JEE Mains Physics · Current Electricity

Cells: EMF, Internal Resistance and Combinations

A real cell is an emf ε in series with an internal resistance r, so it drives I = ε/(R + r) and its terminal voltage ε − Ir falls as it delivers more current.

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Why this matters

Twenty-five PYQs, fifteen of them multiple choice, and six from 2026. Thirteen are about one cell, or two in series, driving a load: internal resistance from two readings, the terminal voltage while a cell delivers or takes current, the load that draws the most power, and the load for which one cell's terminal voltage is zero. Twelve combine cells in series, in parallel or against each other, including four where the same current flows either way.

Concept 1 of 2: EMF, internal resistance and terminal voltage

Inside every real cell there is some resistance. When the cell delivers a current, part of its emf is used up pushing that current through itself, so less is left at the terminals. The more current it gives, the lower its terminal voltage. When a stronger source pushes current backwards through it, as in charging, the terminal voltage is higher than the emf.

Definition

$I = \dfrac{\varepsilon}{R + r}$ . Delivering current: $V = \varepsilon - Ir$ . Being charged: $V = \varepsilon + Ir$ .
With no current (an ideal voltmeter alone across it), $V = \varepsilon$ . Short circuit: $I = \varepsilon/r$ .
r from two loads: $\varepsilon = I_1(R_1 + r) = I_2(R_2 + r)$ . If voltages are given instead, first find each current as V/R.
r from one load and a voltmeter: $r = R\left(\dfrac{\varepsilon}{V} - 1\right)$ .
The power in R is largest when $R = r$ ; then $P_{\max} = \dfrac{\varepsilon^{2}}{4r}$ .
Cells in series: for one cell's terminal voltage to be zero, its $\varepsilon - Ir$ must vanish. Set $I = \varepsilon/r$ for that cell and solve for R.

A real cell

I = \frac{\varepsilon}{R + r}, \qquad V = \varepsilon \mp Ir, \qquad P_{\max} = \frac{\varepsilon^{2}}{4r}\ (R = r)

Worked example

A cell drives 1 A through a 10 Ω resistor and 2 A through a 4 Ω resistor. Find its emf and internal resistance.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q24Moderate

Example 1 · Current Electricity · Cells: EMF, Internal Resistance and Combinations

When an external resistance of

5\Omega

is connected across terminals of a cell, a current of 0.25 A flows through it. When the

5\Omega

resistor is replaced by a

2\Omega

resistor, a current of 0.5 A flows through it. The internal resistance of the cell is

\_\_\_\_

\Omega

Charging adds Ir

When a cell is pushed backwards by a stronger one, its terminal voltage is ε + Ir, higher than its emf. Using ε − Ir for the weaker cell in opposition gives the wrong sign.

Maximum power is at R = r, not R = 0

A short circuit gives the largest current but no power in the load. The power in R peaks when R matches r.

Include r in the total

The current is ε/(R + r). Dividing ε by R alone forgets the cell's own resistance.

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Concept 2 of 2: Cells in series, parallel and opposition

Cells in series stack their emfs and their internal resistances, like resistors in series; a reversed cell subtracts its emf but still adds its resistance. Cells in parallel share the load. Their combined emf is a weighted average, with each cell weighted by 1/r, so a cell with low internal resistance has more say. The combined emf always lies between the separate emfs when the cells face the same way.

Definition

Series: $\varepsilon_{\text{eq}} = \sum \pm\varepsilon_i$ (minus for a reversed cell), $r_{\text{eq}} = \sum r_i$ .
Parallel: $\varepsilon_{\text{eq}} = \dfrac{\sum \pm\varepsilon_i/r_i}{\sum 1/r_i}$ , $\dfrac{1}{r_{\text{eq}}} = \sum \dfrac{1}{r_i}$ . Reversing a cell flips only the sign of its $\varepsilon/r$ term.
n identical cells in parallel: emf ε, internal resistance r/n.
m rows, each of n identical cells in series: $\varepsilon_{\text{eq}} = n\varepsilon$ , $r_{\text{eq}} = nr/m$ , $I = \dfrac{n\varepsilon}{R + nr/m}$ .
n identical cells give the same current in series and in parallel when $\dfrac{n\varepsilon}{R + nr} = \dfrac{\varepsilon}{R + r/n}$ ; clearing the fractions gives $R = r$ .

Two cells in parallel

\varepsilon_{\text{eq}} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2}, \qquad r_{\text{eq}} = \frac{r_1r_2}{r_1 + r_2}

Worked example

A 6 V cell (r = 2 Ω) and a 4 V cell (r = 1 Ω) are joined in parallel, + to +, across a 4 Ω resistor. Find the current. Then find it with the 4 V cell reversed.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q24Moderate

Example 2 · Current Electricity · Cells: EMF, Internal Resistance and Combinations

Two cells of emfs 1 V and 2 V and internal resistance

2\Omega

and

1\Omega

, respectively connected in parallel, gave current of 1A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be

\frac{\alpha}{5}\text{ }A

. The value of

\alpha

\_\_\_\_

Weight the emfs by 1/r

The emf of cells in parallel is not the plain average. A 6 V cell with 2 Ω and a 4 V cell with 1 Ω give 14/3 V, not 5 V.

A reversed cell still adds its resistance

Reversing a cell flips the sign of its emf term only. Its internal resistance stays in the combined internal resistance exactly as before.

Equivalent emf lies between the emfs

For two cells facing the same way in parallel, the combined emf is between the two emfs, and the combined internal resistance is smaller than either. An answer outside that range signals a slip.

Drill 11 more on cells in series, parallel and opposition

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

EMF, internal resistance and terminal voltage
A real cell
$I = \frac{\varepsilon}{R + r}, \qquad V = \varepsilon \mp Ir, \qquad P_{\max} = \frac{\varepsilon^{2}}{4r}\ (R = r)$
Cells in series, parallel and opposition
Two cells in parallel
$\varepsilon_{\text{eq}} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2}, \qquad r_{\text{eq}} = \frac{r_1r_2}{r_1 + r_2}$

Watch out for (6)

Charging adds Ir
→ EMF, internal resistance and terminal voltage
Maximum power is at R = r, not R = 0
→ EMF, internal resistance and terminal voltage
Include r in the total
→ EMF, internal resistance and terminal voltage
Weight the emfs by 1/r
→ Cells in series, parallel and opposition
A reversed cell still adds its resistance
→ Cells in series, parallel and opposition
Equivalent emf lies between the emfs
→ Cells in series, parallel and opposition

Test yourself on Current Electricity

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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