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4.5 Series Circuits

GCSE Physics · AQA · Chapter 4: Electric Circuits

A free GCSE Physics lesson for AQA (8463) from OBA School: current, potential difference and resistance in series circuits; cells in series. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • State that the current is the same through every component in series
  • Explain that the supply p.d. is shared between components in series
  • Work out the total p.d. of cells in series
  • Use Rtotal = R₁ + R₂ and explain why adding resistors in series increases the total resistance
  • Solve series-circuit problems using V = I R

Key words

  • Series: Components connected one after another in a single loop, so the same charge passes through each.
  • Total resistance: The resistance of a single resistor that could replace the combination and give the same current.
  • Share (of p.d.): Each component in series gets part of the supply p.d.; the parts add up to the supply p.d.
  • Cells in series: Cells connected end to end; their p.d.s add if they all face the same way.

Current in a series circuit

In a series circuit there is only one path. Every electron passes through every component, so the current is the same through each component.

Potential difference is shared

The battery gives each coulomb of charge a fixed amount of energy. As the charge passes through each component it transfers some of this energy, so the supply p.d. is shared between the components:

Vtotal = V₁ + V₂ + … In the figure: 6.0 V = 4.0 V + 2.0 V

Because the current is the same in each, and V = I R, the component with the bigger resistance gets the bigger share of the p.d. In the figure, R₁ is twice R₂, so it gets twice the p.d.

Key fact: If a lamp is in series with a variable resistor, increasing the variable resistor’s resistance gives it a bigger share of the p.d., leaving less for the lamp – so the lamp dims.

Cells in series

When cells are joined in series facing the same direction, each one gives the charge an extra push, so the total p.d. is the sum of the individual p.d.s. A cell connected the wrong way round subtracts its p.d.

Total resistance in series

Rtotal = R₁ + R₂ (all in ohms, Ω)

Why does adding a resistor in series increase the total resistance? The supply p.d. is now shared between more resistors, so each one has a smaller p.d. across it and the current through each (the same current) is smaller. The supply p.d. has not changed, so Rtotal = V ÷ I is larger.

Worked example: a full series calculation

A 9.0 V battery is connected to a 12 Ω resistor and a 6.0 Ω resistor in series.
Rtotal = 12 + 6.0 = 18 Ω
I = V ÷ Rtotal = 9.0 ÷ 18 = 0.50 A (the same through both)
V₁₂ = 0.50 × 12 = 6.0 V; V₆ = 0.50 × 6.0 = 3.0 V (check: 6.0 + 3.0 = 9.0 V ✓)

Answer: I = 0.50 A; p.d.s 6.0 V and 3.0 V

Worked example: finding an unknown resistor

A 6.0 V supply drives 0.15 A through a 25 Ω resistor and an unknown resistor X in series.
Rtotal = 6.0 ÷ 0.15 = 40 Ω
X = 40 − 25

Answer: X = 15 Ω

Exam tip: In series problems, find the total resistance first, then the current, then each p.d. Finish by checking the p.d.s add up to the supply.

Common mistakes

  • Thinking the current is bigger before a component than after it. It is the same everywhere in series.
  • Using the supply p.d. with just one resistor: V = I R for one resistor needs the p.d. across that resistor only.
  • Forgetting that a reversed cell subtracts from the total p.d.
  • Saying the biggest resistor gets the smallest p.d. In series it gets the biggest share.

Revision questions

Try these, then check your answers below each one.

Question 1. A 15 Ω resistor and a 25 Ω resistor are connected in series. Calculate their total resistance.

Answer: 40 Ω

Working: Rtotal = 15 + 25 = 40 Ω.

Question 2. A 9.0 V battery is connected to a 10 Ω resistor and a 20 Ω resistor in series. Calculate the current.

Answer: 0.3 A

Working: Rtotal = 10 + 20 = 30 Ω. I = 9.0 ÷ 30 = 0.30 A.

Question 3. Four 1.5 V cells are connected in series, but one of them is the wrong way round. What is the total p.d.? (Choose from: 3.0 V, 6.0 V, 4.5 V, 1.5 V.)

Answer: 3.0 V

Working: 1.5 + 1.5 + 1.5 − 1.5 = 3.0 V.

Related lessons

Continue in the full course

This lesson is section 4.5 (AGCPH0405) of our free GCSE Physics course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Physics course.

 

 

 

 

 

 

 

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