Cells in series and in parallel

How do cells combine?

In plain words: stacking cells end to end, + to −, gives more push (more emf) but also more internal resistance. Putting them side by side, + to +, keeps the push about the same but shares the current, so the internal resistance drops.

Cells in series

Two cells of emfs ε₁, ε₂ and internal resistances r₁, r₂ are joined with the negative terminal of the first to the positive terminal of the second, at B (NCERT page 95, Figure 3.13). With a current I:

• Across the first cell, V_AB = ε₁ − Ir₁; across the second, V_BC = ε₂ − Ir₂ (NCERT Eqs. (3.41)–(3.42), page 95).
• Adding, V_AC = (ε₁ + ε₂) − I(r₁ + r₂) (NCERT Eq. (3.43)). Comparing with one cell, V_AC = ε_eq − Ir_eq (NCERT Eq. (3.44)):

• If instead the two negative terminals are joined, V_BC = −ε₂ − Ir₂, and ε_eq = ε₁ − ε₂ for ε₁ > ε₂ (NCERT Eq. (3.47), page 96).
• The reversed cell’s internal resistance still adds, since the −Ir₂ term is unchanged: r_eq = r₁ + r₂. So n identical cells in series with m of them reversed give ε_eq = (n − 2m)ε and r_eq = nr (reasoned from NCERT page 96).
• For n cells in series the equivalent emf is the sum of the emfs and the equivalent internal resistance the sum of the internal resistances, when the current leaves each cell from its positive electrode. A cell that the current leaves from its negative electrode enters ε_eq with a minus sign (page 96).

Cells in parallel

Two cells are joined with their positive terminals together at B₁ and their negative terminals at B₂ (NCERT page 96, Figure 3.14). I₁ and I₂ leave their positive electrodes:

• At B₁, as much charge flows in as out (conservation of charge): I = I₁ + I₂ (NCERT Eq. (3.48), page 96).
• Both cells have the same terminal potential difference V: V = ε₁ − I₁r₁ = ε₂ − I₂r₂ (NCERT Eqs. (3.49)–(3.50)).
• So I₁ = (ε₁ − V)/r₁ and I₂ = (ε₂ − V)/r₂. Adding them, I = ε₁/r₁ + ε₂/r₂ − V/r₁ − V/r₂ (NCERT Eq. (3.51), page 96).
• Solve this for V and compare with one cell, V = ε_eq − Ir_eq (NCERT Eqs. (3.52)–(3.53)). This gives (NCERT Eqs. (3.54)–(3.55)):

Equivalent emf and internal resistance

NCERT writes the parallel results more simply (NCERT Eqs. (3.56)–(3.57), page 97):

• If the negative terminal of the second cell is joined to the positive terminal of the first, the same equations hold with ε₂ replaced by −ε₂ (page 97).
• For n cells in parallel, 1/r_eq = 1/r₁ + … + 1/r_n and ε_eq/r_eq = ε₁/r₁ + … + ε_n/r_n (NCERT Eqs. (3.58)–(3.59), page 97).
• Identical cells (emf ε, internal resistance r): n in series give nε and nr; n in parallel give ε and r/n (reasoned from NCERT Eqs. (3.45)–(3.46) and (3.58)–(3.59)).
• Mixed grouping: m rows in parallel, each of n identical cells in series, act as one cell of emf nε and internal resistance nr/m, so I = nε/(R + nr/m). With N = mn fixed, treating n as continuous gives the stationary optimum nr/m = R. Actual n and m must be positive integer factors of N: compare the allowed groupings, including all-series and all-parallel boundaries. For N = 6, r = 1 Ω and R = 100 Ω, all six in series give the greatest allowed current even though their 6 Ω internal resistance cannot equal R (reasoned from the same equations).

Solved examples

Common mistakes to avoid

Quick revision summary

Parallel formulas assume positive internal resistances and signed emfs referred to the same terminal orientation. A negative calculated branch current means current enters that cell’s positive terminal.

• A combination of cells acts as one cell of emf ε_eq and internal resistance r_eq.
• Series (+ to −): ε_eq = ε₁ + ε₂ + …, r_eq = r₁ + r₂ + …; a reversed cell’s emf enters with a minus sign, but its internal resistance still adds.
• Parallel (+ to +): 1/r_eq = 1/r₁ + 1/r₂ + …, ε_eq/r_eq = ε₁/r₁ + ε₂/r₂ + ….
• n identical cells: series nε and nr; parallel ε and r/n.
• External resistance below r: all in parallel gives more current; above r: all in series does; at R = r they are equal.

Questions students ask

Why are cells connected in parallel, and does it give more current?

Cells are connected in parallel to lower the internal resistance and share the load, not to raise the emf: n identical cells in parallel act as one cell of emf ε and internal resistance r/n (reasoned from NCERT Eqs. (3.58)–(3.59), page 97), and each supplies only 1/n of the current, so at a fixed total current their combined internal heating is lower. With a fixed load, the total current rises, and total internal heating can increase even while efficiency improves. The current rises noticeably only when the external resistance is not much larger than r: with three 1.2 V cells of 0.30 Ω each, a 0.20 Ω load takes 2.4 A from one cell but 4.0 A from all three in parallel, while a 10 Ω load takes about 0.117 A and 0.119 A.

What happens when two cells of different emf are joined in parallel?

When two cells of different emf are joined in parallel, positive to positive, a current circulates round them even with no load, out of the higher-emf cell and into the other, in the charging direction (this does not mean a primary cell is rechargeable), with I = (ε₁ − ε₂)/(r₁ + r₂) (beyond the NCERT text, reasoned from NCERT Eqs. (3.48)–(3.50), page 96). For cells of 2.1 V, 0.35 Ω and 1.9 V, 0.15 Ω this is 0.20/0.50 = 0.40 A, and both terminals sit at 2.1 − 0.40 × 0.35 = 1.96 V, the ε_eq of NCERT Eq. (3.57). The circulating current wastes energy in both cells all the time, and equal emfs remove this circulating current in the ideal model.

Does adding another cell in series always increase the current?

No, adding a cell in series does not always raise the current, because the cell adds its internal resistance as well as its emf (NCERT Eqs. (3.45)–(3.46), page 95): beyond the NCERT text, reasoned from those equations, the current rises only if the new cell’s own short-circuit current, ε/r, is larger than the current already flowing. Two 1.5 V cells of 0.25 Ω drive 3.0/(4.5 + 0.50) = 0.60 A through a 4.5 Ω bulb, but adding a cell of 1.3 V and 6.0 Ω, whose ε/r is only about 0.22 A, cuts the current to 4.3/11.0 ≈ 0.39 A. So a cell with a high internal resistance should not be added to a series chain of good ones.

Is mixed grouping of cells in the JEE Main and NEET syllabus?

Mixed grouping of cells is not named in either syllabus, but it is covered by what they name: the JEE Main syllabus and the NEET (UG) 2026 syllabus both list “a combination of cells in series and parallel” in Unit 12, and a mixed grouping, rows of cells in series that are then joined in parallel, needs only the two NCERT rules applied one after the other (NCERT Eqs. (3.45)–(3.46) and (3.58)–(3.59), pages 95–97). NCERT Class 12 Physics Part I does not work an example of it; this page gives the result and the condition for the largest current (see mixed grouping).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Three cells, each of emf 2.0 V and internal resistance 0.20 Ω, are in series. Find the equivalent emf and internal resistance.

   Show answerHide answer

   ε_eq = 6.0 V and r_eq = 0.60 Ω (NCERT Eqs. (3.45)–(3.46), page 95).

2. Q2. Two identical cells of emf 1.5 V and internal resistance 1.0 Ω are joined in parallel. Find the equivalent cell.

   Show answerHide answer

   ε_eq = 1.5 V and r_eq = 0.50 Ω (NCERT Eqs. (3.56)–(3.57), page 97).

3. Q3. What does it mean for a cell to enter ε_eq with a minus sign?

   Show answerHide answer

   The current leaves that cell from its negative electrode, as when two negative terminals are joined in series (NCERT page 96).