5.8 - combination circuits

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Dr Pusey www.puseyscience.com

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Page 1: 5.8 - combination circuits

Dr Pusey

www.puseyscience.com

Page 2: 5.8 - combination circuits

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

Page 3: 5.8 - combination circuits

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

Page 4: 5.8 - combination circuits

Network circuit Sometimes, life’s tough….

Network/combination circuit = a circuit with components in both series and parallel

Page 5: 5.8 - combination circuits

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

Page 6: 5.8 - combination circuits

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

Page 7: 5.8 - combination circuits

Example 1

Find the current and potential difference across each individual resistor for the circuit below.

Page 8: 5.8 - combination circuits

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

Page 9: 5.8 - combination circuits

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

Page 10: 5.8 - combination circuits

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

Page 11: 5.8 - combination circuits

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

Page 12: 5.8 - combination circuits

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

Page 13: 5.8 - combination circuits

Example 2

Calculate the total resistance of this circuit.

Page 14: 5.8 - combination circuits

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

Page 15: 5.8 - combination circuits

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

Page 16: 5.8 - combination circuits

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.