Internal resistance and EMF — cell combinations in series and parallel

Question

How do internal resistance and EMF affect the terminal voltage of a cell? How do we combine cells in series and parallel to get maximum current through an external resistance?

(CBSE 12, JEE Main, NEET — cell combinations and internal resistance are tested in numericals every year)

Solution — Step by Step

Every real cell has an internal resistance ( $r$ ) — resistance of the electrolyte and electrodes inside the cell.

V = \varepsilon - Ir

where $\varepsilon$ = EMF (open circuit voltage), $I$ = current flowing, $r$ = internal resistance.

When current flows, some voltage is “lost” inside the cell. The terminal voltage $V$ is always less than $\varepsilon$ during discharge. When $I = 0$ (open circuit), $V = \varepsilon$ — this is how we measure EMF.

Current through the circuit: $I = \frac{\varepsilon}{R + r}$ , where $R$ = external resistance.

$n$ identical cells, each with EMF $\varepsilon$ and internal resistance $r$ , connected in series with external resistance $R$ :

\varepsilon_{eq} = n\varepsilon, \quad r_{eq} = nr

I = \frac{n\varepsilon}{R + nr}

Use series when: External resistance $R$ is large ( $R \gg r$ ). Then $I \approx n\varepsilon / R$ — series connection multiplies the EMF.

If $R \ll r$ : $I \approx \varepsilon / r$ — series connection gives no advantage (internal resistance dominates).

$m$ identical cells, each with EMF $\varepsilon$ and internal resistance $r$ , connected in parallel with external resistance $R$ :

\varepsilon_{eq} = \varepsilon, \quad r_{eq} = \frac{r}{m}

I = \frac{\varepsilon}{R + r/m}

Use parallel when: External resistance $R$ is small ( $R \ll r$ ). Then $I \approx m\varepsilon / r$ — parallel connection reduces internal resistance, allowing more current.

If $R \gg r$ : $I \approx \varepsilon / R$ — parallel connection gives no advantage.

For $n$ cells in series per row and $m$ rows in parallel:

I = \frac{n\varepsilon}{R + nr/m}

Maximum current flows when internal resistance matches external resistance:

R = \frac{nr}{m}

This is the condition for maximum power transfer — a fundamental result in circuit theory.

flowchart TD
    A["Cell combination problem"] --> B{"Compare R with r"}
    B -->|"R >> r (large external resistance)"| C["Series connection<br/>I = nε/(R + nr) ≈ nε/R"]
    B -->|"R << r (small external resistance)"| D["Parallel connection<br/>I = ε/(R + r/m) ≈ mε/r"]
    B -->|"R ≈ r"| E["Mixed grouping<br/>Match R = nr/m for max current"]

Why This Works

Series connection adds EMFs (voltage boost) at the cost of adding internal resistances. Parallel connection keeps the same EMF but divides internal resistance (current capacity boost). The optimal choice depends on the external load:

High-resistance load needs more voltage (series)
Low-resistance load needs more current capacity (parallel)
Intermediate load benefits from a mix that matches impedances

The maximum power transfer theorem ( $R = r_{eq}$ ) follows from calculus — differentiating power with respect to $R$ and setting it to zero gives the matching condition.

Common Mistake

Students assume that connecting more cells always gives more current. If the external resistance is large and you connect cells in parallel, the current barely increases (since $r/m \ll R$ , the denominator hardly changes). Similarly, if $R$ is small and you add cells in series, the added internal resistance eats up the extra EMF. Always compare $R$ with $r$ before choosing the connection type. JEE Main 2024 tested exactly this conceptual distinction.

For maximum current problems, remember the matching rule: total internal resistance = external resistance. If you have $N$ total cells and can arrange them as $n$ in series and $m = N/n$ in parallel, choose $n$ and $m$ such that $nr/m = R$ , i.e., $n/m = R/r$ , i.e., $n^2 = NR/r$ .

Question

Solution — Step by Step

Why This Works

Common Mistake

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