In short

A source of electrical energy — a cell, a battery, a generator — has two characteristic numbers:

  • EMF (\varepsilon, electromotive force): the work done per unit charge by the source's internal mechanism (chemistry, in a cell), equal to the open-circuit terminal voltage. Unit: volts. A fresh Eveready AA cell has \varepsilon \approx 1.50 V; a Duracell alkaline AA, \varepsilon \approx 1.55 V; a 12 V Maruti car battery, \varepsilon \approx 12.6 V when fully charged.
  • Internal resistance (r): the resistance of the electrolyte, electrodes, and connections inside the cell. Unit: ohms. An AA cell has r \sim 0.15\ \Omega when fresh, growing to ohms as the cell ages. A Maruti car battery has r \sim 0.01\ \Omega so it can deliver hundreds of amps to the starter motor.

When the cell drives a current I into an external load, the terminal voltage V (what a voltmeter reads across the cell's terminals in the working circuit) is

\boxed{\;V \;=\; \varepsilon - I\,r\;}.

The drop I r is voltage lost inside the cell to its own internal resistance. If I = 0, V = \varepsilon — the open-circuit reading is the EMF. If I is large (a shorted cell, a starter motor), V collapses toward zero.

Cells in series combine like resistors-in-series: \varepsilon_\text{total} = \varepsilon_{1}+\varepsilon_{2}+\ldots and r_\text{total} = r_{1}+r_{2}+\ldots. Four 1.5 V cells in a torch give 6 V with total internal resistance 0.6 Ω.

Cells in parallel combine like resistors-in-parallel, but with a twist: the cells must have the same EMF or circulating currents will flow. For n identical cells in parallel, \varepsilon_\text{total} = \varepsilon (unchanged) and r_\text{total} = r/n. Parallel trades voltage stability for current capacity — which is why a Maruti's 12 V, 45 Ah battery is actually six 2-volt cells in series internally, and why a solar-street-light pack is many 3.2 V LiFePO_{4} cells in parallel.

Pull a fresh AA cell out of its pack, touch a voltmeter across its two ends, and the meter reads 1.52 V. Now drop the cell into a small torch with a 1 Ω LED load. The torch lights up and the voltmeter across the cell — still the same cell — now reads 1.35 V. You did not change the cell. You did not change the voltmeter. Yet the voltage you measure is different, by a tenth of a volt, just because the cell is now delivering a current of about 1.3 A instead of zero. If the cell really were a perfect 1.5 V source, the meter should read 1.5 V whether a load is connected or not. Something inside the cell is resisting its own output.

Look at the label on a fully charged Maruti 12 V car battery. In the morning, on a voltmeter, it reads 12.6 V with nothing connected. Turn the ignition key, engage the starter motor. The starter draws 200 A for about a second to turn the engine over, and if you have the voltmeter still connected, the terminal voltage plunges from 12.6 V down to about 10.4 V for that second, then springs back. A 2.2 V drop while sourcing 200 A — what you're measuring is the internal resistance of the battery, (12.6 - 10.4)/200 \approx 11\ \text{m}\Omega. On a cold morning in Shimla, that same internal resistance doubles (colder chemistry means sluggish ions), the voltage sag is 4 V, and the starter struggles. This is real physics, and it is worth making precise.

Defining EMF — work per unit charge, not a voltage across a wire

When a current I flows through an external resistor, positive charge moves from the cell's positive terminal, around the external circuit, and back into the negative terminal. Inside the cell, that charge must be pumped uphill — from the negative terminal back out through the positive terminal — because the external circuit only lets charge slide downhill from high potential to low.

Something inside the cell does this pumping work. In a lead-acid battery, it is the chemical reaction

\text{Pb} + \text{PbO}_{2} + 2\,\text{H}_{2}\text{SO}_{4} \longrightarrow 2\,\text{PbSO}_{4} + 2\,\text{H}_{2}\text{O}

which liberates free energy per electron transferred, and uses it to do work against the electric field pushing the electron from the low-potential to the high-potential plate. In an alkaline AA cell, it is the zinc-manganese-dioxide reaction. In a solar cell, it is a photon creating an electron-hole pair in the depletion region of a p-n junction. In a Van de Graaff generator, it is a rubber belt mechanically carrying charge against the field.

EMF is defined as the work done by this internal mechanism per unit positive charge pumped from the negative to the positive terminal:

\varepsilon \;\equiv\; \frac{W_\text{internal}}{q}. \tag{1}

The units of EMF are joules per coulomb, which is volts. But EMF is conceptually not the same kind of quantity as the potential difference across a resistor. EMF is a property of the source, depending on the internal mechanism (chemistry, photons, mechanical drive). Potential difference is a property of the field — an integral of \vec{E}\cdot d\vec{l} between two points.

The numerical coincidence is this: when the cell is disconnected from any external circuit (no current flows, the electron pump is idling), the potential difference between the cell's terminals exactly balances the EMF. The chemistry has pushed enough charge onto the plates that the electrostatic field inside the cell cancels the driving force of the chemistry, and the system sits at equilibrium. Open-circuit terminal voltage equals EMF. This is what a voltmeter reads when there is no load.

Once a current is drawn, this equilibrium is broken, and the terminal voltage drops below the EMF by an amount proportional to the current. That proportionality constant is the internal resistance.

The cell as EMF + internal resistance

A real cell is modelled as an ideal EMF source in series with an internal resistance:

Equivalent circuit of a real cellDashed box enclosing an ideal EMF source epsilon in series with an internal resistance r. Two wires emerge as the cell's positive and negative terminals, labelled A and B. Outside the dashed box a load resistor R and ammeter I are shown in a loop.real cell (dashed box)εrA (+)B (−)RAammeter II →
Model of a real cell: an ideal EMF source $\varepsilon$ in series with an internal resistance $r$, both enclosed inside the dashed box. The external circuit sees only the two terminals A and B, across which the terminal voltage is $V = \varepsilon - I r$. The current $I$ circulates through the whole loop — the same $I$ flows through $\varepsilon$, $r$, and $R$.

The model says: an external circuit cannot tell the difference between (i) a pure 1.5 V source in series with a 0.15 Ω resistor, and (ii) a real AA cell with EMF 1.5 V and internal resistance 0.15 Ω. Whatever load you put across the terminals, the current, voltage, and power in the external circuit come out the same. This is a powerful simplification: it lets you analyse any cell-driven circuit using just two numbers.

Deriving V = \varepsilon - I r from a round trip

Start at terminal B (negative), traverse the cell from B to A, then continue through the external resistor R back to B. Along each element, keep a running tally of the potential:

Step 1. Inside the cell, go from B to A. You pass through the internal resistance r (in the direction of the current I), then through the EMF source in the "uphill" direction.

Going through r in the direction of I drops the potential by I r (Ohm's law inside the cell: the EMF drives current up through r, and the voltage across r pulls down by I r). Going through the EMF source in the uphill direction raises the potential by \varepsilon. Net change from B to A:

V_{A} - V_{B} \;=\; +\varepsilon \;-\; I r. \tag{2}

Why: the EMF source is an electron pump; going "through" it from negative to positive terminal raises a positive charge's potential by \varepsilon. The internal resistance, by contrast, is an ordinary resistor; traversing it in the direction of current flow lowers the potential by I r. Signs are what make circuit analysis work: EMF source in the + direction contributes +\varepsilon; resistor in the direction of current contributes -I R.

Step 2. The terminal voltage is defined as V = V_{A} - V_{B}. So

\boxed{\;V \;=\; \varepsilon - I\,r\;} \tag{3}

immediately. The working voltage between the cell's terminals is the EMF minus the internal ohmic drop.

When the load is known, solve for I directly

Complete the loop by going from A through the external resistor R back to B. The potential drop across R in the direction of current is I R, and it must equal the terminal voltage:

V \;=\; I R. \tag{4}

Why: R is an ordinary external resistor; Ohm's law gives V = I R across it, with V on the positive-terminal side. This is Kirchhoff's voltage law applied to the loop, with one equation written for the cell (3) and one for the external resistor (4).

Combine (3) and (4):

I R \;=\; \varepsilon - I r
I (R + r) \;=\; \varepsilon
I \;=\; \frac{\varepsilon}{R + r}. \tag{5}

Why: the total resistance in the loop is R + r — the external resistor in series with the internal resistance. Dividing by the total resistance gives the current driven by the EMF. This is Ohm's law for the whole loop, with R + r as the total resistance.

Equation (5) is the single most useful formula in this chapter. It tells you the current a real cell will drive through any external load, given the cell's EMF and internal resistance and the load's resistance.

Limiting cases:

Interactive — load resistance and terminal voltage

Interactive: terminal voltage and current vs load resistance for a 1.5 V, 0.15 Ω cell Two curves: terminal voltage V = epsilon R / (R + r) rising with load resistance toward 1.5 V, and current I = epsilon / (R + r) falling with load resistance. Draggable load resistance slider with live readouts. load resistance R (Ω) V (V) and I (A) 1 3 5 7 9 1.0 0.5 1.5 ε = 1.5 V V (V) I (A) drag the red dot
A 1.5 V AA cell with internal resistance 0.15 Ω driving a load $R$. Drag the red dot to change $R$. As $R \to 0$, current spikes (short-circuit, $I \to 10$ A) and terminal voltage collapses; as $R \to \infty$, current falls to zero and terminal voltage rises to $\varepsilon = 1.5$ V. The power delivered to the load, $P_{R} = I^{2}R$, is shown in the readout — it peaks at $R = r = 0.15$ Ω, the maximum-power-transfer condition.

Two curves overlay on the same axes. The red curve is terminal voltage V = \varepsilon R/(R+r), which starts at zero (short circuit) and rises asymptotically to 1.5 V (open circuit). The dark curve is current I = \varepsilon/(R+r), which starts at 10 A (short-circuit current, \varepsilon/r = 1.5/0.15) and falls to zero as R grows. They cross near R = r, and the power delivered to the load — P_R = V I — peaks right there.

Cells in series — EMFs add, internal resistances add

A standard Eveready-powered torch uses four AA cells end-to-end in a tube. Each cell has \varepsilon = 1.5 V, r = 0.15 Ω. The torch sees them as a single source: EMF 6 V, internal resistance 0.6 Ω.

Three cells in series with combined EMF and internal resistanceThree AA cells in a row, each with EMF epsilon and internal resistance r. The combined circuit sees 3 epsilon and 3 r in series.three cells in seriesε₁r₁ε₂r₂ε₃r₃equivalent: ε₁+ε₂+ε₃, r₁+r₂+r₃
Three cells in series, each with its own EMF and internal resistance. Since the same current flows through all three, the EMFs simply add (the uphill voltage gained at each cell accumulates), and the internal resistances add in series.

Derivation. For n cells in series (all oriented the same way, positive-to-negative), going once around the loop picks up \varepsilon_{1}+\varepsilon_{2}+\ldots+\varepsilon_{n} from the EMFs and drops I(r_{1}+r_{2}+\ldots+r_{n}) across the internal resistances. Writing Kirchhoff's voltage law for the whole loop including the external resistor R:

\bigl(\varepsilon_{1}+\ldots+\varepsilon_{n}\bigr) \;=\; I\,(R + r_{1}+\ldots+r_{n}).

Why: this is just Ohm's law for a single equivalent cell with \varepsilon_\text{eq} = \sum \varepsilon_{i} and r_\text{eq} = \sum r_{i}. The identical-current condition is what makes the rule simple; it comes from series connection.

\varepsilon_\text{eq} \;=\; \sum_{i} \varepsilon_{i}, \qquad r_\text{eq} \;=\; \sum_{i} r_{i}. \tag{6}

Opposition (one cell flipped). If one cell is reversed (say \varepsilon_{2} points the wrong way), its EMF contribution becomes -\varepsilon_{2} — but its internal resistance is still +r_{2} (a resistor doesn't care about direction). So with a flipped cell:

\varepsilon_\text{eq} \;=\; \varepsilon_{1} - \varepsilon_{2} + \varepsilon_{3}, \qquad r_\text{eq} \;=\; r_{1}+r_{2}+r_{3}.

If a child in a hurry to change torch batteries inserts one of four AAs backwards, \varepsilon_\text{eq} = 1.5 + 1.5 - 1.5 + 1.5 = 3 V instead of 6 V — the torch will be dim. And the forward cells are wastefully charging (discharging?) the reversed cell through the low internal resistance. That is why torches have clear "+/−" markings inside.

Cells in parallel — how to combine when EMFs differ

Connect two cells in parallel by joining their positive terminals together, and joining their negative terminals together. Now drive a load R from the joined terminals.

For identical cells (same \varepsilon, same r), symmetry demands each cell supplies half the total current. The parallel combination then behaves exactly like a single cell of EMF \varepsilon and internal resistance r/n (for n identical cells). More current can be supplied without the cell's terminal voltage drooping, because the internal resistance r_\text{eq} = r/n is small.

For non-identical cells — different EMFs, different internal resistances — the formula requires more care. Suppose two cells: \varepsilon_{1}, r_{1} and \varepsilon_{2}, r_{2} in parallel, driving a load R.

Let I be the current leaving the parallel combination through R. Let I_{1} be the current from cell 1 into the junction, I_{2} from cell 2. At the junction, conservation of charge gives

I \;=\; I_{1} + I_{2}. \tag{7}

The terminal voltage V is the same for both cells (they share the terminals). For each cell:

V \;=\; \varepsilon_{1} - I_{1} r_{1}, \qquad V \;=\; \varepsilon_{2} - I_{2} r_{2}. \tag{8}

Solve each for the cell current:

I_{1} \;=\; \frac{\varepsilon_{1} - V}{r_{1}}, \qquad I_{2} \;=\; \frac{\varepsilon_{2} - V}{r_{2}}. \tag{9}

Substitute into (7):

I \;=\; \frac{\varepsilon_{1} - V}{r_{1}} + \frac{\varepsilon_{2} - V}{r_{2}} \;=\; \frac{\varepsilon_{1}}{r_{1}} + \frac{\varepsilon_{2}}{r_{2}} \;-\; V\left(\frac{1}{r_{1}} + \frac{1}{r_{2}}\right). \tag{10}

Rearrange for V:

V\left(\frac{1}{r_{1}} + \frac{1}{r_{2}}\right) \;=\; \frac{\varepsilon_{1}}{r_{1}} + \frac{\varepsilon_{2}}{r_{2}} \;-\; I.

Define

\varepsilon_\text{eq} \;\equiv\; \frac{\dfrac{\varepsilon_{1}}{r_{1}} + \dfrac{\varepsilon_{2}}{r_{2}}}{\dfrac{1}{r_{1}} + \dfrac{1}{r_{2}}}, \qquad \frac{1}{r_\text{eq}} \;\equiv\; \frac{1}{r_{1}} + \frac{1}{r_{2}}. \tag{11}

Then V = \varepsilon_\text{eq} - I\,r_\text{eq}, which is exactly the V = \varepsilon - I r relation for a single equivalent cell.

Why: the weighted-average formula \varepsilon_\text{eq} = (\varepsilon_{1}/r_{1} + \varepsilon_{2}/r_{2})/(1/r_{1} + 1/r_{2}) has a physical meaning: it is the EMF weighted by the conductance 1/r_{i} of each cell. A low-internal-resistance cell (big conductance) dominates the average; a high-internal-resistance cell barely shifts it. This is why you cannot usefully parallel-combine an old cell with a fresh one — the fresh cell dominates and ends up supplying most of the current, with extra current circulating wastefully.

For n identical cells, this reduces to \varepsilon_\text{eq} = \varepsilon (EMFs all equal, so the weighted average is trivially \varepsilon) and r_\text{eq} = r/n.

Why circulating currents happen with mismatched cells. If \varepsilon_{1} \neq \varepsilon_{2} and you short-circuit the external load (R = 0), then from (9), with V given by the zero-load equation:

I_\text{circ} \;=\; \frac{\varepsilon_{1} - \varepsilon_{2}}{r_{1} + r_{2}}.

Current flows from the higher-EMF cell through the lower-EMF cell, dissipating power (\varepsilon_{1} - \varepsilon_{2})^{2}/(r_{1}+r_{2}) internally. That is why battery makers insist that cells combined in parallel should be of the same type, the same age, and the same state of charge.

Worked examples

Example 1: The torch with a weak cell

A 4-cell Eveready torch has three fresh 1.5 V cells with r = 0.15\ \Omega each, and one older cell whose EMF has drooped to 1.3 V with r = 0.40\ \Omega. The bulb has resistance 4 Ω. Find (a) the current through the bulb, (b) the terminal voltage across the series combination of cells, (c) the voltage across the weak cell alone, and (d) the power dissipated in the weak cell.

Four cells in series driving a bulb: three fresh, one weakTorch circuit with four cells in series. Three fresh cells each supply 1.5 V and 0.15 ohms internal resistance. One weak cell supplies 1.3 V and 0.40 ohms. External load is a 4 ohm bulb. Total EMF 5.8 V, total internal resistance 0.85 ohms.four cells in series (3 fresh + 1 weak)ε=1.5, r=0.15 ×3weak: 1.3 V, 0.40 ΩR = 4 Ω
Four cells in series drive a 4 Ω bulb. Three are fresh (1.5 V, 0.15 Ω each); one is weak (1.3 V, 0.40 Ω). The series combination acts as one equivalent cell of EMF 5.8 V and internal resistance 0.85 Ω.

Step 1. Combine the cells in series.

\varepsilon_\text{eq} = 1.5 + 1.5 + 1.5 + 1.3 = 5.8 V.

r_\text{eq} = 0.15 + 0.15 + 0.15 + 0.40 = 0.85 Ω.

Why: same current flows through all four cells, so their EMFs and internal resistances combine by direct addition (equation 6). The weak cell pulls the total EMF down by 0.2 V and pushes total internal resistance up by 0.25 Ω above what a four-fresh-cell pack would look like (which would be 6.0 V, 0.6 Ω).

Step 2. Apply Ohm's law to the loop.

I \;=\; \frac{\varepsilon_\text{eq}}{R + r_\text{eq}} \;=\; \frac{5.8}{4 + 0.85} \;=\; \frac{5.8}{4.85} \;\approx\; 1.196\ \text{A}.

Why: equation (5) for the single equivalent cell. The 0.85 Ω internal resistance steals some of the 5.8 V EMF and drops it inside the pack rather than across the bulb.

Step 3. Terminal voltage of the pack.

V_\text{pack} \;=\; \varepsilon_\text{eq} - I\,r_\text{eq} \;=\; 5.8 - (1.196)(0.85) \;=\; 5.8 - 1.02 \;\approx\; 4.78\ \text{V}.

Check: V_\text{pack} = I R = 1.196 \times 4 \approx 4.78 V. ✓

Why: the terminal voltage of the battery pack equals the voltage across the bulb (they're connected directly). Both computations give the same answer — a useful consistency check.

Step 4. Voltage across the weak cell alone.

V_\text{weak} \;=\; \varepsilon_\text{weak} - I\,r_\text{weak} \;=\; 1.3 - (1.196)(0.40) \;=\; 1.3 - 0.478 \;\approx\; 0.82\ \text{V}.

Why: apply V = \varepsilon - I r to the weak cell alone. Its terminal voltage is much lower than its nominal 1.3 V because its high internal resistance drops almost 0.48 V internally when the current flows.

Step 5. Power dissipated inside the weak cell.

P_\text{weak} \;=\; I^{2} r_\text{weak} \;=\; (1.196)^{2}(0.40) \;\approx\; 0.57\ \text{W}.

Power delivered by the weak cell's EMF to the circuit: \varepsilon_\text{weak} I = 1.3 \times 1.196 \approx 1.55 W. Of this, 0.57 W is wasted inside the weak cell; only 0.98 W reaches the external circuit.

Why: the weak cell is doing useful work (still producing 1.3 V of EMF, still sourcing 1.2 A) but a significant fraction of that work is burned up internally as heat. You can feel this if you run the torch for ten minutes and then touch the cells — the weak one is noticeably warmer than the three fresh ones.

Result: I \approx 1.20 A; V_\text{pack} \approx 4.78 V; V_\text{weak} \approx 0.82 V; P_\text{weak} \approx 0.57 W.

What this shows: One weak cell in a series pack doesn't just reduce the EMF proportionally — it makes the whole pack run at a higher effective internal resistance, and it wastefully dissipates current inside the weak cell. This is why mixing old and new cells in the same torch is advised against: the old cell burns more energy than it produces externally, and in extreme cases can overheat or leak.

Example 2: The car starter motor

A Maruti 12 V lead-acid battery with EMF 12.6 V has internal resistance 11 mΩ (fully charged, room temperature). The starter motor has a resistance of 52 mΩ during cranking (the motor is essentially a low-resistance coil). (a) Compute the cranking current. (b) Find the battery's terminal voltage during cranking. (c) Compute the power delivered to the motor versus the power wasted in the battery's internal resistance. (d) If on a cold Shimla morning the battery's internal resistance doubles to 22 mΩ (while EMF stays at 12.6 V), how do these numbers change?

Step 1. Cranking current at room temperature.

I_\text{crank} \;=\; \frac{\varepsilon}{R_\text{motor} + r} \;=\; \frac{12.6}{0.052 + 0.011} \;=\; \frac{12.6}{0.063} \;\approx\; 200\ \text{A}.

Why: 200 A is a very large current, which is why starter circuits use heavy-gauge copper cable (often 25 mm² or more) and have no fuse on the cranking loop. The current is limited by the sum of motor winding resistance and battery internal resistance; the cable resistance is usually below 1 mΩ for a 30 cm run and contributes negligibly.

Step 2. Terminal voltage during cranking.

V_\text{terminal} \;=\; \varepsilon - I\,r \;=\; 12.6 - (200)(0.011) \;=\; 12.6 - 2.2 \;=\; 10.4\ \text{V}.

Why: that 2.2 V sag is what a driver sees on the voltmeter as the key is turned. Car stereos and dashboard electronics briefly lose supply (dropping to 10.4 V) during cranking — which is why cars have a "cranking disconnect" for sensitive electronics on newer models.

Step 3. Power split.

Power delivered to the motor: P_\text{motor} = I^{2} R_\text{motor} = (200)^{2}(0.052) = 2080 W \approx 2.1 kW.

Power wasted in the battery's internal resistance: P_\text{r} = I^{2} r = (200)^{2}(0.011) = 440 W.

Total: \varepsilon I = 12.6 \times 200 = 2520 W. Check: 2080 + 440 = 2520 ✓.

Why: efficiency = P_\text{motor}/P_\text{total} = 2080/2520 \approx 83\%. Most of the battery's chemical energy during cranking goes into the motor; only 17% is wasted inside the battery. A starter engages for about 1 second, so total wasted energy per crank is only 0.4 kJ — the battery warms imperceptibly.

Step 4. Cold morning: r doubled to 22 mΩ.

I_\text{crank} = 12.6/(0.052 + 0.022) = 12.6/0.074 \approx 170 A.

V_\text{terminal} = 12.6 - (170)(0.022) = 12.6 - 3.74 = 8.86 V.

P_\text{motor} = (170)^{2}(0.052) \approx 1500 W.

P_\text{r} = (170)^{2}(0.022) \approx 636 W.

Efficiency: 1500/(1500+636) \approx 70\%.

Why: on the cold Shimla morning the battery only manages 170 A instead of 200, the motor sees 1.5 kW instead of 2.1 kW, and the battery's terminal voltage sags below 9 V — often below the electronics' brown-out threshold, which is why cars beep or reset their clocks when cranked in cold weather. The same battery is now 13% less efficient and delivers 30% less cranking power, purely because the chemical reactions and ionic transport are slower at lower temperature (shorter \tau-like effect inside the electrolyte, modelled as higher r).

Result at 25 °C: I = 200 A, V = 10.4 V, P_\text{motor} = 2.1 kW, efficiency 83%. At cold: I = 170 A, V = 8.9 V, P_\text{motor} = 1.5 kW, efficiency 70%.

What this shows: A lead-acid starter battery is specified by its cold-cranking amps (CCA) rating — the current it can deliver at -18 °C while maintaining at least 7.2 V terminal voltage for 30 seconds. A CCA number is really an indirect specification of r at low temperature. A Maruti Alto might use a 35 Ah battery with 300 CCA; a Mahindra Scorpio needs 700 CCA because its bigger diesel engine is harder to crank.

Example 3: Comparing two parallel cell configurations

You are building a solar-powered LED street light controller. You have six identical 3.2 V LiFePO_{4} cells, each with r = 0.05 Ω, capable of 30 A pulse current. The LED load requires 12 V at a constant 5 A. (a) Configure the cells as three pairs in series, each pair being two cells in parallel (3S2P). (b) Configure the cells as two strings in parallel, each string being three cells in series (2P3S... actually 3S2P written differently). Find \varepsilon_\text{eq}, r_\text{eq}, and the terminal voltage at 5 A for each configuration. Which is better?

Step 1 (Configuration A: 3S2P — three series-pairs, each pair parallel).

Each pair: two cells in parallel. \varepsilon_\text{pair} = 3.2 V (identical cells, weighted average unchanged). r_\text{pair} = 0.05/2 = 0.025 Ω.

Three pairs in series: \varepsilon_\text{eq} = 3 \times 3.2 = 9.6 V. r_\text{eq} = 3 \times 0.025 = 0.075 Ω.

Why: parallel-first halves the effective r of each pair, and then series-stacking triples it. Net result: r_\text{eq} = 3r/2, which is smaller than the six-cells-in-series configuration (6r = 0.30 Ω) by a factor of four.

Step 2 (Configuration B: 3S in parallel with another 3S — "2P3S").

Each string: three cells in series. \varepsilon_\text{string} = 3 \times 3.2 = 9.6 V. r_\text{string} = 3 \times 0.05 = 0.15 Ω.

Two strings in parallel: \varepsilon_\text{eq} = 9.6 V (identical strings). r_\text{eq} = 0.15/2 = 0.075 Ω.

Why: series-first triples the voltage and triples the effective r, then parallel halves it. Net result is the same 3r/2 = 0.075 Ω, which is no accident — both configurations have the same total number of cells wired electrically equivalently.

Step 3. Neither configuration reaches 12 V!

The total EMF available from six cells is 6 \times 3.2 = 19.2 V if you put them all in series. The 3S2P and 2P3S configurations give 3 \times 3.2 = 9.6 V. To drive a 12 V LED, you need a configuration like 4S (four cells in series: 4 \times 3.2 = 12.8 V, with the remaining two cells either parallel to some of those or held in reserve) or a different topology. The configurations I analysed above don't meet the 12 V requirement from two groups of three.

Redo with 4S2P (four-cells-in-series, two such strings in parallel, using eight cells — not available here) or with a DC-DC boost converter from 9.6 V to 12 V.

Why: this example shows that series-parallel choices are constrained by the voltage requirement first. The parallel stage increases current capacity (lowers r_\text{eq}), but does not raise voltage. Always compute \varepsilon_\text{eq} first and check it meets the load's voltage need.

Step 4. Terminal voltage at 5 A (hypothetically, if the load worked at 9.6 V).

V_\text{terminal} = 9.6 - (5)(0.075) = 9.6 - 0.375 = 9.225 V. Only 3.9% sag under 5 A draw.

Result: 3S2P and 2P3S are electrically equivalent (both give 9.6 V, 0.075 Ω) but do not reach the 12 V requirement. A real solar street-light design would use 4S2P (eight cells total) or 4S1P + boost converter.

What this shows: Series-parallel topology choice is driven by voltage and current requirements together. For the same total cell count, you can trade voltage against current-delivery-without-sag. Battery-pack engineers in Indian EV startups like Ola and Ather spend much of their design time on exactly this optimisation — balancing cell count, cost, voltage, internal resistance, and safety.

Common confusions

If you came here to understand EMF, terminal voltage, internal resistance, and the series and parallel rules, you have what you need. What follows is the proof of the maximum-power-transfer theorem, a derivation of the Nernst equation relating EMF to chemistry, and a note on AC impedance of batteries.

Maximum power transfer theorem

A cell with fixed \varepsilon and r drives a variable load R. The power delivered to R is

P_{R}(R) \;=\; I^{2} R \;=\; \frac{\varepsilon^{2}\,R}{(R+r)^{2}}.

To maximise P_{R} with respect to R, differentiate and set to zero:

\frac{dP_{R}}{dR} \;=\; \varepsilon^{2}\,\frac{(R+r)^{2} - R\cdot 2(R+r)}{(R+r)^{4}} \;=\; \varepsilon^{2}\,\frac{r - R}{(R+r)^{3}}.
\frac{dP_{R}}{dR} \;=\; 0 \quad \Longleftrightarrow \quad R \;=\; r.

The maximum power delivered is

P_{R,\text{max}} \;=\; \frac{\varepsilon^{2}}{4 r},

and half the source's power goes to the load, half is wasted in the internal resistance. Efficiency at matched load is therefore 50%.

When is matched load the right design choice? When maximising power delivered is the goal: audio amplifiers driving speakers, antennas matched to transmission lines, certain thermoelectric generators. When is it wrong? When efficiency is the goal: car batteries, power grids, any system where wasted heat is expensive. A car battery in cranking is operating near matched load (efficiency 70–80%) — tolerable for a one-second burst. A car battery in normal running is far from matched load (efficiency 99%+) — essential for long-term operation.

The 50% efficiency cap at matched load is a fundamental thermodynamic consequence of the single-source/single-load topology. Transformers and impedance-matching networks can shift the apparent impedance of a load without changing the efficiency limit — they rearrange where the 50% losses occur, not eliminate them.

Nernst equation — where EMF comes from

The EMF of an electrochemical cell is set by the free-energy change of the cell reaction:

\varepsilon \;=\; -\,\frac{\Delta G}{n F},

where \Delta G is the Gibbs free-energy change per mole of reaction, n is the number of electrons transferred per reaction unit, and F = 96{,}485 C/mol is Faraday's constant (one mole of electron charge). For the alkaline Zn/MnO_{2} reaction, \Delta G \approx -289 kJ/mol with n = 2, giving \varepsilon \approx 289{,}000/(2 \times 96485) \approx 1.50 V — exactly the EMF of a fresh alkaline AA cell.

As the cell discharges, the concentrations of reactants and products change, so \Delta G changes. The Nernst equation quantifies this:

\varepsilon(Q) \;=\; \varepsilon^{\circ} - \frac{R T}{n F}\ln Q,

where \varepsilon^{\circ} is the EMF at standard concentrations, Q is the reaction quotient (concentration of products / concentration of reactants, each raised to their stoichiometric powers), R = 8.314\ \text{J mol}^{-1}\,\text{K}^{-1}, and T is the temperature. The Nernst equation explains the slow EMF droop you see as a torch battery ages: Q changes, \varepsilon drops by tens of millivolts, even before the cell is "dead" in the ordinary sense.

The internal resistance r, by contrast, has kinetic origins (ion mobility through the electrolyte, charge-transfer overpotentials at the electrodes) rather than thermodynamic ones. As a cell discharges, r generally rises — reaction products clog the electrode pores, ionic concentrations become depleted. The combination of falling EMF and rising internal resistance is what makes a cell's capacity fall off sharply near end-of-life.

AC impedance and "beyond DC" internal resistance

The r in V = \varepsilon - I r is really a DC approximation. At higher frequencies, a cell behaves as a complex AC impedance:

Z(\omega) \;=\; r_{0} + \frac{1}{j\omega C_\text{dl}} + r_\text{ct} \parallel (\text{Warburg term}),

where r_{0} is a series ohmic term (electrolyte + wires), C_\text{dl} is the double-layer capacitance at each electrode surface (microfarads per square centimetre), r_\text{ct} is the charge-transfer resistance, and the Warburg term is a diffusion-limited impedance that goes as 1/\sqrt{\omega}. For a lithium-ion cell of a smartphone, a plot of |Z(\omega)| vs frequency shows r \sim 0.05 Ω at 1 kHz, rising to 0.3 Ω at 10 mHz (slow-discharge regime).

The rapid pulse loading a Maruti starter subjects the battery to is closer to the high-frequency regime; the slow trickle from a parked car's clock and alarm is the low-frequency regime. Good battery management systems model both with separate r values.

Indian-context note — battery chemistry in the grid

India's National Solar Mission (2010–) has driven massive deployment of battery storage. As of 2025, utility-scale installations at Bhadla Solar Park (Rajasthan) and Pavagada Solar Park (Karnataka) use lithium-iron-phosphate (LiFePO_{4}) cells with \varepsilon = 3.2 V and r \sim 0.001 Ω per cell (cell-sized LiFePO_{4} is a 100-Ah prismatic, not an AA-size cell). A 1 MWh grid-connected bank is 300,000 cells, carefully graded so that cells combined in parallel within a string are within 10 mV of each other — precisely to avoid the circulating-current wastage derived in the main text.

Where this leads next