Cells in series and parallel
n identical cells in series give nE and nr; in parallel they give E and r/n. Four 1.5 V cells of 0.5 Ω in series drive 6 / (4 + 2) = 1 A through 4 Ω; in parallel only 1.5 / (4 + 0.125) = 0.36 A.
- Four in series1 AE = 6 V, r = 2 Ω
- Four in parallel0.36 AE = 1.5 V, r = 0.125 Ω
- One cell0.33 AE = 1.5 V, r = 0.5 Ω
Four cells in series on 4 Ω
Tap a resistor to choose it, then change its value or add another in series or in parallel with it. The glowing dots are the current: faster where more flows. Brighter resistors turn more power into heat.
Equivalent resistance4 Ωbetween A and B
Current1 Aterminal 4 V
Power6 WP = VI
| R | I | V | Power | |
|---|---|---|---|---|
| R₁ | 4 Ω | 1 A | 4 V | 4 W |
Step by step
- One resistor, so Req = 4 Ω
- Current from the cell: I = E / (Req + r) = 6 / (4 + 2) = 1 A
- Terminal voltage: V = E - Ir = 6 - 1 × 2 = 4 V
- Power: P = EI = 6 × 1 = 6 W, of which 2 W heats the cell.
Why it works
- In series the emfs add when + meets -: E = E₁ + E₂, and the internal resistances add: r = r₁ + r₂.
- If one cell is turned round, its emf subtracts: E = E₁ - E₂.
- Two cells in parallel give E = (E₁r₂ + E₂r₁)/(r₁ + r₂) and 1/r = 1/r₁ + 1/r₂.
- For n identical cells in parallel this becomes E and r/n.
- Series gives more emf, best when the load R is much larger than r. Parallel gives less internal resistance, best when R is small.
Cells in series and parallel: quick answers
What is the emf of cells in series?
The sum of their emfs, with a reversed cell counted as negative. The internal resistances add too.
What is the emf of identical cells in parallel?
The same as one cell, E, but the internal resistance falls to r/n, so they can give a larger current to a small load.
Which arrangement gives more current here?
Series: 1 A through 4 Ω, against 0.36 A in parallel, because the 4 Ω load is much larger than the 0.5 Ω of each cell.