What EMF Really Means
A cell has two electrodes, a positive (P) and a negative (N), immersed in an electrolytic solution. Each electrode exchanges charge with the electrolyte: the positive electrode acquires a potential difference $V_+$ relative to the solution next to it, and the negative electrode develops a potential $-V_-$. When no current flows, the electrolyte is at a single potential throughout, so the potential difference between P and N is $V_+ + V_-$. NCERT defines this quantity as the electromotive force (EMF) of the cell, denoted $\varepsilon$:
$$\varepsilon = V_+ + V_- > 0$$
Physically, EMF is the energy supplied by the source per unit charge that is driven around the circuit. It is the property that keeps current flowing against the resistances in the loop. NCERT is explicit that $\varepsilon$ is the potential difference between the positive and negative electrodes in an open circuit — that is, when no current is flowing through the cell.
"Electromotive force" is not a force
Despite the word "force" in its name, EMF is a potential difference, measured in volts — not a force in newtons. NCERT notes the name was given historically, "at a time when the phenomenon was not understood properly." Examiners exploit this in assertion–reason items.
EMF $\varepsilon$ has units of volts (J/C). It is energy supplied per unit charge, not a mechanical force.
Internal Resistance of a Cell
To maintain a steady current, charge must flow inside the cell as well — from the negative electrode N to the positive electrode P through the electrolyte. The electrolyte is not a perfect conductor; it has a finite resistance $r$, called the internal resistance of the cell. This resistance sits in series with whatever external resistance the cell is connected to, and every electron that travels through the cell must overcome it.
NCERT records two practical facts about $r$. First, internal resistances vary from cell to cell, and in calculations they "may be neglected when the current $I$ is such that $\varepsilon \gg Ir$." Second, the internal resistance of dry cells is much higher than that of the common electrolytic (wet) cells, which is why a torch cell sags badly under load while a car battery holds its voltage. NIOS adds that EMF itself depends on the electrolyte, the electrode material, and the temperature, but not on the size of the cell — a large and a small cell of the same chemistry have equal EMF, though the larger one offers lower internal resistance and lasts longer.
A real cell: EMF $\varepsilon$ with internal resistance $r$ in series, driving current $I$ through an external resistor $R$.
Terminal Voltage: V = ε − Ir
Now connect a resistor $R$ across the cell so a current $I$ flows. The current passes from P to N through $R$, and from N to P through the electrolyte. As it crosses the internal resistance, a potential drop $Ir$ develops. The potential difference now measured between the terminals P and N — the terminal voltage $V$ — is what is left of the EMF after this internal drop:
$$V = \varepsilon - Ir$$
This is the master relation of the topic. NIOS arrives at the same equation by writing $\varepsilon = V + Ir$, and states the conclusion plainly: while drawing current from a cell, the EMF is always greater than the potential difference across the external resistance, unless the internal resistance is zero. Combining $V = \varepsilon - Ir$ with Ohm's law $V = IR$ for the external resistor gives the current in the loop, treated in the next section.
V = ε only when I = 0
Terminal voltage equals EMF in exactly one situation: an open circuit, where $I = 0$ and the $Ir$ drop vanishes. A high-resistance voltmeter across an idle cell reads $\varepsilon$; the same voltmeter reads less the instant the circuit is closed and current flows. Treating the terminal reading as the EMF whenever current flows is a classic error.
Discharging: $V = \varepsilon - Ir < \varepsilon$. Open circuit ($I = 0$): $V = \varepsilon$.
When the current drawn from a battery is 0.5 A, the terminal voltage is 20 V; when the current is 2.0 A, it drops to 16 V. Find the EMF and internal resistance. (NIOS 17.8)
Using $V = \varepsilon - Ir$ for both readings: $20 = \varepsilon - 0.5r$ and $16 = \varepsilon - 2.0r$. Subtracting, $4 = 1.5r$, so $r = 2.67\ \Omega$, and back-substituting gives $\varepsilon = 21.3\ \text{V}$. The terminal voltage falls as more current is drawn — exactly as $V = \varepsilon - Ir$ predicts.
Current Drawn and Maximum Current
Setting $IR = \varepsilon - Ir$ and solving for the current gives the working equation for a single cell driving an external resistance:
$$I = \frac{\varepsilon}{R + r}$$
The internal resistance simply adds to the external resistance — the cell behaves as an ideal EMF source $\varepsilon$ in series with a resistor $r$. The largest current the cell can supply occurs when the external resistance is reduced to zero, i.e. on a short circuit. NCERT gives this maximum directly:
$$I_{\max} = \frac{\varepsilon}{r} \quad (R = 0)$$
In most cells the allowed current is kept far below $I_{\max}$ to prevent permanent damage. This short-circuit value is the only current the internal resistance alone can limit, which makes $r$ the quantity that protects a cell from destroying itself.
Once you can model one real cell as $\varepsilon$ with series $r$, combining several cells is the natural next step. See Cells in Series and Parallel for equivalent EMF and equivalent internal resistance.
The V–I Graph of a Real Cell
Rewriting $V = \varepsilon - Ir$ as a function of current shows it is a straight line: the vertical intercept is the EMF $\varepsilon$ (the value of $V$ at $I = 0$), and the slope is $-r$. A steeper downward slope means a larger internal resistance. The line meets the horizontal axis at $I = \varepsilon/r$, the short-circuit current, where the terminal voltage has fallen to zero.
Terminal voltage versus current: a straight line with intercept $\varepsilon$ and slope $-r$, reaching $V = 0$ at $I = \varepsilon/r$.
This graph is a favourite source of NEET questions: the intercept gives EMF, the magnitude of the slope gives internal resistance, and the $x$-intercept gives the short-circuit current. A cell with $r = 0$ would produce a horizontal line at $V = \varepsilon$ — the ideal source that never sags under load.
Charging vs Discharging
Everything above assumes the cell is discharging — supplying current to an external circuit. When a cell is charging, an external source of higher voltage forces current into the cell, against its EMF. The current through the cell reverses, so the $Ir$ drop now adds to the EMF rather than subtracting from it, and the terminal voltage rises above the EMF:
$$V = \varepsilon + Ir \quad (\text{charging})$$
This is why a battery on charge reads higher than its rated EMF. NCERT's exercise on charging a storage battery (8.0 V EMF, $0.5\ \Omega$ internal resistance, fed by a 120 V supply through a $15.5\ \Omega$ series resistor) turns precisely on this sign reversal. The three regimes are summarised below.
| State of the cell | Current | Terminal voltage V | Comparison with ε |
|---|---|---|---|
| Open circuit | I = 0 | V = ε | $V = \varepsilon$ |
| Discharging (supplying current) | I out of + terminal | V = ε − Ir | $V < \varepsilon$ |
| Charging (current forced in) | I into + terminal | V = ε + Ir | $V > \varepsilon$ |
The sign of Ir flips on charging
Students memorise $V = \varepsilon - Ir$ and apply it blindly to charging problems. During charging the current direction through the cell is reversed, so the relation becomes $V = \varepsilon + Ir$ and the terminal voltage exceeds the EMF. Read whether the cell is sourcing or absorbing energy before fixing the sign.
Discharge: subtract $Ir$. Charge: add $Ir$. The flip follows the reversed current direction.
Maximum Power Transfer
The power delivered to the external resistor is $P = I^2 R = \dfrac{\varepsilon^2 R}{(R + r)^2}$. Treating $R$ as the variable, this power is not largest when $R$ is huge (then $I$ is tiny) nor when $R \to 0$ (then all power is wasted inside $r$). It peaks at an intermediate value, found by differentiating: the power delivered to $R$ is maximum when
$$R = r$$
This is the maximum power transfer condition. At this matched load the power delivered to the external resistor is $P_{\max} = \dfrac{\varepsilon^2}{4r}$, and since an equal amount is dissipated inside the cell, the efficiency of transfer is exactly 50%. A circuit can be tuned either for maximum power (set $R = r$) or for maximum efficiency (make $R \gg r$ so terminal voltage approaches EMF), but not both at once.
Primary vs Secondary Cells
NIOS classifies chemical cells into two families by whether their chemical reaction can be reversed. The distinction governs whether a cell can be recharged.
| Feature | Primary cell | Secondary cell |
|---|---|---|
| Energy conversion | Chemical energy converted directly to electrical energy | Electrical energy stored as a reversible chemical reaction |
| Reversibility | Material is consumed; reaction not reversible | Reaction reverses when current is driven in |
| Rechargeable | Cannot be recharged or reused | Can be charged again and again |
| Examples | Dry cell, Daniel cell, Voltaic cell | Acid accumulator (car / inverter battery) |
A secondary cell is exactly the device that the charging relation $V = \varepsilon + Ir$ describes during its charge phase — current is pushed in, the reaction reverses, and the original substances are restored.
The five lines to carry into the exam hall
- EMF $\varepsilon$ is the open-circuit terminal voltage — energy supplied per unit charge, a potential difference (in volts), not a force.
- Terminal voltage while discharging: $V = \varepsilon - Ir$, so $V < \varepsilon$ whenever current flows; $V = \varepsilon$ only when $I = 0$.
- Current in the loop: $I = \dfrac{\varepsilon}{R + r}$; short-circuit ($R = 0$) gives the maximum current $I_{\max} = \dfrac{\varepsilon}{r}$.
- Charging reverses the current, so $V = \varepsilon + Ir$ and the terminal voltage exceeds the EMF.
- Maximum power to the external resistor occurs when $R = r$, giving $P_{\max} = \dfrac{\varepsilon^2}{4r}$ at 50% efficiency. Secondary cells (not primary) can be recharged.