Cells in series and parallel

A torch needs 3 volts to light its bulb properly, but each cell in your drawer gives only 1.5 volts — so you stack two of them, nose to tail, inside the torch barrel. A power bank, on the other hand, packs several identical cells side by side, all their positive tabs bonded together, so it can push a hefty charging current into your phone for hours without any one cell overheating. Same basic cell, two wildly different wirings, two completely different jobs. The torch wants voltage; the power bank wants stamina. Knowing which wiring to reach for — and being able to predict exactly how much current either arrangement will deliver — is the whole content of this topic. The physics behind both tricks is refreshingly small: every real cell is an ideal EMF source $\varepsilon$ in series with a tiny internal resistance $r$ that quietly eats some of the cell's own voltage whenever current flows. Connect cells in series and both the EMFs and the internal resistances simply add. Connect identical cells in parallel and the EMF does not change at all, while the internal resistance shrinks — exactly like resistors in parallel. Everything else in this article — the derivation, two worked numbers, the traps that cost board marks, and the exam-card summary — grows out of that one asymmetry between how EMF and how internal resistance behave under the two wirings. By the end you should be able to look at any external resistor $R$ and immediately judge whether stacking your cells in series or spreading them in parallel delivers more current — and defend that judgement with the numbers, not a guess.

Key takeaways

A single cell behaves as EMF $\varepsilon$ in series with internal resistance $r$ ; combining cells changes the effective $\varepsilon_{eff}$ and $r_{eff}$ that the external circuit actually sees.
Series ( $n$ identical cells, tail to head): EMFs add and internal resistances add — $\varepsilon_{eff} = n\varepsilon$ , $r_{eff} = nr$ .
Parallel ( $n$ identical cells, like terminals joined): EMF stays the same as one cell, internal resistance drops — $\varepsilon_{eff} = \varepsilon$ , $r_{eff} = r/n$ .
Current delivered to external resistance $R$ is always $I = \varepsilon_{eff}/(R + r_{eff})$ once you know the effective EMF and resistance.
Series wins when $R$ is large compared with $r$ (voltage-limited circuits); parallel wins when $R$ is small compared with $r$ (current-hungry circuits), because it is the internal resistance that then dominates.
Mixed grouping — $n$ cells per row in series, $m$ rows in parallel — gives $\varepsilon_{eff} = n\varepsilon$ and $r_{eff} = nr/m$ , combining both ideas.
Key idea: cells in series add their EMFs and their internal resistances (n and n), while identical cells in parallel keep the same EMF but divide the internal resistance by n, so the better wiring is decided by comparing the external resistance R with the single-cell internal resistance r.

The core insight

For $n$ identical cells of EMF $\varepsilon$ and internal resistance $r$ , series wiring gives $\varepsilon_{eff}=n\varepsilon,\ r_{eff}=nr$ while parallel wiring gives $\varepsilon_{eff}=\varepsilon,\ r_{eff}=r/n$ , and comparing $R$ with $r$ tells you which wiring pushes more current through an external resistor $R$ via $I=\varepsilon_{eff}/(R+r_{eff})$ .

Where it fits in the syllabus

This is the direct sequel to EMF and internal resistance of a single cell, still inside the Current Electricity chapter of CBSE Class 12 Physics (2025-26 prescribed scope: "EMF and internal resistance of a cell $\varepsilon = I(R+r)$ , terminal potential difference; cells in series and parallel"). The board wants you to derive the effective EMF and effective internal resistance for both series and parallel groupings of identical cells, and to apply $I = \varepsilon_{eff}/(R+r_{eff})$ to find current in simple circuits. This section deliberately stays with identical cells and does not chase mixed-EMF parallel combinations or potentiometer-based EMF comparison — the potentiometer principle was dropped from the 2025-26 theory syllabus and is out of scope here.

The physical picture

A real cell is never a perfect voltage source. Chemical reactions inside it push charge around, but the electrolyte itself resists that motion, so every cell is best modelled as an ideal EMF $\varepsilon$ in series with a small internal resistance $r$ . When such a cell drives current $I$ through an external resistor $R$ , some of the EMF is used up just pushing current through the cell's own innards, and Ohm's law for the whole loop reads

\varepsilon = I(R + r)

where: $\varepsilon$ is the cell's EMF (volts, the open-circuit voltage with no current drawn), $I$ is the current in the single-loop circuit (amperes), $R$ is the external resistance (ohms), and $r$ is the cell's internal resistance (ohms). The physical reading: the terminal voltage actually delivered to $R$ is $V = IR = \varepsilon - Ir$ , always a little less than the full EMF, and the "loss" $Ir$ grows with current. Now combine several identical cells. In a series chain, the positive terminal of one cell touches the negative terminal of the next, all the way down the line, so the same current $I$ threads through every cell one after another. Each cell contributes its own $\varepsilon$ in the same direction and its own $r$ in the same current path, so both simply add:

\varepsilon_{eff} = n\varepsilon, \qquad r_{eff} = nr

where: $n$ is the number of identical cells in the row. Physically, series wiring is a chain of voltage sources and resistances laid end to end — you are stacking up push and stacking up friction together. In a parallel bank, all the positive terminals are bonded to one common node and all the negative terminals to another. Every cell now sees the same pair of terminals, so the combination behaves like a single EMF equal to one cell's EMF, but with $n$ equal internal resistances sitting side by side between those two nodes — exactly the parallel-resistor situation:

\varepsilon_{eff} = \varepsilon, \qquad r_{eff} = \dfrac{r}{n}

where: the drop in $r_{eff}$ reflects that the total current now has $n$ parallel paths to flow through inside the battery bank, so each individual cell only has to supply $1/n$ of the total current — its own internal resistance is proportionally less of a bottleneck for the group.

Figure

Two circuit sketches side by side, one showing three cells stacked in a series chain feeding an external resistor, the other showing three cells stacked in a parallel bank feeding the same external resistor, with the effective EMF and effective internal resistance labelled under each

Once you have the effective values for either wiring, the current in the single external loop is found the same way it always is for a simple series circuit:

I = \dfrac{\varepsilon_{eff}}{R + r_{eff}}

where: $\varepsilon_{eff}$ and $r_{eff}$ are whichever pair applies to the wiring in question. This single formula is the workhorse for every problem in this topic — the only thing that changes between series and parallel is which pair of effective values you plug in. For a mixed grouping — $n$ cells per row connected in series, and $m$ identical rows connected in parallel — the two ideas combine: each row behaves like a single series cell with $\varepsilon_{eff,row} = n\varepsilon$ and $r_{eff,row} = nr$ , and then $m$ such rows in parallel behave like the parallel rule applied to that row:

\varepsilon_{eff} = n\varepsilon, \qquad r_{eff} = \dfrac{nr}{m}

The detailed optimisation of choosing $n$ and $m$ for maximum current (which happens when $R$ equals $r_{eff}$ ) goes a little beyond what board exams test, but computing $\varepsilon_{eff}$ and $r_{eff}$ correctly for a stated mixed arrangement, and then applying $I=\varepsilon_{eff}/(R+r_{eff})$ , is squarely within scope.

Derivation AEffective EMF and internal resistance of cells in parallel

The goal is to show, starting only from the definition of a cell as an ideal EMF in series with an internal resistance, and from Kirchhoff's current and voltage rules, why $n$ identical cells wired in parallel behave as a single cell of unchanged EMF $\varepsilon$ but reduced internal resistance $r/n$ — rather than simply guessing the parallel-resistor formula by analogy. This matters because internal resistance is not an ordinary passive resistor sitting alone; it sits in series with its own EMF source, so the reduction needs to be justified from the circuit equations, not assumed.

Step 1. Set up the two parallel branches. Take the simplest case of two identical cells, each of EMF $\varepsilon$ and internal resistance $r$ , with their positive terminals joined at node $A$ and negative terminals joined at node $B$ . Let $I_1$ be the current delivered by cell 1 and $I_2$ by cell 2, both flowing from $B$ to $A$ inside their own cell, and let $V$ be the potential difference $V_A - V_B$ across the common terminals.

Step 2. Apply the terminal-voltage relation to each cell separately. For any real cell delivering current $I$ into its terminals, the terminal voltage is $V = \varepsilon - Ir$ (EMF minus the drop across its own internal resistance). So for the two branches, $V = \varepsilon - I_1 r$ and $V = \varepsilon - I_2 r$ , since both cells share the same terminal voltage $V$ across nodes $A$ and $B$ .

Step 3. Solve for each branch current and add them using Kirchhoff's junction rule. From Step 2, $I_1 = \dfrac{\varepsilon - V}{r}$ and $I_2 = \dfrac{\varepsilon - V}{r}$ . The total current leaving the combination toward the external circuit is $I = I_1 + I_2 = \dfrac{2(\varepsilon - V)}{r}$ , because at node $A$ the two branch currents simply add.

Step 4. Rearrange into the standard single-cell form $V = \varepsilon_{eff} - I r_{eff}$ . From Step 3, $I = \dfrac{2\varepsilon - 2V}{r}$ , so $Ir = 2\varepsilon - 2V$ , giving $2V = 2\varepsilon - Ir$ , or $V = \varepsilon - I\left(\dfrac{r}{2}\right)$ . Comparing with $V = \varepsilon_{eff} - Ir_{eff}$ shows immediately that $\varepsilon_{eff} = \varepsilon$ (unchanged) and $r_{eff} = r/2$ for two cells.

Step 5. Generalise to $n$ identical cells by the same argument. With $n$ branches each obeying $I_k = (\varepsilon - V)/r$ , the total current is $I = n(\varepsilon - V)/r$ , which rearranges the same way to $V = \varepsilon - I(r/n)$ . So for $n$ identical cells in parallel, $\varepsilon_{eff} = \varepsilon$ and $r_{eff} = r/n$ — precisely the parallel-resistor rule applied to the internal resistances, but now derived rather than assumed, because the EMFs being identical and aligned is exactly what allows them to cancel out of the algebra and leave a pure parallel-resistance result.

Worked examples

Board level - series versus parallel current

Three identical cells, each of EMF $1.5\ \text{V}$ and internal resistance $0.2\ \Omega$ , are connected first in series and then in parallel, each time across the same external resistor of $5.4\ \Omega$ . Find the current delivered to $R$ in both cases, and state which wiring gives more current.

Step 1. Series case — combine the effective values. With $n = 3$ , $\varepsilon_{eff} = 3 \times 1.5\ \text{V} = 4.5\ \text{V}$ and $r_{eff} = 3 \times 0.2\ \Omega = 0.6\ \Omega$ .

Step 2. Series case — apply $I = \varepsilon_{eff}/(R+r_{eff})$ . $I_{series} = \dfrac{4.5}{5.4 + 0.6} = \dfrac{4.5}{6.0} = 0.75\ \text{A}$ .

Step 3. Parallel case — combine the effective values. With $n = 3$ identical cells, $\varepsilon_{eff} = 1.5\ \text{V}$ (unchanged) and $r_{eff} = 0.2/3 \approx 0.0667\ \Omega$ .

Step 4. Parallel case — apply the same formula. $I_{parallel} = \dfrac{1.5}{5.4 + 0.0667} = \dfrac{1.5}{5.467} \approx 0.274\ \text{A}$ .

Step 5. Compare. The series combination delivers the larger current, $0.75\ \text{A}$ , against the parallel combination's $\approx 0.274\ \text{A}$ — because here $R = 5.4\ \Omega$ is far larger than a single cell's $r = 0.2\ \Omega$ , exactly the regime in which boosting the EMF (series) beats shrinking the internal resistance (parallel).

Competitive level - choosing the wiring that maximises current

The same three cells (EMF $1.5\ \text{V}$ , internal resistance $0.2\ \Omega$ each) must now be connected — either all in series or all in parallel — across a much smaller external resistor, $R = 0.1\ \Omega$ . Find the current delivered by each wiring, identify which wiring wins this time, and explain the reversal from the board-level example.

Step 1. Series case with the new $R$ . Effective values are unchanged from before: $\varepsilon_{eff} = 4.5\ \text{V}$ , $r_{eff} = 0.6\ \Omega$ . Current: $I_{series} = \dfrac{4.5}{0.1 + 0.6} = \dfrac{4.5}{0.7} \approx 6.43\ \text{A}$ .

Step 2. Parallel case with the new $R$ . Effective values: $\varepsilon_{eff} = 1.5\ \text{V}$ , $r_{eff} = 0.2/3 \approx 0.0667\ \Omega$ . Current: $I_{parallel} = \dfrac{1.5}{0.1 + 0.0667} = \dfrac{1.5}{0.1667} \approx 9.00\ \text{A}$ .

Step 3. Compare and explain the reversal. This time the parallel combination wins, delivering $\approx 9.00\ \text{A}$ against the series combination's $\approx 6.43\ \text{A}$ , because now $R = 0.1\ \Omega$ is smaller than a single cell's internal resistance $r = 0.2\ \Omega$ — the total resistance in the loop is dominated by internal resistance, and it is the parallel wiring that shrinks that internal resistance ( $r/n$ ), while series wiring only makes it worse ( $nr$ ). The lesson generalises: compare $R$ against $r$ first, then pick the wiring, rather than assuming one arrangement is always "stronger." Timing note: in an exam, do not recompute $\varepsilon_{eff}$ and $r_{eff}$ from scratch each time — write both series and parallel effective-value pairs once at the top of your rough work, then just plug the given $R$ into $I=\varepsilon_{eff}/(R+r_{eff})$ for each; this two-line setup takes under a minute and prevents you from silently swapping the $nr$ and $r/n$ rules under time pressure.

Common traps

Assuming parallel wiring always gives more current than series — false whenever $R \gg r$ , as the board-level example shows; parallel only wins when $R$ is comparable to or smaller than $r$ .
Writing $r_{eff} = nr$ for a parallel combination — this is the series rule; the parallel rule for identical cells is $r_{eff} = r/n$ , the same shrinking pattern as resistors in parallel.
Forgetting that the parallel EMF rule $\varepsilon_{eff} = \varepsilon$ (unchanged) holds only because the cells are identical and aligned the same way; mismatched EMFs in parallel is a materially harder problem and not the standard case tested here.
Adding EMFs in series but forgetting internal resistance scales too — students often write $\varepsilon_{eff} = n\varepsilon$ correctly but then keep $r_{eff} = r$ , which overestimates the current a series bank can deliver.
Reversing one cell's polarity inside a series row (a common wiring mistake in a torch) subtracts that cell's EMF instead of adding it — always check every cell in a series row points the same way before summing EMFs.

You should be able to…

State and justify $\varepsilon_{eff} = n\varepsilon$ , $r_{eff} = nr$ for $n$ identical cells in series.
Derive (not just quote) $\varepsilon_{eff} = \varepsilon$ , $r_{eff} = r/n$ for $n$ identical cells in parallel, starting from the terminal-voltage equation of a single cell.
Compute the current delivered to a stated external resistor $R$ for both series and parallel groupings using $I = \varepsilon_{eff}/(R+r_{eff})$ .
Compare $R$ against a single cell's internal resistance $r$ to judge, before calculating, which wiring will deliver more current.
Extend the series and parallel rules to a mixed grouping of $n$ cells per row and $m$ parallel rows.
Spot the common wiring and formula mix-ups — reversed cell polarity, swapped $nr$ / $r/n$ rules, and unjustified use of the identical-EMF parallel shortcut.

Ask the tutor: Want to try a mixed series-parallel grouping problem, or see what happens when the parallel cells are not quite identical?

Why does connecting cells in parallel not just add up their currents automatically, the way series adds voltages?

Because in a parallel bank every cell sees the same pair of terminals and therefore the same terminal voltage $V$ , not a fixed share of the current. Each identical cell then contributes its own share $I_k = (\varepsilon - V)/r$ set by that shared voltage, and the total current is whatever those shares add up to once $V$ settles based on the external resistor $R$ . It genuinely does add up correctly (that is what the derivation on this page shows), but it happens through the terminal-voltage equation, not by simply declaring "current adds" the way EMFs simply add in series.

If I connect cells of different EMFs in parallel, can I still use e_eff = e and r_eff = r/n?

No, not directly — those two formulas assume every cell in the bank has the identical EMF $\varepsilon$ , which is exactly what let the $\varepsilon$ terms cancel cleanly in the derivation above. With unequal EMFs, the terminal voltage settles at a value where a stronger cell can actually push current backward into a weaker one, and you need to solve the full set of Kirchhoff equations for the branch currents rather than use a one-line shortcut. CBSE Class 12 keeps this topic to identical cells for exactly this reason, so treat mismatched-EMF parallel banks as beyond the standard syllabus problem set.

My torch has two cells in series and it still feels dimmer after a few hours — is the series/parallel formula wrong?

No, the formula is fine — what changes over hours of use is $r$ itself. A cell's internal resistance is not perfectly constant; it creeps upward as the chemicals inside get used up, especially as the cell nears the end of its charge. Plug a larger $r$ into $\varepsilon_{eff}=n\varepsilon$ , $r_{eff}=nr$ and $I=\varepsilon_{eff}/(R+r_{eff})$ and you get a smaller current and a dimmer bulb, exactly what you are observing — the two-cells-in-series wiring is still doing its job, it is just working with weaker cells than when they were fresh.

Is there ever a reason to mix series and parallel instead of picking one or the other?

Yes — that is exactly what the mixed-grouping formula on this page is for. A designer who needs both a higher voltage than one cell provides and the current-sharing benefit of parallel wiring will build several series rows and then bank those rows in parallel, giving $\varepsilon_{eff}=n\varepsilon$ (from each row's series length) and $r_{eff}=nr/m$ (from the m rows sharing the load in parallel). Real multi-cell battery packs, including the ones in laptops and power tools, are built this way for precisely that reason, though picking the single best $n$ and $m$ for a given job is a design problem that goes beyond what board exams ask you to compute.