Cells emf and internal resistance

Squeeze a fresh 1.5 volt cell and an old, tired one side by side into a torch, and the torch glows weakly or flickers — even though a voltmeter, touched to the old cell's bare terminals with nothing connected, might still read close to 1.5 volts. Something is lying to you. The cell has not lost its electromotive force; it has gained an invisible internal resistance that eats the voltage the moment current actually has to flow. Every real cell — a torch cell, a car battery, the lithium cell in a phone — is an engine and a leak rolled into one. The engine is chemical: it does work on charge, pushing it around a circuit against the electric field, and the size of that push per unit charge is called the electromotive force, EMF, $\varepsilon$ . The leak is the cell's own internal resistance $r$ , the resistance of the electrolyte and electrode material the current itself must cross before it ever leaves the cell. The two combine into one of the most exam-tested equations in the whole Current Electricity chapter, $\varepsilon = I(R + r)$ , and almost every mark lost on this topic in a board paper comes from confusing EMF (a fixed property of the source) with terminal voltage (what the external circuit actually gets, which drops the moment current is drawn). This article builds that single relation from a plain energy argument, reuses two worked numbers you may already have seen — a 2.0 V cell driving 0.4 A through 4.5 Ω, and a 12 V battery whose short-circuit current pins down its internal resistance — and extends them with a Kirchhoff's-loop cross-check and a slightly harder competitive-style problem involving a charging cell. By the end you should be able to look at any cell-and-resistor circuit and immediately say what the terminal voltage will be, without needing to be told which "case" it is.

Key takeaways

EMF is a source property. $\varepsilon$ is the work done per unit charge by the cell in driving current around the complete circuit; it equals the terminal voltage only when no current flows (open circuit).
Terminal voltage drops under load. With current $I$ flowing through internal resistance $r$ , $V = \varepsilon - Ir$ , and since the same current also flows through the external resistor, $V = IR$ too.
One governing equation. Combining the two gives $\varepsilon = I(R + r)$ , so $I = \dfrac{\varepsilon}{R + r}$ — this single line answers most numerical questions in this topic.
Short circuit gives the ceiling. Setting $R = 0$ gives the largest current the cell can ever deliver, $I_{sc} = \varepsilon / r$ — a fast way to sanity-check any answer.
Charging flips the sign. When an external source forces current backward through the cell (charging it), the terminal voltage exceeds the EMF: $V = \varepsilon + Ir$ .
Internal resistance is not fixed. It grows as a cell ages, discharges, or gets colder — which is why an old battery can read a near-normal open-circuit voltage yet fail to light a torch.
Key idea: because current must also cross the cell's own internal resistance r, the terminal voltage available to a circuit is always less than the EMF while discharging (V = ε − Ir) and more than it while charging (V = ε + Ir), with the single relation ε = I(R + r) governing both.

The core insight

A real cell's terminal voltage differs from its EMF by exactly the drop across its own internal resistance, $V = \varepsilon \mp Ir$ , and the whole simple-circuit behaviour collapses into one equation, $\varepsilon = I(R + r)$ .

Where it fits in the syllabus

This is the EMF-and-internal-resistance strand of the CBSE Class 12 Physics Current Electricity chapter, sitting between Ohm's law/resistivity and the topic of cells in series and parallel that follows it. It is squarely inside the 2025-26 prescribed theory syllabus and is one of the most reliably tested numericals in the chapter. One honest caveat: the classic apparatus for measuring EMF or internal resistance experimentally, the potentiometer, was dropped from the 2025-26 theory syllabus in the recent rationalisation — so this article deliberately treats EMF and internal resistance only through the $\varepsilon = I(R+r)$ circuit relation and Kirchhoff's rules, never through potentiometer balance-point methods, since those no longer belong to the theory paper.

The physical picture

Picture the cell as a black box with two terminals, doing chemical work on every unit of positive charge that passes through it internally, lifting that charge's potential energy the way a pump lifts water. If you could measure the potential difference across the terminals while the switch is open — no current flowing anywhere, so no energy is being spent overcoming any resistance — you would read the EMF directly, because in that condition the full push of the cell appears at the terminals with nothing subtracted from it. That is the operational definition worth memorising:

\varepsilon = V \quad \text{(open circuit, } I = 0\text{)}

where: $\varepsilon$ is the EMF in volts, and $V$ is the terminal potential difference measured with no current drawn. The moment you close the switch and let current $I$ flow, the picture changes. The charge does not only have to push through the external resistor $R$ ; it first has to push through the resistive electrolyte and electrode material inside the cell itself, which is the internal resistance $r$ . That internal push-through costs a voltage drop $Ir$ , exactly like any other resistor, so whatever is left over at the terminals for the external circuit is reduced:

V = \varepsilon - Ir

where: $V$ is the terminal potential difference under load, $I$ is the current the circuit actually carries, and $r$ is the cell's internal resistance in ohms. Physically, $Ir$ is voltage "spent inside the battery" — energy the chemical reaction supplied that never makes it out to light a bulb or turn a motor; it appears instead as heat inside the cell, which is why a cell with a very small $r$ short-circuited by a wire can get noticeably warm.

Figure

A single cell of emf epsilon and internal resistance r connected to an external resistor R, with an arrow showing current I flowing around the loop and labels marking the voltage drop I r inside the cell and the terminal voltage V across the external resistor

Because the same current $I$ that flows through the internal resistance also flows through the external resistor, $V$ can equally be written as $V = IR$ (plain Ohm's law across the outside of the circuit). Setting the two expressions for $V$ equal gives the single governing equation of the whole topic:

\varepsilon = IR + Ir = I(R + r) \qquad \Rightarrow \qquad I = \dfrac{\varepsilon}{R + r}

where: the sum $R + r$ behaves as one total resistance the EMF has to push current through — the external resistor and the cell's own internal resistance sit in the same series loop, so they simply add. Two limiting cases are worth holding in memory as sanity checks. If $R \to \infty$ (open switch, no external path), $I \to 0$ and $V \to \varepsilon$ exactly, matching the open-circuit definition above. If $R = 0$ (a short circuit — the terminals joined directly by a low-resistance wire), the equation gives the largest current the cell can ever produce, $I_{sc} = \varepsilon / r$ , since $r$ is now the only resistance left to limit it. There is also a reverse situation worth knowing precisely because it is the opposite sign of the everyday case: charging a cell. If an external source (a charger) is connected so as to force current into the cell's positive terminal, against the direction the cell's own chemistry would drive it, then the external source has to supply not only the cell's own EMF but also push that same current through the internal resistance in the "wrong" direction. The terminal voltage measured across the cell during charging therefore exceeds the EMF:

V = \varepsilon + Ir \qquad \text{(charging)}

where: the plus sign, rather than the minus sign of ordinary discharge, is the entire content of this special case — everything else about the circuit reasoning is identical.

The physical picture — Kirchhoff's-loop view

The same relation also falls straight out of Kirchhoff's voltage law, which states that the sum of potential changes around any closed loop is zero. Walking around the single loop of cell, internal resistance, and external resistor, and adding up the EMF gained and the two resistive drops lost, gives

\varepsilon - Ir - IR = 0

where: each term is a potential change encountered while traversing the loop once — $\varepsilon$ is the rise supplied by the cell, and $Ir$ , $IR$ are drops across the two resistances in series with it. Rearranging reproduces $\varepsilon = I(R+r)$ exactly, confirming that the cell-with-internal-resistance relation is nothing more than a one-loop application of Kirchhoff's voltage law, the same rule that governs every more complicated multi-loop circuit later in the chapter.

Derivation ATerminal voltage of a cell under load, from energy conservation

The goal of this derivation is to obtain the relation between a cell's EMF, its internal resistance, the current it delivers, and the terminal potential difference actually available to an external circuit, starting from nothing more than the definition of EMF as work done per unit charge and the fact that the same charge, and hence the same current, must pass through both the cell's internal resistance and the external resistor connected across its terminals, since they form a single series loop with no branch points.

Step 1. Start from the definition of EMF. By definition, the EMF $\varepsilon$ is the total work $W$ done by the cell's chemical energy per unit charge $q$ carried once around the complete circuit: $\varepsilon = W/q$ . This work goes toward moving charge through every resistance the charge encounters on its full trip around the loop, both inside the cell and outside it.

Step 2. Identify every resistance the charge crosses. In a single cell connected to one external resistor $R$ , a unit of charge circulating once around the loop passes through exactly two resistances in series: the external resistor $R$ , and the cell's own internal resistance $r$ . No other resistive element exists in this simple circuit.

Step 3. Write the work done per unit charge against each resistance. Work done per unit charge moving current $I$ through a resistance $X$ is, from the definition of potential difference, simply $IX$ . So the work per unit charge spent crossing the external resistor is $IR$ , and the work per unit charge spent crossing the internal resistance is $Ir$ .

Step 4. Apply conservation of energy. The total work the cell supplies per unit charge, $\varepsilon$ , must exactly equal the sum of the work per unit charge spent on the two resistances, since energy is neither created nor destroyed as the charge completes the loop: $\varepsilon = IR + Ir$ .

Step 5. Factor and identify the terminal voltage. Writing $\varepsilon = I(R + r)$ makes the total effective resistance of the loop explicit. Separately, the potential difference actually delivered to the external circuit — the terminal voltage $V$ — is, by Ohm's law applied to the external resistor alone, $V = IR$ . Substituting $IR = \varepsilon - Ir$ from Step 4 gives $V = \varepsilon - Ir$ , showing directly that the terminal voltage falls short of the EMF by exactly the internal voltage drop $Ir$ , and that this shortfall grows with the current drawn.

Worked examples

Board level - current and terminal voltage of a loaded cell

A cell of EMF $2.0\ \text{V}$ and internal resistance $0.5\ \Omega$ is connected to an external resistor of $4.5\ \Omega$ . Find (a) the current in the circuit, and (b) the terminal potential difference of the cell.

Step 1. Identify the quantities. EMF $\varepsilon = 2.0\ \text{V}$ , internal resistance $r = 0.5\ \Omega$ , external resistance $R = 4.5\ \Omega$ .

Step 2. Find the current using $I = \dfrac{\varepsilon}{R + r}$ : $I = \dfrac{2.0}{4.5 + 0.5} = \dfrac{2.0}{5.0} = 0.4\ \text{A}$ .

Step 3. Find the terminal potential difference using $V = \varepsilon - Ir$ : $V = 2.0 - (0.4)(0.5) = 2.0 - 0.2 = 1.8\ \text{V}$ .

Step 4. Cross-check with $V = IR$ : $V = 0.4 \times 4.5 = 1.8\ \text{V}$ , which matches Step 3 exactly, confirming the answer is internally consistent. The current is $I = 0.4\ \text{A}$ and the terminal potential difference is $V = 1.8\ \text{V}$ , which is $0.2\ \text{V}$ less than the EMF — the amount "lost" driving current through the cell's own $0.5\ \Omega$ .

Competitive level - internal resistance from short-circuit data, then reversing to find an unknown resistor and comparing with charging

A battery of EMF $12\ \text{V}$ is momentarily short-circuited and found to deliver a current of $30\ \text{A}$ . (a) Find its internal resistance. (b) This same battery is then connected to an external resistor so that its terminal voltage reads $10.5\ \text{V}$ ; find the current drawn and the value of that resistor. (c) If instead a battery charger forces $3.75\ \text{A}$ backward into this cell, find the terminal voltage during charging.

Step 1. Use the short-circuit condition to get $r$ . In a short circuit $R = 0$ , so $I_{sc} = \varepsilon / r$ , giving $r = \dfrac{\varepsilon}{I_{sc}} = \dfrac{12}{30} = 0.4\ \Omega$ .

Step 2. Use $V = \varepsilon - Ir$ to find the current for the stated terminal voltage: $10.5 = 12 - I(0.4)$ , so $I(0.4) = 1.5$ , giving $I = 3.75\ \text{A}$ .

Step 3. Find the external resistor from $R = V/I$ : $R = \dfrac{10.5}{3.75} = 2.8\ \Omega$ .

Step 4. Now switch to the charging case with the same $3.75\ \text{A}$ magnitude, but current forced the opposite way, so the sign in front of $Ir$ flips: $V = \varepsilon + Ir = 12 + (3.75)(0.4) = 12 + 1.5 = 13.5\ \text{V}$ . The internal resistance is $r = 0.4\ \Omega$ ; discharging through a $2.8\ \Omega$ resistor at $3.75\ \text{A}$ gives terminal voltage $10.5\ \text{V}$ ; charging at the same $3.75\ \text{A}$ instead gives terminal voltage $13.5\ \text{V}$ , exactly $1.5\ \text{V}$ above the EMF, the mirror image of the $1.5\ \text{V}$ shortfall in part (b). Timing note: in an exam, part (a) alone is a 30-second plug-in; do not spend time re-deriving $I_{sc} = \varepsilon/r$ from scratch each time — recognise "short circuit" instantly as $R = 0$ and move straight to the formula. Parts (b) and (c) together should take under three minutes if you keep $\varepsilon$ and $r$ written at the top of your working rather than re-substituting them from the question each step.

Common traps

Treating EMF and terminal voltage as interchangeable words for "the voltage of the battery" — they are numerically equal only at open circuit ( $I = 0$ ); the instant any current flows, $V < \varepsilon$ while discharging.
Forgetting the sign flip during charging: while a cell is being charged, terminal voltage exceeds EMF ( $V = \varepsilon + Ir$ ), the mirror image of the discharging relation, and many students apply the discharge formula blindly in both cases.
Assuming internal resistance can be ignored "because it's usually small" — the numericals in this topic exist specifically to test whether $Ir$ is accounted for; dropping it changes the answer, not just its precision.
Miscalculating or misusing short-circuit current: $I_{sc} = \varepsilon / r$ is the ceiling, the largest current the cell can ever supply to any external circuit with $R \geq 0$ ; if a computed current for some $R > 0$ comes out larger than $I_{sc}$ , that is a sure sign of an arithmetic slip.
Mixing up which resistance goes where in $\varepsilon = I(R+r)$ : $R$ and $r$ are in series and simply add for the purpose of finding $I$ , but only $R$ (not $R+r$ ) is used when converting back to terminal voltage via $V = IR$ .

You should be able to…

Define EMF as the open-circuit terminal voltage and distinguish it clearly from terminal voltage under load.
State and apply $\varepsilon = I(R+r)$ and $V = \varepsilon - Ir$ to find any one unknown given the other three quantities.
Derive the terminal-voltage relation from energy conservation and, separately, recognise it as a one-loop application of Kirchhoff's voltage law.
Compute short-circuit current $I_{sc} = \varepsilon/r$ and use it as a sanity-check ceiling on any current calculated for the same cell.
Explain and correctly apply the sign flip that occurs when a cell is being charged, $V = \varepsilon + Ir$ .
State, without needing to derive it, that potentiometer methods for measuring EMF are outside the current syllabus.

Ask the tutor: Want to try a multi-cell circuit — cells in series or parallel feeding one resistor, or a two-loop Kirchhoff's-rules problem — to see how this single-cell relation extends?

If the open-circuit voltage of my old torch cell still reads close to 1.5 V, why does the torch glow so dimly?

An old cell's chemistry has usually not lost much of its EMF, but its internal resistance $r$ has grown a lot as the electrolyte degrades and the electrodes corrode. A voltmeter draws almost no current, so it barely disturbs the open-circuit reading and still shows close to the full EMF. But the torch bulb draws real current, and with $r$ now large, most of the EMF gets dropped as $Ir$ inside the cell itself instead of reaching the bulb, so the terminal voltage — and the glow — collapses even though $\varepsilon$ looks fine on paper.

Why does a short-circuited battery get warm, sometimes dangerously so?

When $R = 0$ , the current $I_{sc} = \varepsilon/r$ is the largest the cell can produce, and every bit of the electrical energy the cell supplies has nowhere to go except straight into the internal resistance itself, since there is no external resistor to share the load. Power dissipated inside the cell is $I^2 r$ , and because $I$ is large in a short circuit, this heating can be substantial — which is exactly why short-circuiting a real battery (for example touching both terminals with a metal wire) is genuinely dangerous and can even cause a fire.

Is the potentiometer method for measuring EMF still something I need to know for boards?

No. The potentiometer principle, comparing EMFs, and measuring internal resistance using a potentiometer have all been removed from the 2025-26 CBSE theory syllabus in the recent rationalisation; the potentiometer survives only as apparatus you might see named in the practicals list, not as theory to be tested. For this topic, stick to the circuit relation $\varepsilon = I(R + r)$ and Kirchhoff's rules — that is what boards will actually examine.

How is internal resistance different from the resistance of the wires connecting the cell to the circuit?

Internal resistance is a property of the cell itself — it comes from the resistance of the electrolyte and the electrode material that the current must cross while still inside the cell, before it ever reaches the external terminals. Ordinary connecting-wire resistance, by contrast, is part of the external circuit and, in most textbook problems, is either given separately and added into $R$ , or is idealised as zero. In the equation $\varepsilon = I(R + r)$ , $r$ specifically refers only to what is inside the cell.