Series, Parallel & Grouped Cells

R adds in series, 1/R in parallel; cells combine too.

10 minUpdated 22 October 2026series resistorsparallel resistorsgrouping of cells

Lesson at a glanceShow the key points

R_eq = V/i; series R = ΣRᵢ (32.13); parallel 1/R = Σ1/Rᵢ (32.14).
Parallel current divides inversely as resistance: i₁ = R₂i/(R₁ + R₂).
Cells in series: ℰ₁ + ℰ₂, r₁ + r₂ (|ℰ₁ − ℰ₂| if one is reversed).
Cells in parallel: ℰ₀ = (ℰ₁r₂ + ℰ₂r₁)/(r₁ + r₂), r₀ = r₁r₂/(r₁ + r₂).

Any two-terminal network of resistors acts like one resistor. Series and parallel rules handle most of them; the same ideas tell us how to group cells.

Equivalent resistance

Apply potential difference V across a network’s two end points; if it draws current i, its equivalent resistance is R_eq = V/i.

Series

Resistors are in series when the same current passes through each. Round the loop: iR₁ + iR₂ + iR₃ − V = 0, so

Series(32.13)

R_{eq} = R_1 + R_2 + R_3 + \dots

Parallel

Resistors are in parallel when the same potential difference is across each. The current splits: i = i₁ + i₂ + i₃ = V/R₁ + V/R₂ + V/R₃, so

Parallel(32.14)

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots

two resistors: R₁R₂/(R₁ + R₂), always less than the smaller one

Series

Same current. Voltages add. R adds.

Parallel

Same voltage. Currents add. 1/R adds.

Interactive

Resistors in series and in parallel

Series: one path, the same current through each, voltages add, R = R₁ + R₂. Parallel: the same voltage across each, the current splits in inverse ratio of the resistances (i₁/i₂ = R₂/R₁), and the combination is smaller than either resistor.

equivalent resistance2.00Ω

battery current6.00A

R₁: current, voltage2.00 A, 12.0 V

R₂: current, voltage4.00 A, 12.0 V

Connection

R₁6Ω

R₂3Ω

Battery

V

12V

Worked example

A small network

A 12 Ω and a 6 Ω resistor are in parallel between A and C; a 2 Ω resistor joins C to B. Find the resistance between A and B.

How current divides

Across two parallel resistors V is common, so i₁R₁ = i₂R₂: the current divides in inverse ratio of the resistances.

i_1 = \frac{R_2}{R_1 + R_2}\,i, \qquad i_2 = \frac{R_1}{R_1 + R_2}\,i

Worked example

Splitting 3 A

3.0 A enters a 4 Ω and a 12 Ω resistor joined in parallel. How much goes through each?

Grouping cells

Series: two cells (ℰ₁, r₁) and (ℰ₂, r₂) driving R. Round the loop, i = (ℰ₁ + ℰ₂)/(R + r₁ + r₂): they act as one cell of emf ℰ₁ + ℰ₂ and internal resistance r₁ + r₂. Reverse one cell and the emf becomes |ℰ₁ − ℰ₂|.

Parallel: applying the loop law to each cell’s loop through R and eliminating gives one equivalent cell:

Cells in parallel

\mathcal E_0 = \frac{\mathcal E_1r_2 + \mathcal E_2r_1}{r_1 + r_2}, \qquad r_0 = \frac{r_1r_2}{r_1 + r_2}

two identical cells: same ℰ, half the internal resistance

Interactive

Cells in series and in parallel

Each cell has ℰ = 1.5 V and r = 0.5 Ω. In series the emfs and internal resistances add (ℰ₀ = nℰ, r₀ = nr). In parallel the emf stays ℰ but the internal resistance drops to r/n. Series wins when the load R is large compared with r; parallel wins when R is small.

combined emf4.50V

combined internal resistance1.50Ω

current in R1.29A

the other way: parallel0.69A

Cells joined in

Number of cells

n

Load

R

2.0Ω

Worked example

Two unequal cells

Cells of (2.0 V, 0.50 Ω) and (1.5 V, 0.25 Ω) drive a 4.0 Ω resistor. Find the current with the cells (a) in series, (b) in parallel.

What to remember

1R_eq = V/i for the network.
2Series: R adds (32.13). Parallel: 1/R adds (32.14).
3Parallel current divides inversely as the resistances.
4Cells in series: emfs and r's add. In parallel: ℰ₀ = (ℰ₁r₂ + ℰ₂r₁)/(r₁ + r₂), r₀ = r₁r₂/(r₁ + r₂).

Check yourself

Two 10 Ω resistors in parallel give…

In a parallel pair, which resistor carries more current?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PlayCircuit Construction Kit: DCPhET · CU Boulder ReadResistors in series and parallelOpenStax University Physics Vol. 2 §10.2

PreviousKirchhoff's Laws Next Wheatstone Bridge, Ammeters & Voltmeters

Series, Parallel & Grouped Cells

R adds in series, 1/R in parallel; cells combine too.

10 minUpdated 22 October 2026series resistorsparallel resistorsgrouping of cells

Lesson at a glanceShow the key points

R_eq = V/i; series R = ΣRᵢ (32.13); parallel 1/R = Σ1/Rᵢ (32.14).
Parallel current divides inversely as resistance: i₁ = R₂i/(R₁ + R₂).
Cells in series: ℰ₁ + ℰ₂, r₁ + r₂ (|ℰ₁ − ℰ₂| if one is reversed).
Cells in parallel: ℰ₀ = (ℰ₁r₂ + ℰ₂r₁)/(r₁ + r₂), r₀ = r₁r₂/(r₁ + r₂).

Any two-terminal network of resistors acts like one resistor. Series and parallel rules handle most of them; the same ideas tell us how to group cells.

Equivalent resistance

Apply potential difference V across a network’s two end points; if it draws current i, its equivalent resistance is R_eq = V/i.

Series

Resistors are in series when the same current passes through each. Round the loop: iR₁ + iR₂ + iR₃ − V = 0, so

Series(32.13)

R_{eq} = R_1 + R_2 + R_3 + \dots

Parallel

Resistors are in parallel when the same potential difference is across each. The current splits: i = i₁ + i₂ + i₃ = V/R₁ + V/R₂ + V/R₃, so

Parallel(32.14)

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots

two resistors: R₁R₂/(R₁ + R₂), always less than the smaller one

Series

Same current. Voltages add. R adds.

Parallel

Same voltage. Currents add. 1/R adds.

Interactive

Resistors in series and in parallel

equivalent resistance2.00Ω

battery current6.00A

R₁: current, voltage2.00 A, 12.0 V

R₂: current, voltage4.00 A, 12.0 V

Connection

R₁6Ω

R₂3Ω

Battery

V

12V

Worked example

A small network

A 12 Ω and a 6 Ω resistor are in parallel between A and C; a 2 Ω resistor joins C to B. Find the resistance between A and B.

How current divides

Across two parallel resistors V is common, so i₁R₁ = i₂R₂: the current divides in inverse ratio of the resistances.

i_1 = \frac{R_2}{R_1 + R_2}\,i, \qquad i_2 = \frac{R_1}{R_1 + R_2}\,i

Worked example

Splitting 3 A

3.0 A enters a 4 Ω and a 12 Ω resistor joined in parallel. How much goes through each?

Grouping cells

Parallel: applying the loop law to each cell’s loop through R and eliminating gives one equivalent cell:

Cells in parallel

\mathcal E_0 = \frac{\mathcal E_1r_2 + \mathcal E_2r_1}{r_1 + r_2}, \qquad r_0 = \frac{r_1r_2}{r_1 + r_2}

two identical cells: same ℰ, half the internal resistance

Interactive

Cells in series and in parallel

combined emf4.50V

combined internal resistance1.50Ω

current in R1.29A

the other way: parallel0.69A

Cells joined in

Number of cells

n

Load

R

2.0Ω

Worked example

Two unequal cells

Cells of (2.0 V, 0.50 Ω) and (1.5 V, 0.25 Ω) drive a 4.0 Ω resistor. Find the current with the cells (a) in series, (b) in parallel.

What to remember

1R_eq = V/i for the network.
2Series: R adds (32.13). Parallel: 1/R adds (32.14).
3Parallel current divides inversely as the resistances.
4Cells in series: emfs and r's add. In parallel: ℰ₀ = (ℰ₁r₂ + ℰ₂r₁)/(r₁ + r₂), r₀ = r₁r₂/(r₁ + r₂).

Check yourself

Two 10 Ω resistors in parallel give…

In a parallel pair, which resistor carries more current?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PlayCircuit Construction Kit: DCPhET · CU Boulder ReadResistors in series and parallelOpenStax University Physics Vol. 2 §10.2

PreviousKirchhoff's Laws Next Wheatstone Bridge, Ammeters & Voltmeters