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Physics Lesson 15.4.4 - Series and Parallel Circuits

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Welcome to our Physics lesson on Series and Parallel Circuits, this is the fourth lesson of our suite of physics lessons covering the topic of Electric Circuits. Series and Parallel Circuits. Short Circuits, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Series and Parallel Circuits

Now, let's extend the concept of series and parallel combination of resistors to include the other quantities such as potential difference and current as well. Before continuing with numerical examples, first let's prove the correctness of the two formulae used in the combination of resistors which we have provided in article 14.2.

Let's consider two resistors connected in series in a circuit, as shown in the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Circuits. Series and Parallel Circuits. Short Circuits

Since there is the same current I flowing through both resistors, we can write

Itot = I1 = I2 = I

Also, we have

emf = ∆V1 + ∆V2

Applying Ohm's Law, in the last equation, we obtain

I ∙ Req = I ∙ R1 + I ∙ R2

Simplifying the current I from both sides, we obtain the known formula of series combination of resistors

Req = R1 + R2

Example 1

What is the value of resistance R2 in the circuit shown below? Neglect the resistance of conductor and that of battery.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Circuits. Series and Parallel Circuits. Short Circuits

Solution 1

We have the following clues:

emf = 24V
I = 3A
R1 = 7Ω
R2 = ?

First, we calculate the total (equivalent) resistance using Ohm's Law formula for the whole circuit. We have:

Req = emf/I
= 24V/3A
= 8Ω

Since the resistors are connected in series, we have:

Req = R1 + R2
R2 = Req - R1
= 8Ω - 7Ω
= 1Ω

Now, let's consider two resistors connected in parallel, as shown in the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Circuits. Series and Parallel Circuits. Short Circuits

The current I divides in the two branches, and the corresponding values of current in each branch are I1 and I2 respectively. The potential difference in each resistor is the same as that in the entire branch. If the resistor of wire and that of battery are neglected, we obtain:

emf = ∆V1 = ∆V2 = ∆V

Also, the equation for currents is

I = I1 + I2

Applying Ohm's Law in the last equation, we obtain

emf/Req = ∆V1/R1 + ∆V2/R2
∆V/Req =∆V/R1 + ∆V/R2

Simplifying ΔV from both sides, we obtain for parallel combination of resistors the known equation:

1/Req =1/R1 + 1/R2

Example 2

What is the current in the main branch and the current flowing in each parallel branch of the circuit shown below? Use two methods for solving this exercise. Physics Tutorials: This image provides visual information for the physics tutorial Electric Circuits. Series and Parallel Circuits. Short Circuits

Solution 2

Method 1

This method consist in calculating the equivalent resistance, which helps finding the current in the main branch first. Then, since potential difference in each branch of the parallel setup is considered as equal the electromotive force of battery, we uses Ohm's Law to calculate the current in each resistor.

The equivalent resistance of the (parallel) circuit is:

1/Req = 1/R1 + 1/R2
= 1/6 + 1/12
= 2/12 + 1/12
= 3/12

Thus,

Req = 12/3 Ω
=4Ω

Now, let's calculate the current I in the main branch using Ohm's Law. We have

I = emf/Req
= 18 V/4 Ω
= 4.5 A

Since the electromotive force of battery corresponds to the potential difference in each wire of the parallel branch, we obtain for the currents I1 and I2 flowing in these wires:

I1 = emf/R1
= 18 V/6 Ω
= 3 A

and

I2 = emf/R1
= 18 V/12 Ω
= 1.5 A

Method 2

This method consists in finding the currents in each wire directly by applying Ohm's Law. Then, we add these currents to find the current in the main branch. Thus, we have:

I1 = emf/R1
= 18 V/6 Ω
= 3A

and

I2 = emf/R2
=18 V/12 Ω
= 1.5A

Therefore, the current in the main branch is

I = I1 + I2
= 3A + 1.5A
= 4.5A

As you can see, the results are the same in both cases.

You have reached the end of Physics lesson 15.4.4 Series and Parallel Circuits. There are 5 lessons in this physics tutorial covering Electric Circuits. Series and Parallel Circuits. Short Circuits, you can access all the lessons from this tutorial below.

More Electric Circuits. Series and Parallel Circuits. Short Circuits Lessons and Learning Resources

Electrodynamics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
15.4Electric Circuits. Series and Parallel Circuits. Short Circuits
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
15.4.1Electrical Circuits. Circuit Components
15.4.2Sources of Electricity
15.4.3Combination of Cells
15.4.4Series and Parallel Circuits
15.4.5Short Circuits

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