Internal resistance and EMF — cell combinations in series and parallel

Question

Two cells of EMF $\varepsilon_1 = 2$ V and $\varepsilon_2 = 4$ V with internal resistances $r_1 = 1\,\Omega$ and $r_2 = 2\,\Omega$ are connected in series with an external resistance $R = 7\,\Omega$ . Find the current in the circuit and the terminal voltage of each cell.

(CBSE 12, JEE Main & NEET — appears almost every year)

Solution — Step by Step

When cells are in series (aiding — positive terminal of one to negative of next):

\varepsilon_{\text{total}} = \varepsilon_1 + \varepsilon_2 = 2 + 4 = 6 \text{ V}

r_{\text{total}} = r_1 + r_2 = 1 + 2 = 3\,\Omega

I = \frac{\varepsilon_{\text{total}}}{R + r_{\text{total}}} = \frac{6}{7 + 3} = \frac{6}{10} = \mathbf{0.6 \text{ A}}

Terminal voltage = EMF minus voltage drop across internal resistance:

V_1 = \varepsilon_1 - Ir_1 = 2 - 0.6 \times 1 = \mathbf{1.4 \text{ V}}

V_2 = \varepsilon_2 - Ir_2 = 4 - 0.6 \times 2 = \mathbf{2.8 \text{ V}}

Check: $V_1 + V_2 = 1.4 + 2.8 = 4.2$ V, and $IR = 0.6 \times 7 = 4.2$ V. Consistent.

Why This Works

Every real cell has internal resistance — the electrolyte and electrodes resist current flow. The EMF is the maximum potential difference (when no current flows). Under load, some voltage drops inside the cell itself, reducing what’s available externally.

graph TD
    A["Cell Combination Problem"] --> B{"Series or Parallel?"}
    B -->|"Series"| C["ε_total = ε₁ + ε₂<br/>r_total = r₁ + r₂"]
    B -->|"Parallel (same EMF)"| D["ε_total = ε<br/>1/r_total = 1/r₁ + 1/r₂"]
    B -->|"Parallel (different EMF)"| E["Use Kirchhoff's laws<br/>or equivalent cell formula"]
    C --> F["I = ε_total / (R + r_total)"]
    D --> F
    E --> G["ε_eq = (ε₁/r₁ + ε₂/r₂) /<br/>(1/r₁ + 1/r₂)"]
    G --> F
    F --> H["V_terminal = ε - Ir"]

Series combination increases the total EMF — useful when you need higher voltage (like stacking batteries in a torch). Parallel combination keeps the same EMF but reduces internal resistance — useful when you need higher current capacity.

Alternative Method — Kirchhoff’s Loop Rule

Apply Kirchhoff’s voltage law around the loop: sum of all potential changes = 0.

Starting from one point and going around: $+\varepsilon_1 + \varepsilon_2 - Ir_1 - Ir_2 - IR = 0$

6 - I(1 + 2 + 7) = 0 \implies I = 0.6 \text{ A}

This method generalises to any circuit — even when cells oppose each other.

For NEET: when two cells of different EMFs are connected in parallel across a resistance, use the equivalent cell formula: $\varepsilon_{\text{eq}} = \dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$ and $r_{\text{eq}} = \dfrac{r_1 r_2}{r_1 + r_2}$ . This saves time compared to solving simultaneous Kirchhoff equations.

Common Mistake

When cells are connected in series opposing each other (positive to positive), the EMFs subtract: $\varepsilon_{\text{total}} = |\varepsilon_1 - \varepsilon_2|$ . Students forget to check the polarity and add them blindly. Always trace the circuit in the direction of current and note whether each EMF adds or opposes.

Question

Solution — Step by Step

Why This Works

Alternative Method — Kirchhoff’s Loop Rule

Common Mistake

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