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NDA Physics · Electricity and Magnetism

Cells, EMF and Kirchhoff's Laws

A cell's EMF is the full push it can give; its internal resistance drops some of that, leaving the terminal voltage V = ε − Ir. Kirchhoff's two laws — junction (charge conservation) and loop (energy conservation) — let you solve any circuit.

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Why this matters

NDA tests this lightly — just 3 PYQs — but the ideas are load-bearing: EMF vs terminal voltage, internal resistance, how cells and bulbs combine to change brightness, and Kirchhoff's loop rule as a statement of energy conservation. Learn the concepts even where the bank is thin; they underpin every multi-cell circuit.

Concept 1 of 3: EMF, internal resistance and terminal voltage

A real cell isn't a perfect battery — it has its own small internal resistance. The EMF (ε) is the full voltage it would give with no current flowing; once current flows, some voltage is 'lost' inside the cell across its internal resistance, so the voltage you actually get at the terminals is a little less.

Definition

EMF (ε) is the energy a cell gives per unit charge — the open-circuit (no-current) voltage. A real cell has internal resistance r. When it drives current II, the terminal voltage is V=ε−IrV = \varepsilon - I r — the EMF minus the internal drop. Connected to an external resistance RR: I=εR+rI = \dfrac{\varepsilon}{R + r}.

Terminal voltage and circuit current

V=ε−Ir,I=εR+rV = \varepsilon - I r, \qquad I = \dfrac{\varepsilon}{R + r}
  • ε\varepsilonEMF of the cell (volt)
  • rinternal resistance (Ω)
  • Rexternal resistance (Ω)
  • Vterminal voltage (volt)
real cellεrRI = ε / (R + r)V = ε − Ir(terminal voltage)

Some EMF is lost across the internal resistance r, so the terminal voltage V = ε − Ir is a little less than the EMF whenever current flows.

Worked example

A cell of EMF 1.5 V and internal resistance 0.5 Ω drives a current through a 2.5 Ω resistor. Find the current and the terminal voltage.
Practice this conceptself-check · 3 quick reps

Terminal voltage drops as current rises

EMF is fixed, but the terminal voltage V = ε − Ir falls as the cell delivers more current. A heavily loaded battery (large I) shows a noticeably lower terminal voltage — that's the internal resistance at work.
Drill 1 more on emf, internal resistance and terminal voltage

Concept 2 of 3: Kirchhoff's two laws

Kirchhoff's rules are just conservation laws in disguise. At any junction, the charge that flows in must flow out — nothing piles up (charge conservation). Around any closed loop, all the voltage gains and drops must cancel to zero — because a charge returning to its start has the same energy it began with (energy conservation).

Definition

  • Junction (current) rule — the sum of currents into a junction equals the sum out: ∑Iin=∑Iout\sum I_\text{in} = \sum I_\text{out}. This is conservation of charge.
  • Loop (voltage) rule — around any closed loop, the algebraic sum of EMFs and potential drops is zero: ∑ε+∑(−IR)=0\sum \varepsilon + \sum (-IR) = 0. This is conservation of energy.

Kirchhoff's laws

∑junctionI=0,∑loop(ε−IR)=0\sum_\text{junction} I = 0, \qquad \sum_\text{loop} (\varepsilon - IR) = 0

Worked example

At a junction, currents of 3 A and 2 A flow in along two wires, and current flows out along a third wire. What is the outgoing current?
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q67Moderate

Example 2 · Electricity and Magnetism · Cells, EMF and Kirchhoff's Laws

"The sum of emfs and potential differences around a closed loop equals zero" is a consequence of

Loop rule = energy; junction rule = charge

Don't mix them up. The LOOP (voltage) rule comes from energy conservation; the JUNCTION (current) rule comes from charge conservation. The distractors 'Ohm's law' and 'conservation of momentum' are both wrong for the loop rule.

Concept 3 of 3: Combining cells and bulb brightness

Cells in series add their EMFs — two equal cells give double the voltage. A bulb's brightness is its power, P = V²/R, so the bulb that has the MOST voltage across it (and isn't sharing current) glows brightest. Bulbs in parallel each get the full supply voltage; bulbs in series have to split it.

Definition

Cells in series add EMF: two cells of EMF ε give 2ε. Bulb brightness = power dissipated in it (P=V2/RP = V^2/R). Bulbs in parallel each receive the full supply voltage (brighter); bulbs in series share the voltage (dimmer). So the brightest single bulb is the one across the highest voltage carrying its own current — e.g. two cells in series feeding bulbs in parallel.

Worked example

All cells and bulbs are identical (EMF ε, bulb resistance R; ignore internal resistance). Which glows brightest: (a) 1 cell + 1 bulb, or (b) 2 cells in series + 2 bulbs in parallel?
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2024 · Q69Hard

Example 3 · Electricity and Magnetism · Cells, EMF and Kirchhoff's Laws

In which one among the following situations, the bulb would glow the most ? (Consider all batteries are the same)

Parallel bulbs each get full voltage — series bulbs split it

More cells in series AND bulbs in parallel both raise the voltage across each bulb. The brightest arrangement maximises per-bulb voltage: two cells in series feeding bulbs in parallel beats any series-bulb arrangement.

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

    Terminal voltage and circuit current

    V=ε−Ir,I=εR+rV = \varepsilon - I r, \qquad I = \dfrac{\varepsilon}{R + r}
  • Kirchhoff's two laws

    Kirchhoff's laws

    ∑junctionI=0,∑loop(ε−IR)=0\sum_\text{junction} I = 0, \qquad \sum_\text{loop} (\varepsilon - IR) = 0

Watch out for (3)

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