Series & Parallel Circuits (AQA AS Physics): Revision Note

Exam code: 7407

Dan Mitchell-Garnett

Written by: Dan Mitchell-Garnett

Reviewed by: Tim

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Series & parallel circuits

Current

  • In a series circuit, the current is the same for all components

  • In a parallel circuit, the current is split across the different branches (or junction)

    • The total current into a junction must equal the total current out of a junction

      • The amount of current in each branch depends on the total resistance of the components within that branch

Potential difference

  • In a series circuit, the e.m.f of the power supply is shared amongst all the components in different amounts, depending on their resistance

  • In a parallel circuit, the voltage of all the components in each branch is equal to the e.m.f of the power supply

  • Cells can also be connected in series or parallel

  • The total voltage of the combined cells can be calculated in the same way as voltage

    • If the cells are connected in series, the total voltage between the ends of the chain of cells is the sum of the potential difference across each cell

    • If the cells are connected in parallel, the total voltage across the arrangement is the same as for one cell

Diagram comparing three 2-volt cells in series, producing 6 volts, with three 2-volt cells in parallel, producing 2 volts across the arrangement.
Cells in series add their voltages, while identical cells in parallel give the same voltage as one cell
  • A summary of the current, voltage and resistance within a series and parallel circuit are summarised below:

Table of Voltage, Current & Resistance in Series & Parallel Circuits

Series and parallel circuit comparison table with circuit diagrams. For three resistors connected in series, series voltage Vin equals V1 plus V2 plus V3, current is equal throughout, and Rtotal equals R1 plus R2 plus R3. For three resistors connected in parallel, parallel voltage is equal across branches, currents add, and reciprocal resistances add.
Table of Voltage, Current & Resistance in Series & Parallel Circuits

Examiner Tips and Tricks

The best way to practise these calculations is to understand why you have made a mistake. Common mistakes are:

  • Thinking the current is the same through every branch in a parallel circuit

  • Thinking the voltage is the same through all components in a series circuit

  • Not taking into account multiple resistors

  • Not calculating the total resistance using the appropriate parallel or series resistance equation

  • The sum of the voltages of all the components in a series circuit not adding up to the e.m.f of the supply

  • The current into a junction not being equal to the current out of a junction (Kirchhoff’s first law)

Don’t be afraid to annotate circuit diagrams to help with this. The more information you have about all the components, the easier it is to calculate values that are missing

Conservation of charge & energy

Conservation of charge

  • Charge is never used up or lost in a circuit - this is known as conservation of charge

  • As a result of this, for current in a parallel circuit:

    The sum of the currents entering a junction always equal the sum of the currents out of the junction

  • This is sometimes known as Kirchhoff's first law

  • In a circuit:

    • A junction is a point where at least three circuit paths meet

    • A branch is a path connecting two junctions

  • If a circuit splits into two branches, then the current before the circuit splits should be equal to the current after it has split

  • In the circuit below, I = I1 + I2 + I3, where I represents the current in the circuit before it branches, and I1, I2 and I3 represent the current in the respective three branches:

Diagram of a junction showing incoming current I splitting into three outgoing branch currents I₁, I₂ and I₃. The textbox states "Kirchhoff’s first law: I equals I₁ plus I₂ plus I₃".
The current I into the junction is equal to the sum of the currents out of the junction
  • The charge is conserved on both sides of the junction

  • In a series circuit, the current is the same through all components

Series circuit with two lamps and three ammeters, each labelled 0.3 amperes, showing that the current is the same at every point.
The current is the same at each point in a series circuit
  • In a parallel circuit, the current divides at the junctions and each branch has a different value

  • Kirchhoff’s first law applies at each junction

A 3-ampere current splits at the left junction into 2 amperes through the upper branch and 1 ampere through the lower, then recombines at the right junction. The textbox states “Kirchhoff’s First Law: 3 = 2 + 1”.
The current divides at each junction in a parallel circuit

Conservation of energy

  • Energy is never used up or lost in a circuit - this is known as conservation of energy

  • As a result of this, for voltage in any circuit:

    The total e.m.f. in a closed circuit equals the sum of the potential differences across each component

  • This is sometimes known as Kirchhoff's second law

  • Each closed circuit can be treated like a series circuit

  • Below is a circuit explaining Kirchhoff’s second law with the sum of the voltages in the closed series circuit equal to the total e.m.f:

Two series cells labelled E₁ and E₂ power two series resistors, with voltmeters V₁ and V₂ across them and current I. The textbox states “Kirchhoff's Second Law: E₁ + E₂ = V₁ + V₂”.
The sum of the voltages is equal to the total e.m.f from the batteries
  • In a series circuit, the voltage is split across all components depending on their resistance

    • The sum of the voltages is equal to the total e.m.f of the power supply

  • In a parallel circuit, the voltage is the same across each closed loop

    • The sum of the voltages in each closed circuit loop is equal to the total e.m.f of the power supply:

Two cells, E₁ and E₂, are supplying two parallel loops. An upper loop with two resistors and voltmeters V₁ and V₂, and a lower loop with two lamps and V₃ and V₄. The textbox states: “Kirchhoff’s Second Law: E₁ + E₂ = V₁ + V₂ = V₃ + V₄”.
The sum of the e.m.fs in each closed loop is equal to the total e.m.f of the power supply
  • A closed-circuit loop acts as its own independent series circuit and each one separates at a junction

    • A parallel circuit is made up of two or more of these loops

Two cells, E₁ and E₂, are supplying two parallel loops. An upper loop labelled loop 1 with two resistors and voltmeters V₁ and V₂, and a lower loop labelled loop 2 with two lamps and V₃ and V₄. The textbox states: “Kirchhoff’s Second Law: E₁ + E₂ = V₁ + V₂ = V₃ + V₄”.
Each circuit loop acts as a separate, independent series circuit
  • This makes parallel circuits incredibly useful for home wiring systems

    • A single power source supplies all lights and appliances with the same voltage

    • If one light breaks, voltage and current can still flow through for the rest of the lights and appliances

Worked Example

For the circuit below, state the readings of ammeters A1, A2 and A3.

Ammeter A1 is on the same branch as the cell. A 0.270 ampere supply current splits into a lamp branch containing ammeter A3 and a lower branch where 80 milliamperes splits into branches reading 27 milliamperes and ammeter A2.

Answer:

Worked-example solution using Kirchhoff’s First Law.
Step 1: 0.270 A and A1 are connected in series. The current entering the cell equals the current leaving it. A1 is also 0.270 A. 
Step 2: From Kirchhoff's first law, the total current into the junction must equal the total current out of it. 80 mA = 27 mA + A₂, so A₂ = 53 mA.
Step 3: Kirchhoff's first law at a different junction. 270 mA = A₃ + 80 mA, so A₃ = 190 mA.

Worked Example

For the circuit below, state the readings of the voltmeters V1, V2 and V3. All the lamps and resistors have the same resistance.

A 24-volt supply feeds a 2-volt lamp and three parallel branches. The top branch contains a resistor with V1 and a 12-volt lamp. The middle branch contains an 8-volt resistor and two lamps with V2 across one of the lamps. The bottom branch contains a resistor with V3.

Answer:

Worked-example solution applying Kirchhoff’s Second Law.
Step 1: Kirchhoff's second law states that the sum of the three components is equal to the e.m.f of the supply, so 2 V + V₁ + 12 V = 24 V and V₁ = 24 − 12 − 2 = 10 volts.
Step 2: Kirchhoff's second law in this loop, 2 V + 8 V + V of both lamps = 24 V so V of both lamps = 14 volts. Since both lamps have the same resistance R and the current I through them both is the same, they share the voltage left equally. two equal lamps share 14 volts, so V₂ = 7 volts.
Step 3: Kirchhoff's second law in this loop V₃ = 24 − 2 = 22 volts.

Examiner Tips and Tricks

Junctions only appear in parallel circuits. As circuits become more complex, you may find it hard to tell which currents go into a junction and which come out. So draw arrows on the diagram for the current at each junction, from positive to negative.

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Dan Mitchell-Garnett

Author: Dan Mitchell-Garnett

Expertise: Physics Content Creator

Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to Save My Exams. Here, he carries on his passion for writing challenging physics questions and helping young people learn to love physics.

Tim

Reviewer: Tim

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.