac{2.0}{3.5 + 0.5} = \dfrac{2.0}{4.0} = 0.5\ \text{A}$ .

Step 2. Find the terminal p.d. using $V = \varepsilon - Ir$ : $V = 2.0 - (0.5)(0.5) = 2.0 - 0.25 = 1.75\ \text{V}$ . Check by the other route, $V = IR = 0.5 \times 3.5 = 1.75\ \text{V}$ — the two methods agree, so $1.75\ \text{V}$ reaches the external circuit while $0.25\ \text{V}$ is spent inside the cell.

Step 3. Use the second reading to re-derive $r$ independently, treating $\varepsilon$ and $r$ as unknowns satisfied by two equations: $\varepsilon = I_1(R_1+r) = 0.5(4+r)$ and $\varepsilon = I_2(R_2+r) = 0.25(9+r)$ , using $R_1 = 4\ \Omega$ , $I_1 = 0.5\ \text{A}$ as an equivalent first reading. Setting them equal: $0.5(4+r) = 0.25(9+r) \Rightarrow 2 + 0.5r = 2.25 + 0.25r \Rightarrow 0.25r = 0.25 \Rightarrow r = 1\ \Omega$ .

Step 4. Substitute back to find $\varepsilon$ : $\varepsilon = 0.5(4+1) = 2.5\ \text{V}$ . This two-reading method (vary the load, measure the current twice, solve simultaneously) gives $\varepsilon = 2.5\ \text{V}$ and $r = 1\ \Omega$ , and is the standard experimental way to determine a cell's EMF and internal resistance without needing an open-circuit measurement.

Common traps

Treating EMF and terminal voltage as interchangeable. EMF is the cell's maximum possible p.d., measured at zero current; terminal voltage is what a load actually receives while current flows, and on discharge it is always the smaller of the two.
Forgetting EMF is an open-circuit quantity. A voltmeter reads close to true EMF only when it draws negligible current from an otherwise idle cell; as soon as a load current flows, terminal voltage sags below EMF by $Ir$ .
Flipping the sign during charging. On charge, $V = \varepsilon + Ir$ (terminal voltage exceeds EMF), not $V = \varepsilon - Ir$ — many students apply the discharge formula automatically in both cases.
Assuming a short circuit gives infinite current. $R = 0$ still leaves the cell's own $r$ in the loop, capping the current at $I_{sc} = \varepsilon / r$ rather than letting it run away.

Ask the tutor: Ask the AI tutor for a fresh two-reading EMF-and-internal-resistance problem to solve step by step, or to explain the charging-versus-discharging sign flip with a fresh circuit.

If EMF and terminal voltage are almost equal for a fresh, lightly loaded cell, why do textbooks make such a fuss about the difference

Because the gap grows exactly when it matters most — under heavy load or as the cell ages and its internal resistance rises. A fresh cell with a tiny load and small $r$ shows $Ir$ so small that $V \approx \varepsilon$ , which is why casual voltmeter readings can be mistaken for EMF. But connect a low-resistance load, or use an old, worn-out cell whose $r$ has crept up, and $Ir$ becomes significant — the torch dims, the motor stalls, and the distinction is no longer academic. Exams test the general relation precisely because it explains both the ordinary case and the extreme one.

How do I know whether to use plus I r or minus I r in a given problem

Ask which way the current is being pushed through the cell relative to its own EMF. If the cell itself is driving the current out into an external circuit (normal discharge, the usual torch-and-battery setup), the cell's own energy is being spent, and terminal voltage falls short: $V = \varepsilon - Ir$ . If an external source (like a charger) is forcing current backward into the cell against its own EMF, that source must supply extra voltage to overcome both the EMF and the internal resistance, so $V = \varepsilon + Ir$ . Reading the circuit diagram to see who is doing the "pushing" settles the sign every time.

Why does the two-reading method work — why do two different external resistances let me find both epsilon and r

Because each reading gives one linear equation in the two unknowns $\varepsilon$ and $r$ , namely $\varepsilon = I(R+r)$ , and one equation cannot pin down two unknowns on its own. A single measurement of $I$ and $R$ only tells you a relationship between $\varepsilon$ and $r$ , not their individual values. Change the external resistance and measure a second current, and you get a second independent equation; solving the two simultaneously (as in the worked example) isolates $r$ first, then $\varepsilon$ . This is exactly the classroom experiment where a rheostat is varied and the ammeter reading recorded at two or more settings.

EMF and internal resistance

The core idea

Worked example

Common traps

Before this

Where this leads