Litmux Studio

LessonSeries and parallel circuits · 14 min

Series circuits

Parts are in series when they are connected end to end, so there is only one path for the current. A fuse, a switch and the lamp they control; an LED and its resistor; the cells in a torch - all in series.

R1, R2 and R3 in series across 12 V. One loop, one current - 2 mA through every part.
R1, R2 and R3 in series across 12 V. One loop, one current - 2 mA through every part.

The three rules of a series circuit

  1. The same current flows through every part (there is nowhere else for it to go).
  2. The resistances add up: the current has to get through all of them.
  3. The voltages add up to the supply, and each part's share is in proportion to its resistance (V = I × R with the same I).
Rtotal = R1 + R2 + R3 + ... I = Vsupply / Rtotal
Rtotal
the single resistance that would draw the same current
I
the one current through every part
Worked example

R1 = 1 kΩ, R2 = 2 kΩ and R3 = 3 kΩ are in series across 12 V. Find the current and each voltage.

  1. Rtotal = 1 + 2 + 3 = 6 kΩ.
  2. I = 12 V / 6 kΩ = 2 mA.
  3. V1 = 2 mA × 1 kΩ = 2 V, V2 = 2 × 2 = 4 V, V3 = 2 × 3 = 6 V.
  4. Check: 2 + 4 + 6 = 12 V - the whole supply, shared out.

Answer: 2 mA everywhere; 2 V, 4 V and 6 V.

InteractiveTry it: a series circuit
Total resistance-Ω
Current-mA
Voltage across R1-V
Voltage across R2-V
Voltage across R3-V

Make R2 ten times bigger than the others: it takes almost all the voltage.

Circuit for: Try it: a series circuit
The circuit at the starting values

Where you meet it

Cells in a torch

Two 1.5 V cells in series give 3 V: in series, voltages add.

LED and resistor

The resistor sets the one current that flows through both.

Switches and fuses

Always in series with what they switch or protect.

Current sensing

A shunt resistor in series sees the same current as the load.

Practice 1470 Ω, 1 kΩ and 2.2 kΩ are in series. What is the total resistance?

Show the answer

Answer: 3670 Ω

470 + 1000 + 2200 = 3670 Ω.

Practice 2Those three resistors are across 11 V. What voltage is across the 2.2 kΩ?

Hint

First the current, I = V / Rtotal.

Show the answer

Answer: 6.594 V

I = 11 / 3670 = 3.0 mA. V = 0.0030 × 2200 = 6.59 V.

Practice 3In a series circuit of a battery and three lamps, one lamp's filament breaks. The others:

  1. Get brighter
  2. Stay as they were
  3. Go out
  4. Get dimmer but stay on
Show the answer

Answer: C. Go out

The only path is broken, so the current stops everywhere.

Challenge 4Two 12 V 5 W lamps are connected in series across 12 V. Assuming each keeps a fixed resistance, what power does each lamp give?

Hint

Each lamp's resistance is V² / P. Then the series current.

Show the answer

Answer: 1.25 W

R = 144 / 5 = 28.8 Ω each, 57.6 Ω in all. I = 12 / 57.6 = 0.208 A. P = 0.208² × 28.8 = 1.25 W each: half the voltage gives a quarter of the power. (A real filament, cooler, would have a lower resistance and glow a little more.)