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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

Intuition

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}
  • \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.
  1. Total resistance = external + internal = 2.5+0.5=3 Ω2.5 + 0.5 = 3\,\Omega.
  2. Current I=ε/(R+r)=1.5/3=0.5I = \varepsilon/(R+r) = 1.5/3 = 0.5 A.
  3. Terminal voltage V=ε−Ir=1.5−0.5×0.5=1.25V = \varepsilon - Ir = 1.5 - 0.5\times0.5 = 1.25 V.
Answer:I = 0.5 A, terminal voltage = 1.25 V.
Practice this conceptself-check · 3 quick reps

Try it yourself

A battery of EMF 12 V has terminal voltage 11.4 V when it supplies 3 A. What is its internal resistance?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    What is the terminal voltage of a cell on open circuit (no current)?
  2. 2.
    Cell of EMF 2 V, internal 1 Ω, across a 3 Ω resistor — current?
  3. 3.
    Why is terminal voltage less than EMF when current flows?

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

Intuition

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?
  1. Junction rule: total current in = total current out.
  2. In = 3+2=53 + 2 = 5 A.
  3. So the single outgoing wire carries 5 A.
Answer:5 A.
Practice this conceptself-check · 3 quick reps

Try it yourself

"The sum of EMFs and potential differences around any closed loop is zero." Which conservation principle is this a direct consequence of?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Kirchhoff's junction rule expresses conservation of…
  2. 2.
    Kirchhoff's loop rule expresses conservation of…
  3. 3.
    Currents 4 A and 1 A enter a junction; 2 A leaves on one wire. What leaves on the other?

From the bank · past-year question

Example 2Electricity and MagnetismMODERATE
"The sum of emfs and potential differences around a closed loop equals zero" is a consequence of

[Q67 · Sep · 2019]

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

Intuition

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?
  1. (a) One bulb across one cell: voltage ε, power P=ε2/RP = \varepsilon^2/R.
  2. (b) Two cells in series give 2ε; bulbs in parallel each get the full 2ε.
  3. Each bulb's power = (2ε)2/R=4ε2/R(2\varepsilon)^2/R = 4\varepsilon^2/R.
  4. (b) is four times brighter per bulb than (a).
Answer:(b) — each bulb dissipates 4ε²/R, the brightest of the two.
Practice this conceptself-check · 3 quick reps

Try it yourself

Two identical bulbs are connected in series across one cell, versus a single bulb across the same cell. Which single bulb is brighter, and why?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Two equal cells in series give what EMF (each ε)?
  2. 2.
    Do bulbs in parallel or series each get the full supply voltage?
  3. 3.
    Brightness of a bulb corresponds to which quantity?

From the bank · past-year question

Example 3Electricity and MagnetismHARD
In which one among the following situations, the bulb would glow the most ? (Consider all batteries are the same) (a) 1 battery, 2 bulbs in series (b) 2 batteries in series, 2 bulbs in series (c) 1 battery, 1 bulb (d) 2 batteries in series, 2 bulbs in parallel

[Q69 · Sep · 2024]

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)

Mastery check — 1 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Electricity and MagnetismHARD
An electric circuit is given with V1=1V_1 = 1 V and Resistance R=1000 ΩR = 1000\,\Omega. The current through R is close to 1 mA and VAB=1V_{AB} = 1 V. The circuit is changed by adding V2=5V_2 = 5 V in parallel with internal resistance 0.1 Ω0.1\,\Omega for both batteries. The current through R is about:

[Q52 · Apr · 2024]

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