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Syllabus Points
circuit analysis and design involve calculation of the potential difference across, the current in, and the power supplied to, components in series, parallel, and series/parallel circuits
this includes applying the relationships for series and parallel components
Learning Goals
Be able to use the equations below to determine potential difference, current, power and resistance in network circuits
Series𝐼 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉𝑡 = 𝑉1 + 𝑉2 +⋯𝑉𝑛𝑅𝑡 = 𝑅1 + 𝑅2+. . . 𝑅𝑛
Parallel𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐼𝑡 = 𝐼1 + 𝐼2 +⋯𝐼𝑛1
𝑅𝑡=
1
𝑅1+
1
𝑅2+⋯
1
𝑅𝑛
Network circuit Sometimes, life’s tough….
Network/combination circuit = a circuit with components in both series and parallel
Problem solving a network circuit Only way to solve is to simplify the circuit down into an
equivalent series circuit
Step 1: Reduce any parallel sections down into one ‘equivalent’ resistor
Step 2: Treat the circuit as one big ‘series’ circuit
Problem solving a network circuit It helps to create a table of information (as there are a
lot of numbers to deal with), see example below
Some people (like Dr Pusey) find it easier to write all three variables on the diagram. Do what suits you best!
DON’T ADD NUMBERS TO GET THE TOTAL
This will take practice to get better at!Component Voltage (V) Current (A) Resistance (Ω)
1
2
3
Total
Example 1
Find the current and potential difference across each individual resistor for the circuit below.
Example 1 Create a table and fill in what you know.
Add in extra rows for any components in parallel (e.g. 2//3 for this example).
Component Voltage (V)
Current (A)
Resistance (Ω)
1 5
2 8
3 8
2//3
4 6
Total 60
Example 1 Find equivalent resistance of parallel components 2//31
𝑅𝑇=
1
𝑅2+
1
𝑅31
𝑅𝑇=1
8+1
8=1
4𝑅𝑇 = 4 Ω
Component Voltage (V) Current (A)
Resistance (Ω)
1 5
2 8
3 8
2//3 4
4 6
Total 60
R2//3= 4 Ω
Example 1Find the total resistance of the circuit (now that we’ve simplified it into a series circuit):
𝑅𝑇 = 𝑅1 + 𝑅2//3 + 𝑅4𝑅𝑇 = 5 + 4 + 6 = 15 Ω
Now we can find the total current
𝐼 = 𝑉 𝑅 = 6015 = 4 𝐴
In a series circuit current = same at each point
Component Voltage (V)
Current (A)
Resistance (Ω)
1 4 5
2 8
3 8
2//3 4 4
4 4 6
Total 60 4 15
R2//3= 4 Ω
Example 1 We can now find V1, V2//3 and V4
using:𝑉 = 𝐼𝑅
E.g.𝑉1 = 4 × 5 = 20 𝑉
Component Voltage (V)
Current (A)
Resistance (Ω)
1 20 4 5
2 8
3 8
2//3 16 4 4
4 24 4 6
Total 60 4 15
R2//3= 4 Ω
Example 1 Finally looking at the parallel components separately again.We know each branch of parallel receives the total voltage so both R2 and R3 will get a voltage of 16 V.Then we can find current using: 𝐼 = 𝑉 𝑅
Component Voltage (V)
Current (A)
Resistance (Ω)
1 20 4 5
2 16 2 8
3 16 2 8
2//3 16 4 4
4 24 4 6
Total 60 4 15
Example 2
Calculate the total resistance of this circuit.
Example 2
Only looking at resistance for this question, no info given about I or V.Need to simplify parallel bits into ‘equivalent’ resistances.
1
𝑅1/2=
1
𝑅1+
1
𝑅2=
1
10+1
4
𝑅1/2 = 2.86 Ω
Component
Resistance (Ω)
1 10
2 4
3 3
4 8
5 1
1//2 2.86
4//5 0.89
Total
Example 2
Now we can find the total resistance of the circuit.𝑅𝑇 = 𝑅1/2 + 𝑅3 + 𝑅4/5
𝑅𝑇 = 2.86 + 3 + 0.89 = 6.75 Ω
Component
Resistance (Ω)
1 10
2 4
3 3
4 8
5 1
1//2 2.86
4//5 0.89
Total 6.75
Network Circuits – tips Draw a BIG circuit diagram (if one is not provided)
Take the time to organise yourself, use a table (like in the examples)
Turn any parallel components into ‘equivalent’ series components, to make one big series circuit (which is easier to analyse)
Use Ohm’s Law (V=IR)
You will need to alter your approach for different problem situations.
ResourcesAV
Khan Academy - Circuits (part 1) (11:40)
Khan Academy - Circuits (part 2) (11:08)
Khan Academy - Circuits (part 3) (12:21)
Khan Academy - Circuits (part 4) (7:07)
Physics EH – Series and Parallel Circuits (5:12)
Further Reading
Physics Classroom – Combination circuits