10 min readConcept sheetA 3-mark derivation of equivalent emf and internal resistance, and 2 and 3-mark numericals

Combination of cells in series and in parallel

In series, emfs and internal resistances add. In parallel, ε_eq/r_eq = ε₁/r₁ + ε₂/r₂ and 1/r_eq = 1/r₁ + 1/r₂.

Two cells in seriesε_eq = ε₁ + ε₂ (ε₁ − ε₂ if one is reversed); r_eq = r₁ + r₂

Two cells in parallelε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂); r_eq = r₁r₂/(r₁ + r₂)

Parallel, in short1/r_eq = 1/r₁ + 1/r₂ and ε_eq/r_eq = ε₁/r₁ + ε₂/r₂

n identical cells (ε, r) with a load RSeries: I = nε/(R + nr). Parallel: I = ε/(R + r/n)

Derivation: two cells in series

Cell 1 (ε₁, r₁) has its positive terminal at A and its negative terminal at B. At B it is joined to the positive terminal of cell 2 (ε₂, r₂), whose negative terminal is at C. A current I flows.

1
PD across cell 1: V(A) − V(B) = ε₁ − Ir₁.
2
PD across cell 2: V(B) − V(C) = ε₂ − Ir₂.
3
Add: V(A) − V(C) = (ε₁ + ε₂) − I(r₁ + r₂).
4
A single equivalent cell would give V(A) − V(C) = ε_eq − Ir_eq.
5
Compare: ε_eq = ε₁ + ε₂ and r_eq = r₁ + r₂.
6
If cell 2 is joined the other way round, its emf opposes: ε_eq = ε₁ − ε₂, but r_eq is still r₁ + r₂.

Derivation: two cells in parallel

Cells (ε₁, r₁) and (ε₂, r₂) have their positive terminals joined at B₁ and their negative terminals at B₂. They send currents I₁ and I₂ out of B₁, so the external current is I = I₁ + I₂. Let V be the PD between B₁ and B₂.

1
Cell 1: V = ε₁ − I₁r₁, so I₁ = (ε₁ − V)/r₁.
2
Cell 2: V = ε₂ − I₂r₂, so I₂ = (ε₂ − V)/r₂.
3
Add: I = I₁ + I₂ = (ε₁/r₁ + ε₂/r₂) − V(1/r₁ + 1/r₂).
4
Solve for V: V = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) − I r₁r₂/(r₁ + r₂).
5
Compare with a single cell: V = ε_eq − Ir_eq.
6★
So, ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) and r_eq = r₁r₂/(r₁ + r₂). In short: 1/r_eq = 1/r₁ + 1/r₂ and ε_eq/r_eq = ε₁/r₁ + ε₂/r₂.

★ key step

Series or parallel: which gives more current?

Series groupingThe emfs add, so it gives a larger emf. The internal resistances also add. It is best when the external resistance is large compared with r. For n identical cells: I = nε/(R + nr).

Parallel groupingIdentical cells give the same emf but a smaller internal resistance, r/n. It is best when the external resistance is small compared with r. For n identical cells: I = ε/(R + r/n).

Check: when R is much larger than nr, series gives about nε/R, n times one cell. When R is much smaller than r/n, parallel gives about nε/r.

Worked example: two unlike cells, in series and in parallel

Problem

Cell 1 has ε₁ = 2.0 V, r₁ = 1.0 Ω. Cell 2 has ε₂ = 1.0 V, r₂ = 0.50 Ω. Find the current in two cases. (a) They are joined in series, supporting each other, across a 4.5 Ω resistor. (b) They are joined in parallel, like terminals together, across a 0.50 Ω resistor. In (b), also find the current from each cell.

Working

(a) Series: ε_eq = 2.0 V + 1.0 V = 3.0 V; r_eq = 1.0 Ω + 0.50 Ω = 1.5 Ω.
I = ε_eq/(R + r_eq) = 3.0 V ÷ (4.5 Ω + 1.5 Ω) = 3.0 V ÷ 6.0 Ω = 0.50 A.
(b) Parallel: ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) = (2.0 × 0.50 + 1.0 × 1.0) ÷ (1.0 + 0.50) = 2.0 ÷ 1.5 = 4/3 V ≈ 1.33 V.
r_eq = r₁r₂/(r₁ + r₂) = (1.0 × 0.50) ÷ 1.5 = 1/3 Ω ≈ 0.33 Ω.
I = ε_eq/(R + r_eq) = (4/3) ÷ (0.50 + 1/3) = (4/3) ÷ (5/6) = 8/5 = 1.6 A.
Terminal voltage V = IR = 1.6 A × 0.50 Ω = 0.80 V.
I₁ = (2.0 − 0.80) ÷ 1.0 = 1.2 A; I₂ = (1.0 − 0.80) ÷ 0.50 = 0.40 A. Check: 1.2 A + 0.40 A = 1.6 A.

Answer

(a) 0.50 A (ε_eq = 3.0 V, r_eq = 1.5 Ω). (b) 1.6 A (ε_eq ≈ 1.33 V, r_eq ≈ 0.33 Ω), shared as 1.2 A from cell 1 and 0.40 A from cell 2.

Worked example: a cell joined the wrong way

Problem

Four cells, each of emf 1.5 V and internal resistance 0.25 Ω, are meant to be joined in series across a 4.0 Ω bulb. By mistake one cell is reversed. Find the current, and compare it with the current when all are joined correctly.

Working

All correct: ε_eq = 4 × 1.5 V = 6.0 V; r_eq = 4 × 0.25 Ω = 1.0 Ω.
I = 6.0 V ÷ (4.0 Ω + 1.0 Ω) = 1.2 A.
One reversed: it cancels one forward cell, so ε_eq = (3 − 1) × 1.5 V = 3.0 V.
The internal resistances still add: r_eq = 1.0 Ω.
I = 3.0 V ÷ (4.0 Ω + 1.0 Ω) = 0.60 A.

Answer

0.60 A with one cell reversed, half of the correct value of 1.2 A

How to draw the cell combinations

Draw each cell as a long thin line (+) and a short thick line (−). Draw its internal resistance r as a small resistor beside it.
Series: cells in one line, the − of the first touching the + of the next. Label the points A, B and C, and ε₁, r₁, ε₂, r₂.
Parallel: two cells side by side, + terminals joined at B₁ and − terminals at B₂. Label I₁ and I₂ in the cells and I = I₁ + I₂ in the external wire.
Join the combination to a resistor R and show the current direction with arrows.

In real life

A TV remote or torch holds two 1.5 V cells in series to get 3 V. Home inverter and solar setups join batteries in series for a higher voltage, and in parallel for a larger current and a longer backup.

Exam angle

3 marks: “Two cells of emfs ε₁ and ε₂ and internal resistances r₁ and r₂ are connected in parallel, with like terminals together. Derive the expressions for the equivalent emf and internal resistance.” Write V = ε₁ − I₁r₁ and V = ε₂ − I₂r₂; add I = I₁ + I₂; solve for V and compare with V = ε_eq − Ir_eq. Result: ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂), r_eq = r₁r₂/(r₁ + r₂).

2 marks: “n identical cells, each of emf ε and internal resistance r, are joined in series across a resistor R. If one cell is joined the wrong way, find the current.” The reversed cell cancels one forward cell, so ε_eq = (n − 2)ε. The internal resistances still add to nr. I = (n − 2)ε/(R + nr).

1 marks: “Two identical cells, each of emf ε and internal resistance r, are joined in parallel. Write the equivalent emf and internal resistance.” ε_eq = ε and r_eq = r/2.

Remember it

Series: everything adds (ε and r). Parallel: add the 1/r terms, and ε_eq/r_eq = Σ ε/r.