Series & Parallel Circuits

Resistors in Series

In a series circuit:
- The current is the same at any point
- The potential difference is split across all components depending on their resistance

When two or more components are connected in series:

The combined resistance of the components is equal to the sum of individual resistances

Three resistors connected in series will have a total resistance of R = R₁ + R₂ + R₃

Resistors in series diagram, downloadable AS & A Level Physics revision notes

Resistors in series

The equation for the total resistance of resistors in series is derived using the above rules for current and potential difference:

3-6-derivation-resistors-in-series_edexcel-a-level-physics-rn

The equation for the combined resistance of resistors in series is therefore:

Resistors in series equation, downloadable AS & A Level Physics revision notes

This means the total resistance increases
- The combined resistance is more than the resistance of any of the individual components
- For example, if two resistors of equal resistance are connected in series, then the combined resistance will double

5-2-2-resistors-in-series_sl-physics-rn

Connecting more resistors in series increases the overall resistance

Resistors in Parallel

In a parallel circuit:
- The total current is equal to the sum of the currents in each parallel branch of the circuit
- The potential difference is the same across each loop

In a parallel circuit, the combined resistance of the components requires the use of reciprocals

The reciprocal of the combined resistance of two or more resistors is the sum of the reciprocals of the individual resistances

Two resistors connected in parallel will have a total resistance of $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

Resistors in parallel diagram, downloadable AS & A Level Physics revision notes

Resistors in parallel

The equation for the total resistance of resistors in parallel is derived using the above rules for current and potential difference:

3-6-derivation-resistors-in-parallel-edexcel-a-level-physics-rn

The equation for the combined resistance of resistors in parallel is therefore:

Resistors in parallel equation, downloadable AS & A Level Physics revision notes

This means the total resistance decreases
- The combined resistance is less than the resistance of any of the individual components
- For example, if two resistors of equal resistance are connected in parallel, then the combined resistance will halve

Connecting more resistors in parallel decreases the overall resistance

Summary of the rules in series and parallel circuits

The table below summarises the rules for calculating current, potential difference and resistance within series and parallel circuits

Series and Parallel Circuits Summary

2-5-8-series-and-parallel-circuits-summary-table

Worked example

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

WE - Kirchoff's first law question image, downloadable AS & A Level Physics revision notes

Answer:

Step 1: Recall the rules for current in series & parallel

In a series circuit: The current is the same at any point $I_{i n} = I_{1} = I_{2} = I_{3}$
In a parallel circuit: The total current is equal to the sum of the currents in each parallel branch of the circuit $I_{i n} = I_{1} + I_{2} + I_{3}$

Step 2: Determine the current in ammeter A₁

Ammeter A₁ is in series with ammeter A
Therefore, the current in A₁ = the current in A

A₁ = 0.270 A

Step 3: Determine the current in ammeter A₃

The current from ammeter A₁ enters splits into two branches
One branch has a current of 80 mA = 0.080 A
The current in the other branch is the reading of ammeter A₃

0.270 = 0.080 + A₃

A₃ = (0.270 – 0.080) = 0.190 A

Step 4: Determine the current in ammeter A₂

The current of 0.080 A splits into two branches
One branch has a current of 27 mA = 0.027 A
The current in the other branch is the reading of ammeter A₂

0.080 = 0.027 + A₂

A₂ = (0.080 – 0.027) A = 0.053 A

The current readings of the three ammeters are:
- A₁ = 0.270 A = 270 mA
- A₂ = 0.053 A = 53 mA
- A₃ = 0.19 A = 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.

series-and-parallel-resistors

Answer:

Step 1: Recall the rules for p.d. in series & parallel

In a series circuit: The potential difference is split across all components (equally if the resistance is equal) $V_{i n} = V_{1} + V_{2} + V_{3}$
In a parallel circuit: The potential difference is the same across each loop $V_{i n} = V_{1} = V_{2} = V_{3}$

Step 2: Determine the p.d. in voltmeter V₁

The potential difference provided by the cell is the total input p.d., so, each loop must sum to 24 V
The first loop consists of the 2 V lightbulb, the 12 V lightbulb and resistor V₁

2 + 12 + V₁ = 24 V

V₁ = 24 – 2 – 12 = 10 V

Step 3: Determine the p.d. in voltmeter V₂

The second loop consists of the 2 V lightbulb, the 8 V resistor, lightbulb V₂, and the 8 V lightbulb

2 + 8 + V₂ + 8 = 24 V

V₂ = 24 – 2 – 8 – 8 = 6 V

Step 3: Determine the p.d. in voltmeter V₃

The third loop consists of the 2 V lightbulb and resistor V₃

2 + V₃ = 24 V

V₃ = 24 – 2 = 22 V

The potential difference readings of the three voltmeters are:
- V₁ = 10 V
- V₂ = 6 V
- V₃ = 22 V

Exam Tip

The equations for current, potential difference and resistance in series and parallel circuits is given in your data booklet. However, it is much quicker if you remember them!

Series & Parallel Circuits (SL IB Physics)

Revision Note