# Lesson 27: Series and Parallel Circuits

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Courage doesn’t always roar. Sometimes courage is the little voice at the end of the day that says I’ll try again tomorrow.

Mary Anne Radmacher

The purpose of an electric circuit is to provide a closed pathway for electrical energy and electrons to travel along. There are three kinds of electrical circuits: series circuits, parallel circuits, and combined circuits. Combined circuits contain both series and parallel components and are not addressed in this activity.

#### Series Circuits:

In a series circuit, the electrons can only flow along one pathway, or loop; each electron follows along behind the electron in front of it.

**RULES FOR SERIES CIRCUITS:**

The current in a series circuit will be the same everywhere. | I_{T} = I_{1} = I_{2} = I_{3} = … |

The total resistance in a series circuit is the sum of the individual resistances. | R_{T} = R_{1} + R_{2} + R_{3} + … |

The total potential difference in a series circuit is the sum of the individual potential differences. | V_{T} = V_{1} + V_{2} + V_{3} + … |

In a circuit, each resistor or load, is given a number so that it can be identified. The term “total” refers to the power source.

When you are asked to “solve a circuit,” you must calculate the value of the current, resistance and potential difference for all of the loads in the circuit and for the power source. This is done by applying the three rules above. Once you have found two of the three values for any load, you can apply Ohm’s Law to determine the third value.

Learn more about series circuits.

#### Example 1: A Series Circuit

Determine the potential difference, resistance, and current for each component in the series circuit below.

V_{1} = 1 V | I_{1} = | R_{1} = |

V_{2} = 8 V | I_{2} = | R_{2} = |

V_{3} = | I_{3} = | R_{3} = 6 Ω |

V_{T} = 12 V | I_{T} = | R_{T} = |

**Given:**

All of the information that you have been given is in the table.

**Required:**

You are required to find all of the values that are not given in the table.

**Solution:**

You will need the following equations to solve the circuit:

I_{T} = I_{1} = I_{2} = I_{3} = …

R_{T} = R_{1} + R_{2} + R_{3} + …

V_{T} = V_{1} + V_{2} + V_{3} + …

V = IR

As values are calculated, they should be entered into the table.

Since 3 of 4 potential difference values are given, the 4^{th} can be found using V_{T} = V_{1} + V_{2} + V_{3} + …Rearranged we get:

V_{3} = V_{T} – V_{1} – V_{2}

V_{3} = 12 V – 1 V – 8 V = 3 V

V_{1} = 1 V | I_{1} = | R_{1} = |

V_{2} = 8 V | I_{2} = | R_{2} = |

V_{3} = 3 V | I_{3} = | R_{3} = 6 Ω |

V_{T} = 12 V | I_{T} = | R_{T} = |

Since V_{3} and R_{3} are known, we can use Ohm’s Law to find I_{3}.

V_{1} = 1 V | I_{1} = | R_{1} = |

V_{2} = 8 V | I_{2} = | R_{2} = |

V_{3} = 3 V | I_{3} = 0.5 A | R_{3} = 6 Ω |

V_{T} = 12 V | I_{T} = | R_{T} = |

Since the current in a series circuit is the same at all points in the circuit, then I_{1}, I_{2}, and I_{T} must all equal 0.5 A.

V_{1} = 1 V | I_{1} = 0.5 A | R_{1} = |

V_{2} = 8 V | I_{2} = 0.5 A | R_{2} = |

V_{3} = 3 V | I_{3} = 0.5 A | R_{3} = 6 Ω |

V_{T} = 12 V | I_{T} = 0.5 A | R_{T} = |

Now Ohm’s Law can be applied to find R_{1}, R_{2}, and R_{T}.

V_{1} = 1 V | I_{1} = 0.5 A | R_{1} = 2 Ω |

V_{2} = 8 V | I_{2} = 0.5 A | R_{2} = 16 Ω |

V_{3} = 3 V | I_{3} = 0.5 A | R_{3} = 6 Ω |

V_{T} = 12 V | I_{T} = 0.5 A | R_{T} = 24 Ω |

Since this is a series circuit, R_{T }can also be found by adding up all of the individual resistances (R_{T} = R_{1} + R_{2} + R_{3}). This also gives 24 Ω. It is good practice to get into the habit of checking your work in this manner to ensure you have solved the circuit correctly.

### Practice Questions

- Solve the following series circuit for all of the values missing from the table.

V_{1} = | I_{1} = | R_{1} = 32 Ω |

V_{2} = | I_{2} = 2.5 A | R_{2} = |

V_{3} = 30 V | I_{3} = | R_{3} = |

V_{T} = 120 V | I_{T} = | R_{T} = |

**Answer**

**Given:**

All of the information that you have been given is in the table.

**Required:**

You are required to find all of the values that are not given in the table.

**Solution:**

You will need the following equations to solve the circuit:

I_{T} = I_{1} = I_{2} = I_{3} = …

R_{T} = R_{1} + R_{2} + R_{3} + …

V_{T} = V_{1} + V_{2} + V_{3} + …

V = IR

As values are calculated, they should be entered into the table.

Since I_{T} = I_{1} = I_{2} = I_{3}, all of the currents must equal 2.5 A.

V_{1} = | I_{1} = 2.5 A | R_{1} = 32 Ω |

V_{2} = | I_{2} = 2.5 A | R_{2} = |

V_{3} = 30 V | I_{3} = 2.5 A | R_{3} = |

V_{T} = 120 V | I_{T} = 2.5 A | R_{T} = |

R_{3}, R_{T}, and V_{1} can now be found using Ohm’s Law.

V_{1} =I_{1}R_{1} = (2.5 A)(32 Ω) = 80 V

V_{1} = 80 V | I_{1} = 2.5 A | R_{1} = 32 Ω |

V_{2} = | I_{2} = 2.5 A | R_{2} = |

V_{3} = 30 V | I_{3} = 2.5 A | R_{3} = 12 Ω |

V_{T} = 120 V | I_{T} = 2.5 A | R_{T} = 48 Ω |

By rearranging V_{T} = V_{1} + V_{2} + V_{3}, V_{2} can be found.

V_{2} = V_{T} – V_{1} – V_{3}

V_{2} = 120 V – 80 V – 30 V

V_{2} = 28 V

V_{1} = 80 V | I_{1} = 2.5 A | R_{1} = 32 Ω |

V_{2} = 10 V | I_{2} = 2.5 A | R_{2} = |

V_{3} = 30 V | I_{3} = 2.5 A | R_{3} = 12 Ω |

V_{T} = 120 V | I_{T} = 2.5 A | R_{T} = 48 Ω |

R_{2} can now be found using Ohm’s Law or by rearranging R_{T} = R_{1} + R_{2} + R_{3}.

R_{2} = R_{T} – R_{1} – R_{3}

R_{2} = 48 Ω – 32 Ω – 12 Ω = 4 Ω

V_{1} = 80 V | I_{1} = 2.5 A | R_{1} = 32 Ω |

V_{2} = 10 V | I_{2} = 2.5 A | R_{2} = 4 Ω |

V_{3} = 30 V | I_{3} = 2.5 A | R_{3} = 12 Ω |

V_{T} = 120 V | I_{T} = 2.5 A | R_{T} = 48 Ω |

#### Parallel Circuits:

In a parallel circuit, the electrons have more than one pathway, or loop, to follow around the circuit. Each electron may, or may not, follow the electron in front of it.

The electricity can flow through branch 1, **OR** through branch 2, **OR **through branch 3. However, any given electron will only flow through one of the three branches. This means that the current gets divided up.

The total current in a parallel circuit is equal to the sum of the currents in each of the branches.

I_{T} = I_{1} + I_{2} + I_{3} + …

All of the energy being carried by the electrons must be used up by the time the electron returns to the power source. Regardless of the branch chosen, there is only one resistor or load, where the energy can be used.

As a result, the potential difference across each branch in a parallel circuit will be the same.

V_{T} = V_{1} = V_{2} = V_{3} = …

Learn more about parallel circuits.

### Enrichment

Get yourself a drink of your favourite beverage and six drinking straws. Wrap a piece of tape around all six straws to tie them together to make one very thick straw. Insert the thick straw into your beverage and try to drink it. What do you notice? Now carefully remove the tape and connect the straws together in such a way that you make one very long straw out of them. Then set your beverage on the floor, insert the long straw and try to drink it. What do you notice?

Your straws are acting as resistors that resist the flow of your beverage from the cup to your mouth. When the resistors, in this case the straws, are connected in parallel (when they are taped together), there are many pathways from the cup to your mouth. The current, in this case the beverage, can flow through any one of these pathways. This lowers the overall resistance and makes it very easy for the beverage to flow.

When the resistors, or straws, are connected in series (when they are inserted into each other to make a very long straw) there is only one pathway from the cup to your mouth. The current, the beverage, must flow through all of the resistors, the straws, one after another. This increases the overall resistance and makes it very easy difficult for the beverage to flow.

The total resistance in a parallel circuit can be found using the following equation:

#### Example 2: A Parallel Circuit

Determine the potential difference, resistance, and current for each component in the series circuit below.

V_{1} = | I_{1} = | R_{1} = 24 Ω |

V_{2} = | I_{2} = | R_{2} = 8 Ω |

V_{3} = | I_{3} = | R_{3} = 6 Ω |

V_{T} = 24 V | I_{T} = | R_{T} = |

**Given:**

All of the information that you have been given is in the table.

**Required:**

You are required to find all of the values that are not given in the table.

**Solution:**

You will need the following equations to solve the circuit:

V_{T} = V_{1} = V_{2} = V_{3}

I_{T} = I_{1} + I_{2} + I_{3}

V = IR

As values are calculated, they should be entered into the table.

Since the potential difference across each branch is the same, then V_{1}, V_{2}, and V_{3} must all equal 24 V.

V_{1} = 24 V | I_{1} = | R_{1} = 24 Ω |

V_{2} = 24 V | I_{2} = | R_{2} = 8 Ω |

V_{3} = 24 V | I_{3} = | R_{3} = 6 Ω |

V_{T} = 24 V | I_{T} = | R_{T} = |

Ohm’s Law can be used to find I_{1}, I_{2}, and I_{3}.

V_{1} = 24 V | I_{1} = 1 A | R_{1} = 24 Ω |

V_{2} = 24 V | I_{2} = 3 A | R_{2} = 8 Ω |

V_{3} = 24 V | I_{3} = 4 A | R_{3} = 6 Ω |

V_{T} = 24 V | I_{T} = | R_{T} = |

I_{T} can be found using:

I_{T} = I_{1} + I_{2} + I_{3}

I_{T} = 1A + 3 A + 4 A = 8 A

V_{1} = 24 V | I_{1} = 1 A | R_{1} = 24 Ω |

V_{2} = 24 V | I_{2} = 3 A | R_{2} = 8 Ω |

V_{3} = 24 V | I_{3} = 4 A | R_{3} = 6 Ω |

V_{T} = 24 V | I_{T} = 8 A | R_{T} = |

R_{T} can be found using Ohm’s Law:

V_{1} = 24 V | I_{1} = 1 A | R_{1} = 24 Ω |

V_{2} = 24 V | I_{2} = 3 A | R_{2} = 8 Ω |

V_{3} = 24 V | I_{3} = 4 A | R_{3} = 6 Ω |

V_{T} = 24 V | I_{T} = 8 A | R_{T} = 3 Ω |

#### Practice Questions

- Solve the following parallel circuit for all of the values missing from the table.

V_{1} = | I_{1} = | R_{1} = 6 Ω |

V_{2} = | I_{2} = 0.5 A | R_{2} = |

V_{3} = | I_{3} = | R_{3} = 12 Ω |

V_{T} = | I_{T} = | R_{T} = 2 Ω |

**Answer**

**Given:**

All of the information that you have been given is in the table.

**Required:**

You are required to find all of the values that are not given in the table.

**Solution:**

You will need the following equations to solve the circuit:

V_{T} = V_{1} = V_{2} = V_{3}

I_{T} = I_{1} + I_{2} + I_{3}

V = IR

As values are calculated, they should be entered into the table.

Here you have no choice but to solve for R_{2} first by rearranging

R_{2 }= 4Ω

V_{1} = | I_{1} = | R_{1} = 6 Ω |

V_{2} = | I_{2} = 0.5 A | R_{2} = 4 Ω |

V_{3} = | I_{3} = | R_{3} = 12 Ω |

V_{T} = | I_{T} = | R_{T} = 2 Ω |

Since I_{2} and R_{2} are known, Ohm’s Law can be used to find V_{2}.

V_{2} = I_{2}R_{2} = (0.5 A)(4 Ω) = 2 V

And since in a parallel circuit V_{T} = V_{1} = V_{2} = V_{3}, therefore V_{T}, V_{1}, and V_{3} are equal to 2 V.

V_{1} = 2 V | I_{1} = | R_{1} = 6 Ω |

V_{2} = 2 V | I_{2} = 0.5 A | R_{2} = 4 Ω |

V_{3} = 2 V | I_{3} = | R_{3} = 12 Ω |

V_{T} = 2 V | I_{T} = | R_{T} = 2 Ω |

The remaining currents can all be found using Ohm’s Law.

V_{1} = 2 V | I_{1} = 0.333 A | R_{1} = 6 Ω |

V_{2} = 2 V | I_{2} = 0.5 A | R_{2} = 4 Ω |

V_{3} = 2 V | I_{3} = 0.167 A | R_{3} = 12 Ω |

V_{T} = 2 V | I_{T} = 1.0 A | R_{T} = 2 Ω |

This final result should be verified using I_{T} = I_{1} + I_{2} + I_{3}.