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A new torch battery reads 1.5 V on a voltmeter, but less when the torch is switched on. That drop is internal resistance at work, and it drives a whole family of exam questions: terminal voltage, combinations of cells, grouping for maximum current and maximum power. This post works through each with numbers, then lists the mistakes to avoid.
What do EMF and internal resistance mean?
NCERT's Current Electricity chapter defines the EMF ε as the potential difference between the cell's terminals in open circuit, when no current flows. The name is historical: ε is a potential difference, not a force. The electrolyte has its own resistance r, the internal resistance. With a current I flowing:
V = ε − Ir and I = ε/(R + r)
The largest current the cell could give, with R = 0, is ε/r, though NCERT notes that real cells are usually rated for much less to avoid damage. Internal resistance, EMF and combinations of cells are in the JEE (Main) 2026 and NEET (UG) 2026 syllabi and the CBSE 2026-27 curriculum. Check the current year's syllabus.
Example 1. A cell of EMF 1.5 V and internal resistance 0.50 Ω drives a 2.5 Ω bulb. I = 1.5/(2.5 + 0.50) = 0.50 A. Terminal voltage V = 1.5 − 0.50 × 0.50 = 1.25 V. Check: IR = 0.50 × 2.5 = 1.25 V.
When a cell is being charged, current flows into its positive terminal and the sign flips: V = ε + Ir.
How do cells combine?
NCERT derives both rules.
Series (positive of one to negative of the next): ε_eq = ε₁ + ε₂ and r_eq = r₁ + r₂. A cell connected the wrong way round enters with a negative EMF.
Parallel (positives joined, negatives joined): 1/r_eq = 1/r₁ + 1/r₂ and ε_eq/r_eq = ε₁/r₁ + ε₂/r₂.
Example 2. Cells of 2.0 V, 1.0 Ω and 1.5 V, 0.50 Ω in parallel. r_eq = (1.0 × 0.50)/1.5 = 0.33 Ω. ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) = (2.0 × 0.50 + 1.5 × 1.0)/1.5 = 2.5/1.5 = 1.67 V. The combined EMF lies between the two values, never above the larger.
How should identical cells be grouped for the most current?
Example 3. You have 12 identical cells, each 1.5 V and 0.50 Ω, and a 1.5 Ω load. Arrange them in m parallel rows, each row with n cells in series, so nm = 12. Each row has EMF nε and resistance nr; m rows in parallel give EMF nε and resistance nr/m. So
I = nε/(R + nr/m)
Trying the factor pairs:
n = 12, m = 1: I = 18/(1.5 + 6.0) = 2.4 A
n = 6, m = 2: I = 9.0/(1.5 + 1.5) = 3.0 A
n = 4, m = 3: I = 6.0/(1.5 + 0.67) ≈ 2.8 A
n = 3, m = 4: I = 4.5/(1.5 + 0.375) = 2.4 A
The maximum, 3.0 A, comes when the internal resistance of the group, nr/m, equals the load. That is no accident.
Why is power greatest when R = r?
The power delivered to the load is P = I²R = ε²R/(R + r)². Differentiate with respect to R and set it to zero: the numerator gives (R + r)² − 2R(R + r) = 0, so R = r. Then P_max = ε²/(4r).
Example 4. A source with ε = 12 V and r = 2.0 Ω. At R = 2.0 Ω, I = 12/4.0 = 3.0 A and P = 3.0² × 2.0 = 18 W, matching ε²/4r = 144/8 = 18 W. The source also dissipates I²r = 18 W inside itself, so efficiency is only 50%. In practice, sources are usually run with R much larger than r, giving up maximum power for higher efficiency.
What mistakes show up most?
Students often treat a battery as a fixed source of current. In Engelhardt and Beichner (2004), that idea was the one students used most often in interviews about circuits, and many held several misconceptions even after instruction. In reality the current depends on the whole circuit, R + r. Other common slips:
Using V = ε when current flows. Only in open circuit does the voltmeter read ε.
Adding EMFs in parallel. See the callout.
Ignoring a reversed cell in a series string.
Forgetting the charging case, where V = ε + Ir.
Treating maximum power as maximum efficiency. At R = r, half the energy is lost inside the source.
To measure EMF without drawing current, see our guide to the meter bridge and potentiometer; for multi-loop circuits, Kirchhoff's law mistakes.
Practice routine
Three terminal-voltage problems, including one charging case.
Two parallel combinations with unequal cells, checking that ε_eq lies between the EMFs.
One grouping problem like Example 3 with 16 cells.
Plot P against R for ε = 12 V, r = 2 Ω at R = 0.5, 1, 2, 4 and 8 Ω, and confirm the peak at 2 Ω.
If a grouping problem goes wrong, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can see which line of your set-up differs.
Start with item 4 of the routine: five quick calculations that make the R = r result stick.
