K326 / K327 / 2027
D.C. circuits overview

Lesson 2 of 4

Series circuits

  • I can state and apply the equal-current rule for a series circuit.Syllabus

    Syllabus K326 / K327, 13(b). State and apply the equal-current rule for a series circuit.

  • I can state and apply the sum of series potential differences.Syllabus

    Syllabus K326 / K327, 13(c). State and apply the sum of series potential differences.

  • I can recall and apply effective resistance for a number of resistors in series and in parallel.Syllabus

    Syllabus K326 / K327, 13(f). Recall and apply effective resistance for a number of resistors in series and in parallel.

Make a guess. You are not marked.

Two identical lamps are connected in series with a cell and both glow. One lamp is unscrewed from its holder. What happens to the other lamp?

  1. It glows the same as before.
  2. It goes out.
  3. It glows brighter.
Show the answer

It goes out.

A series circuit has one path. Removing a lamp breaks that path, so no current flows anywhere in the circuit and the other lamp goes out.

Series components carry the same current. Their individual potential differences add to the p.d. across the complete path.

For each resistor, V = IR uses the voltage across that resistor and the current through it. Here Vs means the voltage across the complete external network.

Why the current is the same

In a steady unbranched circuit, charge does not continually pile up at one component. The same amount of charge per second passes each position. A resistor transfers energy; it does not consume part of the current before the next resistor.

I = I1 = I2 = ...Same unbranched path: the same current passes every series component.

Each coulomb undergoes successive energy transfers as it passes the components. The work per coulomb across the whole path is the sum of the work per coulomb in its parts, so the potential differences add.

Vs = V1 + V2 + ...Equal current does not require equal voltages across unequal resistances.

Add the series resistances

For two resistors carrying the same current, V1 = IR1 and V2 = IR2. Adding gives Vs = I(R1 + R2). The single equivalent resistance must therefore be their sum.

Rseries = R1 + R2 + ...Adding a positive resistance in series increases the total resistance.

One path has the same current; component voltage drops add

These supplied calculation models use fixed resistances, negligible wire resistance and an ideal source. Current arrows indicate direction, not a current scale.

One path: same current through both resistors

One path: same current through both resistorsAn ideal 6.0 V source drives a single closed path through R1: 2 ohm between A and B and R2: 4 ohm between B and C. The same 1.0 A passes through both. The p.d. across A/B is 2.0 V and across B/C is 4.0 V. The bottom switch is closed.+-6.0 VR1: 2 ohmR2: 4 ohm2.0 V4.0 VABCI = 1.0 ASwitch closed

The drops 2.0 V and 4.0 V add to 6.0 V. Current is 1.0 A in the whole unbranched path.

Open the only path

Open the only pathA 6.0 V ideal source and two series resistors have their only path opened by the bottom switch. Steady current is zero throughout the circuit and each resistor has zero p.d. The source still has 6.0 V. In this ideal model, the right switch contact is 6.0 V above the left one, so the open gap has a 6.0 V p.d.+-6.0 VR1: 2 ohmR2: 4 ohm0 V0 VCurrent = 0 AOpen switch: 6.0 V across the gap

Zero steady current does not mean zero source voltage. The only path is broken.

The closed 6.0 V circuit has one path through 2.0 Ω and 4.0 Ω. Its current is 1.0 A, with voltage drops of 2.0 V and 4.0 V. The separate open-switch view has zero steady current while the ideal source still maintains its voltage.

One current, two voltage drops

2.0 Ω and 4.0 Ω across 6.0 V

Assume fixed resistances, negligible wire resistance and an ideal source whose terminal p.d. stays at 6.0 V.

  1. Total resistance = 2.0 + 4.0 = 6.0 Ω.
  2. Shared current = 6.0/6.0 = 1.0 A.
  3. Across 2.0 Ω: V = 1.0 x 2.0 = 2.0 V.
  4. Across 4.0 Ω: V = 1.0 x 4.0 = 4.0 V.

The drops add to 6.0 V. One coulomb transfers 2.0 J in the first resistor and 4.0 J in the second. The charge passing them has not decreased.

Try it step by step

Work it out in steps: Total resistance in ohm, then Current in A, then p.d. across the first resistor in V.

  1. Problem 1. A 3.0 ohm resistor and a 9.0 ohm resistor are connected in series across a 6.0 V supply. Find the current and the p.d. across the 3.0 ohm resistor.
  2. Problem 2. A 5.0 ohm resistor and a 15 ohm resistor are in series across a 10 V supply. Find the current and the p.d. across the 5.0 ohm resistor.
  3. Problem 3. An 8.0 ohm resistor and a 4.0 ohm resistor are in series across a 9.0 V supply. Find the current and the p.d. across the 8.0 ohm resistor.
Show the answer

Problem 1.

Total resistance in ohm: R = 3.0 + 9.0 = 12 ohm

Current in A: I = V / R = 6.0 / 12 = 0.50 A

p.d. across the first resistor in V: V = IR = 0.50 x 3.0 = 1.5 V (2 sf)

Problem 2.

Total resistance in ohm: R = 5.0 + 15 = 20 ohm

Current in A: I = 10 / 20 = 0.50 A

p.d. across the first resistor in V: V = 0.50 x 5.0 = 2.5 V (2 sf)

Problem 3.

Total resistance in ohm: R = 8.0 + 4.0 = 12 ohm

Current in A: I = 9.0 / 12 = 0.75 A

p.d. across the first resistor in V: V = 0.75 x 8.0 = 6.0 V (2 sf)

At fixed supply voltage, a larger total resistance means a smaller current. These example values describe a model; practical components must also have suitable ratings and remain close to the assumed temperature.

What an open switch changes

Opening the only path stops the steady current throughout this simple circuit. Each fixed resistor then has V = IR = 0 across it, but the source does not lose its e.m.f. In the ideal open-switch arrangement, the source voltage appears across the gap.

Zero current does not always mean zero voltage. An open gap can have a p.d. across it even though there is no conducting path through it.

Check your understanding

Fixed 3.0 ohm and 6.0 ohm resistors are in series across an ideal 9.0 V source. What is the potential difference across the 6.0 ohm resistor?

  1. 4.5 V
  2. 6.0 V
  3. 9.0 V
Show the answer

6.0 V

Total resistance is 3.0 + 6.0 = 9.0 ohm, so the shared current is 9.0/9.0 = 1.0 A. The 6.0 ohm resistor has V = IR = 1.0 x 6.0 = 6.0 V; the other has 3.0 V.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A 4.0 Ω resistor and an 8.0 Ω resistor are connected in series to a 6.0 V battery of negligible internal resistance. Calculate the current in the circuit and the potential difference across the 8.0 Ω resistor. [3]

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