EMF and Internal Resistance
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Direct answer
Every real cell keeps a small resistance r inside, so the voltage at its terminals falls as soon as current flows: V = ε - Ir while discharging, V = ε on open circuit, and V = ε + Ir while being charged, with the circuit current I = ε/(R + r). The emf itself is the chemistry's work per coulomb, unaffected by the load. This bookkeeping decides three NEET favourites: maximum power reaches the external resistance exactly when R = r; the efficiency of transfer is η = R/(R + r), only 50 per cent at maximum power; and identical cells combine as ε_eq = nε, r_eq = nr in series but ε, r/n in parallel.
What you must remember
- Terminal voltage: V = ε - Ir while discharging; the drop Ir is spent inside the cell; a battery tester reads full voltage only because it draws negligible current.
- Charging case: V = ε + Ir — the charger pushes against both the emf and the internal drop.
- Circuit current: I = ε/(R + r); short-circuiting the cell gives I = ε/r, large because r is small — why even a low-voltage cell is dangerous when shorted.
- Maximum power transfer: P_max = ε^2/4r when R = r; beyond this point delivered power falls while efficiency keeps rising — the two optima are different.
- Efficiency: η = R/(R + r); power transfer capability and efficiency pull in opposite directions, and questions probe which one is being maximised.
- Cells in series: ε_eq = nε, r_eq = nr — higher voltage, higher internal resistance.
- Cells in parallel (identical): ε_eq = ε, r_eq = r/n — same voltage, lower internal resistance, longer life.
A 12-volt cell under three loads
Take a cell of ε = 12 V and r = 1 Ω. Across R = 5 Ω: I = 12/6 = 2 A, terminal voltage V = ε - Ir = 10 V, and of the 24 W generated, 20 W reaches the load while 4 W heats the cell — efficiency 5/6 ≈ 83 per cent. Now replace the load with R = 1 Ω (equal to r): I = 6 A, V = 6 V, load power 36 W — the maximum possible, since P(R) = ε^2 R/(R + r)^2 peaks at R = r — but efficiency has crashed to 50 per cent, and 36 W also burns inside the cell. Finally connect R = 11 Ω: current drops to 1 A, terminal voltage rises to 11 V, efficiency climbs to about 92 per cent, though the delivered power falls to 11 W. Read the pattern once and the whole topic becomes a graph in the head: terminal voltage rises toward ε, delivered power peaks at R = r, and efficiency climbs monotonically.
Where the marks leak
The distinction between emf and terminal voltage is the first leak: emf is measured by a potentiometer at null, while any voltmeter across the terminals reads V = ε - Ir, always a little low. The second leak is the maximum-power condition misapplied to efficiency: at R = r only half the generated power reaches the load, so power transmission deliberately runs far from R = r while signal matching (an amplifier to a speaker) sits exactly there. Third, the parallel-cells trap: identical cells in parallel do not multiply the current by n; they divide the internal resistance by n while emf stays put, and mixed groupings (m rows of n cells) give ε_eq = nε with r_eq = mr/n. Finally, a charging cell absorbs energy: V = ε + Ir means the terminal voltage exceeds the emf, and candidates who pattern-match the discharge formula contradict the stem's data.
Frequently asked questions
What is the difference between emf and terminal voltage?
Emf is the energy supplied per coulomb by the cell's chemistry with no current flowing; terminal voltage V = ε - Ir is what the external circuit receives once the internal drop is paid — a potentiometer at null reads the former, any voltmeter the latter.
When is the power delivered to a resistor maximum?
When the external resistance equals the internal resistance, R = r, giving P_max = ε^2/4r at 50 per cent efficiency.
Why does a voltmeter slightly under-read a cell's emf?
The voltmeter draws a small current, so the terminal voltage falls below ε by the Ir drop — only a null method such as a potentiometer reads true emf.
How do identical cells in series and in parallel differ?
Series gives ε_eq = nε with r_eq = nr (higher voltage); parallel keeps ε_eq = ε but drops the internal resistance to r/n (steadier current, longer life).
What happens if a cell is short-circuited?
R = 0 limits the current to ε/r, large because r is small; all the power then dissipates inside the cell as heat.