Complex Circuits: Combining Series and Parallel. Get a calculator and your PJ.

Complex circuits contain resistors that are in both series and parallel, so both sets of rules will apply. Lets summarize them quickly. Series Circuit 1.As you add resistors RT increases. 2.Voltage is used up through the circuit, but is spread out across resistors. 3.Current is the same at all locations. Parallel Circuit 1.As you add resistors RT decreases. 2.Voltage is used up through the circuit and each resistor gets the full amount. 3.Current is summative and dependent on resistance.

Remember equivalent resistance and how it can allow you to simplify a circuit?

How you simplify the circuit will depend on its configuration. 4 Ω 6 Ω 2 Ω

Complex circuits contain resistors that are in both series and parallel. R 2 =3 Ω R 3 =6 Ω R 1 =4 Ω 24 V A1A1 A2A2 A3A3 A4A4 A5A5 V1V1 V2V2 V3V3

When this occurs, the first thing that needs to be done is creating an equivalent circuit. R 2 =3 Ω R 3 =6 Ω R 1 =4 Ω 24 V A1A1 A2A2 A3A3 A4A4 A5A5 V1V1 V2V2 V3V3

When this occurs, the first thing that needs to be done is creating an equivalent circuit. R 2 =2 Ω R 1 =4 Ω 24 V

Now we can start to solve the circuit using the chart. R 2 =2 Ω R 1 =4 Ω 24 V ResistorRIV 1 2 Total 4 Ω 2 Ω 24 V6 Ω 8 V 4 A 16 V

Now we have to apply this information, this is where it gets tricky. Lets define what happens at the first resistor. R 2 =2 Ω R 1 =4 Ω 24 V ResistorRIV 1 2 Total 4 Ω 2 Ω 24 V6 Ω 8 V 4 A 16 V 4 A current 16 Volts DROPPED! 8 Volts are left over for the rest of the circuit.

So what happens when we look back at the original circuit? R 2 =3 Ω R 3 =6 Ω R 1 =4 Ω 24 V 4 A current 16 Volts DROPPED! 8 Volts are left over for the rest of the circuit. ResistorRIV 2 3 Total 3 Ω 6 Ω 8 V2 Ω 8 V 2.67 A 1.33 A 4 A 8 V 8 Volts DROPPED! 2.67 A 1.33 A 4 A current

Now apply what we just did to the meters in the original diagram. R 2 =3 Ω R 3 =6 Ω R 1 =4 Ω 24 V A1A1 A2A2 A3A3 A4A4 A5A5 V1V1 V2V2 V3V3

28 V R 2 =6 Ω R 3 =12 Ω R 1 =6Ω A1A1 VBVB V1V1 V3V3 A2A2 A3A3 A4A4 Now you try a circuit with the exact same configuration. V2V2

Lets look at a different configuration. R 3 =6 Ω 12 V R 1 =3 Ω R 2 =3 Ω A1A1 V2V2 V3V3 A3A3 A4A4 A5A5 A6A6 V1V1 A2A2 VBVB

Lets solve the simplifies circuit. R 3 =6 Ω 12 V R 1,2 =6 Ω ResistorRIV 1/2 3 Total 6 Ω 12 V 3 Ω 12 V 2 A 4 A 12 V

Lets apply this info to our circuit. R 3 =6 Ω 12 V R 1,2 =6 Ω ResistorRIV 1/2 3 Total 6 Ω 12 V 3 Ω 12 V 2 A 4 A 12 V 12 V Dropped

Lets expand and determine what is happening in this branch. R 3 =6 Ω 12 V R 1 =3 Ω R 2 =3 Ω

Now apply what we just did to the meters in the original diagram. 12 V R 1 =3 Ω R 2 =3 Ω A1A1 V2V2 V3V3 A3A3 A4A4 A5A5 A6A6 V1V1 A2A2 VBVB R 3 =6 Ω

Now you try a circuit with the exact same configuration. 28 V R 1 =4 Ω R 2 =6 Ω A1A1 V2V2 V3V3 A3A3 A4A4 A5A5 A6A6 V1V1 A2A2 VBVB R 3 =12 Ω

Making comparisons within complex circuits.

Which bulb will glow the brightest?

12 V R 1 =6 Ω R 2 =3 Ω R 3 =3 ΩR 4 =3 Ω A1A1 V2V2 V3V3 A2A2 A3A3 A4A4 A5A5 A6A6 V1V1

R 1 =5 Ω 9 V R 2 =3 Ω R 3 =6 Ω R 4 =9 Ω R 5 =3 Ω A1A1 V1V1 V3V3 A2A2 A3A3 A4A4 A5A5 V2V2

R 2 =6 Ω R 3 =10 Ω R 6 =0.5 Ω R 7 =5.4 Ω R 1 =4.6 Ω 56 V R 4 =6 Ω R 5 =2 Ω A1A1 V1V1 V2V2 V3V3 A2A2 A5A5 A6A6 A4A4 A3A3

R 2 =12 ΩR 3 =12 Ω R 4 =10 Ω R 5 =6 Ω R 6 =2 Ω R 7 =0.5 Ω R 8 =4 Ω R 9 =16 Ω R 1 =12.8 Ω 100 V A1A1 V2V2 A2A2 A3A3 A4A4 V1V1