Cells in Series and Parallel

Why combine cells at all?

A single dry cell gives about 1.5 V and can safely supply only a modest current — tens of milliamperes for a penlight cell, a few amperes for a D cell before its internal resistance starts to eat away the terminal voltage. Real circuits frequently need more voltage (a torch with a 6 V bulb, a toy that needs 9 V) or more current (a device that draws several amperes without heating up the source). This is where we combine cells.

Two combinations cover most exam problems:

• Series — cells are joined end-to-end, $+$ of one to $-$ of the next. Voltages add.
• Parallel — all $+$ terminals are tied together, all $-$ terminals are tied together. Current capacity adds; voltage stays the same.

A third setup — mixed grouping — combines the two for maximum power transfer at a given load. We'll cover all three, with full derivations using the relation $V = \varepsilon - Ir$ from Section 6.

Cells in Series — Aiding

Consider two cells with EMFs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ connected in series (positive of the first to negative of the second), supplying a current $I$ through an external resistance $R$.

The terminal voltage is

$$V_{AB} = (\varepsilon_1 + \varepsilon_2) - I(r_1 + r_2).$$

Comparing with $V = \varepsilon_{\text{eq}} - I r_{\text{eq}}$:

$$\varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2, \qquad r_{\text{eq}} = r_1 + r_2.$$

For $n$ identical cells each of EMF $\varepsilon$ and internal resistance $r$:

$$\varepsilon_{\text{eq}} = n\varepsilon, \qquad r_{\text{eq}} = nr.$$

When series is useful: when the external resistance $R$ is much larger than the internal resistance of one cell.

Series with cells opposing

If one cell is inserted in the opposite orientation, the EMFs subtract:

$$\varepsilon_{\text{eq}} = \varepsilon_1 - \varepsilon_2, \qquad r_{\text{eq}} = r_1 + r_2.$$

Internal resistances always add in series; only EMFs carry sign.

If $\varepsilon_1 < \varepsilon_2$, the net EMF is negative in the chosen direction, meaning the stronger cell drives current backward through the weaker one. The weaker one is being charged.

Sign convention trick: When walking around a loop, count an EMF as positive if you cross it from $-$ to $+$, and negative if you cross it from $+$ to $-$.

Cells in Parallel

Now connect two cells with the $+$ terminals tied together and the $-$ terminals tied together.

If $I_1$ and $I_2$ are the branch currents, the common terminal voltage is

$$V_{AB} = \varepsilon_1 - I_1 r_1 = \varepsilon_2 - I_2 r_2.$$

Also, total current is

$$I = I_1 + I_2.$$

Solving, the equivalent cell is:

$$\varepsilon_{\text{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}, \qquad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}.$$

Equivalently,

$$\frac{\varepsilon_{\text{eq}}}{r_{\text{eq}}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}.$$

[JEE Tip] For non-identical cells in parallel, the equivalent EMF is a weighted average, with smaller internal resistance having more influence.

Identical cells in parallel — a clean result

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

$$\varepsilon_{\text{eq}} = \varepsilon, \qquad r_{\text{eq}} = \frac{r}{m}.$$

So parallel combination keeps the voltage the same but reduces the effective internal resistance.

When parallel is useful: when the external resistance $R$ is much smaller than the internal resistance of a single cell.

Mixed grouping — best of both worlds

Suppose $N$ identical cells are arranged as $m$ rows in parallel, each row containing $n$ cells in series, so that $mn = N$.

Then:

$$\varepsilon_{\text{eq}} = n\varepsilon, \qquad r_{\text{eq}} = \frac{nr}{m}.$$

Current through an external resistance $R$ is

$$I = \frac{n\varepsilon}{R + \dfrac{nr}{m}} = \frac{N\varepsilon}{mR + nr}.$$

For maximum current (and maximum power transfer), choose the grouping so that

$$R = \frac{nr}{m} = r_{\text{eq}}.$$

[JEE Tip] Given $N$ cells and load $R$, test divisor pairs $(m,n)$ and choose the one for which $r_{\text{eq}}$ is closest to $R$.

Quick comparison and decision guide

Configuration	$\varepsilon_{\text{eq}}$	$r_{\text{eq}}$	When to use
$n$ identical cells in series	$n\varepsilon$	$nr$	$R \gg r$
$m$ identical cells in parallel	$\varepsilon$	$r/m$	$R \ll r$
Mixed grouping	$n\varepsilon$	$nr/m$	when $R \approx nr/m$

Rule of thumb:

• Need more voltage → use series.
• Need more current capacity → use parallel.
• Need maximum power for a fixed load → use mixed grouping with $R \approx r_{\text{eq}}$.

Common traps to avoid

1. In series, EMFs add algebraically but internal resistances always add positively.
2. In parallel, the simple result $\varepsilon_{\text{eq}} = \varepsilon$ holds only for identical cells.
3. Terminal voltage is not the same as EMF when current flows.
4. Maximum current and maximum efficiency are not the same thing.
5. For mixed grouping, always use integer divisor pairs of the total number of cells.

Example 1: Three identical cells in series

Three identical cells, each of EMF 1.5 V and internal resistance 0.4 Ω, are connected in series with an external resistance of 3.0 Ω. Find the current and terminal voltage of the combination.

Solution: For 3 identical cells in series, $$\varepsilon_{\text{eq}} = 3 \times 1.5 = 4.5\,\text{V}, \qquad r_{\text{eq}} = 3 \times 0.4 = 1.2\,\Omega.$$

Current: $$I = \frac{4.5}{3.0 + 1.2} = \frac{4.5}{4.2} \approx 1.07\,\text{A}.$$

Terminal voltage: $$V = IR = 1.07 \times 3.0 \approx 3.21\,\text{V}.$$

Answer: $I \approx 1.07$ A, $V \approx 3.21$ V.

Example 2: One cell reversed in series

Two cells, $\varepsilon_1 = 2.0$ V, $r_1 = 0.5\,\Omega$ and $\varepsilon_2 = 1.5$ V, $r_2 = 0.3\,\Omega$, are connected in series with the second cell reversed across an external resistance of 2.2 Ω. Find the current and state whether cell 2 is charging or discharging.

Solution: For opposing connection, $$\varepsilon_{\text{eq}} = 2.0 - 1.5 = 0.5\,\text{V}, \qquad r_{\text{eq}} = 0.5 + 0.3 = 0.8\,\Omega.$$

Thus, $$I = \frac{0.5}{2.2 + 0.8} = \frac{0.5}{3.0} \approx 0.167\,\text{A}.$$

The stronger 2.0 V cell drives current into the positive terminal of cell 2, so cell 2 is being charged.

Answer: $I \approx 0.167$ A; cell 2 is charging.

Example 3: Three identical cells in parallel

Three identical cells, each of EMF 2 V and internal resistance 0.6 Ω, are connected in parallel across a 1.0 Ω resistor. Find the current through the resistor and through each cell.

Solution: For identical cells in parallel, $$\varepsilon_{\text{eq}} = 2\,\text{V}, \qquad r_{\text{eq}} = \frac{0.6}{3} = 0.2\,\Omega.$$

Current through load: $$I = \frac{2}{1.0 + 0.2} = \frac{2}{1.2} \approx 1.67\,\text{A}.$$

By symmetry, each cell supplies: $$\frac{I}{3} \approx \frac{1.67}{3} \approx 0.56\,\text{A}.$$

Answer: Load current $\approx 1.67$ A; current from each cell $\approx 0.56$ A.

Example 4: Two non-identical cells in parallel

Two cells in parallel: $\varepsilon_1 = 6$ V, $r_1 = 2\,\Omega$, and $\varepsilon_2 = 4$ V, $r_2 = 1\,\Omega$. Find the equivalent EMF and internal resistance, and the current through an external 3 Ω resistor.

Solution: Equivalent EMF: $$\varepsilon_{\text{eq}} = \frac{6 \times 1 + 4 \times 2}{2 + 1} = \frac{14}{3} \approx 4.67\,\text{V}.$$

Equivalent internal resistance: $$r_{\text{eq}} = \frac{2 \times 1}{2 + 1} = \frac{2}{3} \approx 0.67\,\Omega.$$

Current through the load: $$I = \frac{4.67}{3 + 0.67} \approx 1.27\,\text{A}.$$

Answer: $\varepsilon_{\text{eq}} \approx 4.67$ V, $r_{\text{eq}} \approx 0.67\,\Omega$, $I \approx 1.27$ A.

Example 5: Mixed grouping — 12 cells

12 identical cells ($\varepsilon = 2$ V, $r = 0.5\,\Omega$ each) are arranged as $m = 3$ rows in parallel, each row having $n = 4$ cells in series. Find the equivalent EMF, internal resistance, and the current through an external $R = 5\,\Omega$.

Solution: $$\varepsilon_{\text{eq}} = n\varepsilon = 4 \times 2 = 8\,\text{V}$$ $$r_{\text{eq}} = \frac{nr}{m} = \frac{4 \times 0.5}{3} \approx 0.67\,\Omega$$

Current: $$I = \frac{8}{5 + 0.67} \approx 1.41\,\text{A}.$$

Answer: $\varepsilon_{\text{eq}} = 8$ V, $r_{\text{eq}} \approx 0.67\,\Omega$, $I \approx 1.41$ A.

Example 6: Find the optimum grouping

You have 24 identical cells, each $\varepsilon = 1.5$ V, $r = 0.3\,\Omega$. You want the maximum current through an external resistance $R = 0.5\,\Omega$. What is the best arrangement $(m,n)$, and what current does it deliver?

Solution: Optimization condition: $$R = \frac{nr}{m} \Rightarrow 0.5 = \frac{0.3n}{m} \Rightarrow \frac{n}{m} = \frac{5}{3}.$$

With $mn = 24$, the nearest integer divisor pair is $(m,n) = (4,6)$.

Then, $$r_{\text{eq}} = \frac{6 \times 0.3}{4} = 0.45\,\Omega,$$ which is close to $R = 0.5\,\Omega$.

Current: $$I = \frac{24 \times 1.5}{4 \times 0.5 + 6 \times 0.3} = \frac{36}{3.8} \approx 9.47\,\text{A}.$$

Answer: Best grouping is 4 rows in parallel, each with 6 cells in series; $I \approx 9.47$ A.

Example 7: Power efficiency in series vs parallel

Four identical cells ($\varepsilon = 1.2$ V, $r = 0.5\,\Omega$) power a 1 Ω resistor. Compute the current, the power delivered to the resistor, and the efficiency in each arrangement: (a) all four in series, (b) all four in parallel.

Solution: (a) Series: $$\varepsilon_{\text{eq}} = 4.8\,\text{V}, \qquad r_{\text{eq}} = 2.0\,\Omega$$ $$I_s = \frac{4.8}{1 + 2} = 1.6\,\text{A}$$ $$P_R = I_s^2 R = 1.6^2 \times 1 = 2.56\,\text{W}$$ $$\eta_s = \frac{R}{R + r_{\text{eq}}} = \frac{1}{3} \approx 33\%$$

(b) Parallel: $$\varepsilon_{\text{eq}} = 1.2\,\text{V}, \qquad r_{\text{eq}} = \frac{0.5}{4} = 0.125\,\Omega$$ $$I_p = \frac{1.2}{1 + 0.125} \approx 1.07\,\text{A}$$ $$P_R = I_p^2 R \approx 1.14\,\text{W}$$ $$\eta_p = \frac{1}{1.125} \approx 89\%$$

Answer:

• Series: $I = 1.6$ A, $P_R = 2.56$ W, $\eta \approx 33\%$
• Parallel: $I \approx 1.07$ A, $P_R \approx 1.14$ W, $\eta \approx 89\%$

Example 8: Terminal voltage vs EMF

Six identical cells ($\varepsilon = 1.5$ V, $r = 0.2\,\Omega$) are connected to a 3 Ω resistor. Find the terminal voltage across the combination in (a) series and (b) parallel.

Solution: (a) Series: $$\varepsilon_{\text{eq}} = 9\,\text{V}, \qquad r_{\text{eq}} = 1.2\,\Omega$$ $$I = \frac{9}{3 + 1.2} = 2.14\,\text{A}$$ $$V = IR = 2.14 \times 3 = 6.43\,\text{V}$$

(b) Parallel: $$\varepsilon_{\text{eq}} = 1.5\,\text{V}, \qquad r_{\text{eq}} = \frac{0.2}{6} \approx 0.033\,\Omega$$ $$I = \frac{1.5}{3 + 0.033} \approx 0.495\,\text{A}$$ $$V = IR = 0.495 \times 3 = 1.48\,\text{V}$$

Answer:

• Series: $V \approx 6.43$ V
• Parallel: $V \approx 1.48$ V

Example 9: Two different EMFs in parallel with a common load

Two cells, $\varepsilon_1 = 10$ V, $r_1 = 1\,\Omega$ and $\varepsilon_2 = 8$ V, $r_2 = 2\,\Omega$, are connected in parallel across $R = 3\,\Omega$. Find the current in each branch.

Solution: Let $V$ be the common terminal voltage.

$$I_1 = \frac{10 - V}{1}, \qquad I_2 = \frac{8 - V}{2}, \qquad \frac{V}{3} = I_1 + I_2.$$

So, $$\frac{V}{3} = (10 - V) + \frac{8 - V}{2}.$$

Multiplying by 6: $$2V = 60 - 6V + 24 - 3V$$ $$11V = 84 \Rightarrow V = \frac{84}{11} \approx 7.64\,\text{V}.$$

Thus, $$I_1 = 10 - 7.64 = 2.36\,\text{A}$$ $$I_2 = \frac{8 - 7.64}{2} = 0.18\,\text{A}$$

Answer: $I_1 \approx 2.36$ A, $I_2 \approx 0.18$ A.

Example 10: Maximum current from given cells

A device needs the maximum possible current from $N = 20$ identical cells ($\varepsilon = 2$ V, $r = 0.8\,\Omega$) delivered into a load $R = 2\,\Omega$. What grouping should be used, and what is that maximum current?

Solution: Optimization condition: $$\frac{m}{n} = \frac{r}{R} = \frac{0.8}{2} = 0.4, \qquad mn = 20.$$

Two nearby divisor pairs are $(m,n) = (2,10)$ and $(4,5)$.

For $(2,10)$: $$I = \frac{20 \times 2}{2 \times 2 + 10 \times 0.8} = \frac{40}{12} = 3.33\,\text{A}.$$

For $(4,5)$: $$I = \frac{20 \times 2}{4 \times 2 + 5 \times 0.8} = \frac{40}{12} = 3.33\,\text{A}.$$

Both give the same current.

Answer: Either $(m,n) = (2,10)$ or $(4,5)$; maximum current is $3.33$ A.

Example 11: Deciding series vs parallel

You have 4 cells ($\varepsilon = 1.5$ V, $r = 1.0\,\Omega$ each) and a load $R = 0.1\,\Omega$. Should you connect all in series, all in parallel, or mixed, to maximize current?

Solution: Check all three possibilities.

All in series: $$\varepsilon_{\text{eq}} = 6\,\text{V}, \qquad r_{\text{eq}} = 4\,\Omega$$ $$I = \frac{6}{4.1} \approx 1.46\,\text{A}$$

All in parallel: $$\varepsilon_{\text{eq}} = 1.5\,\text{V}, \qquad r_{\text{eq}} = 0.25\,\Omega$$ $$I = \frac{1.5}{0.35} \approx 4.29\,\text{A}$$

Mixed (2 series in each row, 2 rows in parallel): $$\varepsilon_{\text{eq}} = 3\,\text{V}, \qquad r_{\text{eq}} = 1\,\Omega$$ $$I = \frac{3}{1.1} \approx 2.73\,\text{A}$$

Answer: All parallel gives the maximum current.

Example 12: Parallel cells where one cell charges the other

In a parallel combination, cell 1 has $\varepsilon_1 = 6$ V, $r_1 = 0.5\,\Omega$ and cell 2 has $\varepsilon_2 = 4$ V, $r_2 = 0.5\,\Omega$, connected across a 2 Ω resistor. Find the current through the load and through each cell.

Solution: Let the common terminal voltage be $V$.

$$I_1 = \frac{6 - V}{0.5} = 2(6 - V)$$ $$I_2 = \frac{4 - V}{0.5} = 2(4 - V)$$ $$I_{\text{load}} = \frac{V}{2}$$

Using KCL: $$I_1 + I_2 = I_{\text{load}}$$ $$2(6 - V) + 2(4 - V) = \frac{V}{2}$$ $$20 - 4V = \frac{V}{2}$$ $$40 - 8V = V$$ $$9V = 40 \Rightarrow V = \frac{40}{9} \approx 4.44\,\text{V}$$

Then, $$I_1 = 2(6 - 4.44) = 3.11\,\text{A}$$ $$I_2 = 2(4 - 4.44) = -0.89\,\text{A}$$ $$I_{\text{load}} = \frac{4.44}{2} = 2.22\,\text{A}$$

The negative sign for $I_2$ means cell 2 is being charged.

Answer: $I_{\text{load}} \approx 2.22$ A, $I_1 \approx 3.11$ A, $I_2 \approx -0.89$ A (cell 2 charges).