# Electric Potential Difference (Voltage). Ohm's Law | iCalculator™

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15.3Electric Potential Difference (Voltage). Ohm's Law

In this Physics tutorial, you will learn:

• What is the scientific definition of electric potential?
• Why potential difference is more important that the electric potential itself?
• Why the potential difference produced by a battery is different from the potential difference produced by a charged object?
• What is the electromotive force?
• What are the common things and the differences between electromotive force and potential difference as concepts?
• What does Ohm's Law for a part of a circuit say?
• What is the general form of Ohm's Law for the whole circuit?

## Introduction

Do you think the amount of extra charges available at any part of a circuit is enough to produce a sustainable electricity in a circuit? Why? What else do you think is necessary for this?

Is there any kind of relationship between the current flowing through a circuit and the resistance provided by the operating devices in that circuit?

What parameters are stamped on the back plate of an electric appliance? Do you know their meaning?

In this tutorial, we will elaborate further the explanation of the potential difference concept discussed in the previous section. In addition, we will relate this concept to other quantities involved in the study of an electric circuit. Therefore, we consider this tutorial very useful in the practical aspect.

## Recap on Electric Potential and Potential Difference of a Charged Object

In the previous chapter (section), we have explained the concept of electric potential produced by a charged object. Electric potential was defined as the electric potential energy per unit charge. The equation of electric potential was

V = k ∙ Q/r

where Q is the charge that produces the electric field and r is the distance from that charge, at which we want to calculate the electric potential V.

We also gave the unit of electric potential, i.e. Volt [V]. Given that the electric potential of a trial charge Q0 located inside the electric field produced by the charge Q at the distance r from it, is

V = W/Q0

where W is the electric potential energy of the trial charge at the given distance, we obtain for the unit of electric potential

[V] = [j/c]

However, the definition of the electric potential given earlier (electric potential energy per unit charge) looks more a mathematical than a physical definition. It does not gives so much info about the importance of electric potential as a quantity. Therefore, in many textbooks, a more practical definition for the electric potential is used. Thus,

"Electric potential V, is the work done by an electric force to bring a positive test charge Q0 from infinity at a given distance r from another positive charge Q."

Now, the purpose of the electric potential existence is much clearer. Electric field produced by a charged object is different at different distances from the object, but it is constant at all points that are at the same distance from the charged object. This property leads to the concept of equipotential surfaces, as shown in the figure below.

From the figure, it is obvious that VB = VC as both these points have the same distance from the charge that generates the field. We say two test charges located at the points B and C are on an equipotential surface.

On the other hand, the point A is on another equipotential surface, which has a different value from that of the points B and C as the distance from the central charge to each of these points is different. As a result, the positive charge at middle of the sketch, which produces the field E, is able to move a positive test charge from A to B (or a negative test charge from B to A). However, it cannot move a charge from B to C or from C to B, as the potential in these points is equal and the work in the direction of the field lines is zero.

Thus, it is clear that a charged object can do work on a test charge only if it changes the value of electric potential of the test charge. In this way, a potential difference ΔV = V2 - V1 is produced, where V1 and V2 are the values of initial and final electric potential of the test charge. If the corresponding distances from the centre of the field are R1 and R2 respectively, we obtain for the potential difference in the two given points:

∆V = V2 - V1
= k ∙ Q/r2 - k ∙ Q/r1
= k ∙ Q ∙ (1/R2 - 1/R1 )

Here, we must consider all possible cases.

1. If a positive test charge comes from the infinity (R1 = ∞) to a distance R2 from the point charge which produces the field, it (the test charge) experiences a potential difference
∆V = k ∙ Q ∙ (1/R2 - 1/)
= k ∙ Q/r2
2. The work done from the central charge in this case is negative as it acts against the field lines. Thus, we have
W = -Q ∙ ∆V
= -Q ∙ V2
3. If a negative test charge comes from the infinity (R1 = ∞) to a distance R2 from the point charge which produces the field, it (the test charge) experiences a potential difference
∆V = -k ∙ Q ∙ (1/R2 -1/)
= -k ∙ Q/r2
4. The work done from the central charge in this case is positive as it acts against the field lines. Thus, we have
W = Q ∙ ∆V
= Q ∙ V2
(the sign minus in the potential difference comes due to the interaction of two opposite charges).
5. If a positive charge is brought from a distance R1 to a distance R2 from the point charge that produces the field, the potential difference experienced by the testy charge is
∆V = k ∙ Q ∙ (1/R2 -1/R1 )
6. If a positive test charge is brought from a distance R1 to a distance R2 from the point charge which produces the field, the potential difference experienced by the test charge is
∆V = -k ∙ Q ∙ (1/R2 - 1/R1 )
= k ∙ Q ∙ (1/R1 - 1/R2 )

### Example 1

A positive test charge is brought from R1 = 3 m to R2 = 1 m away from a 6 μC point charge that produces an electric field around it. What is the potential difference between the two given points? Take k = 9 × 109 N ∙ m2 / C2.

### Solution 1

Using the equation of the potential difference between two points, we obtain

∆V = V2 - V1
= k ∙ Q/r2 - k ∙ Q/r1
= k ∙ Q ∙ (1/R2 - 1/R1 )
= [9 × 109 ∙ 6 × 10-6 (1/1-1/3)]V
= 36000 V

## Potential Difference Produced by a Battery. Electromotive Force

In simple words, potential difference is a change in the charges ability to do work in two different positions of the circuit. Usually, one of these positions is the initial part (input) of any electrical appliance installed in the circuit while the other position is the ending part (output) of the same appliance. In many cases, we assume all appliances as resistors. The potential difference in a circuit is also known as voltage.

Electric charges flow from places of higher potential in places of lower potential. This flow stops when the potential is balanced.

This potential difference is not casual; rather, it is produced by an electric source such as a battery or an electric outlet. Here, we will focus only on the battery, as it produces direct current, which is a one-directional current (it flows in one direction only).

We say a battery is a source of electricity but not because it generates free electrons; rather, a battery simply pushes the electrons of the conducting wire by making them flow throughout the circuit. As a result, a potential difference between the terminals of the battery is produced, which is the main factor that causes the flow of charges throughout the circuit. Without this potential difference, no charge would flow through the circuit, regardless the number of free charges available.

A battery is composed by two or more cells. A cell is a single unit device, which converts chemical energy into electric energy. A battery usually consists of a group of cells as shown in the figure below.

Henceforth we will use the symbol of cell to represent an electric source when dealing with electrical circuits.

By definition, electromotive force (emf) is the energy supplied to the unit charge by a cell or battery (electric source).

Mathematically, we have

emf = Wsource/Q

There is a close similarity between the concepts of potential difference and electromotive force, as both are calculated by dividing energy and charge. (Also, they are both measured in Volts). However, there is a crucial difference between these two concepts. Electromotive force exists only at the source; it represents the maximum potential difference a battery can generate while potential difference can be calculated between any two points of the circuit (usually between the extremities of an electric consumer - a resistor for example - installed in the circuit. Numerically, electromotive force is greater than the actual potential difference as electromotive force represents the maximum potential difference possible.

A number of features for electromotive force and potential difference are given in the table below. In this way, you will be able to understand better the difference between these two concepts.

Let's elaborate further some of the above features. Thus, emf exists even no current is flowing through the circuit as it is an intrinsic feature of the source. For example, a cell has a pre-defined emf = 1.5 V (you can see this value stamped on the back of a cell). However, when the circuit is OFF (open), there is no potential difference between the extremities of a circuit component as no current is flowing through it.

Another feature that requires explication is the fourth one. Emf is always constant (it is pre-defined since the cell has been produced in factory), while potential difference varies in dependence of how many electrical components we have connected in the circuit.

Last, we will discuss in the future chapters the interaction between gravity, electricity and magnetism. Such interactions involve the electromotive force of the source but not the potential difference of various circuit components.

### Example 2

A 1.5 V cell causes a 0.3 C of charge to flow through a circuit. What is the heat energy dissipated by a resistor connected in this circuit if 80% of energy produced by the battery is converted into heat by the resistor?

### Solution 2

First, we calculate the electric energy produced by the battery. This represents the total energy in the circuit (Wsource). Thus, we write

emf = Wsource/Q
Wsource = emf ∙ Q
= 1.5 V ∙ 0.3C
= 0.45 J

Only 80% of this amount is converted into heat by the resistor. Therefore, we have

Heat energy = 80% of Wsource
= 0.80 ∙ 0.45 J
= 0.36 J

As you see, this is a very small value; it means that it is not a good idea to use a cell as an electric source to supply a resistor with energy.

## Ohm's Law for a Part of the Circuit

There is a direct relationship between the potential difference across a circuit component and the current flowing through it. This means higher the potential difference across a component, larger the current flowing through it. We write

∆V ∝ I

Since all circuit components have their own resistance which opposes the current flow, we must include it in the calculations. Resistance acts as a proportionality constant obtain

∆V = I ∙ R

The above formula when rearranged to isolate R, gives the mathematical expression of Ohm's Law:

I = ∆V/R

The definition of Ohm's Law is as follows:

The current passing through a circuit component is directly proportional to the potential difference applied across it.

### Example 3

The potential difference across a 20 Ω resistor is 12 V. What is the current flowing through the resistor?

### Solution 3

Applying Ohm's Law

I = ∆V/R

we obtain for the current flowing through the resistor:

I = 12 V/20 Ω
= 0.6 A

## Ohm's Law for the Whole Circuit

The battery has its own internal resistance, which we denote as r. We can draw this conclusion when we touch a battery immediately after it has been operating. We can feel the hotness of battery, which indicates the conversion of electricity into heat inside the battery.

In addition, from the previous tutorial, we know that a conductor has its own internal resistance, which depends from the type of material, length and thickness of conductor. This causes a drop in the potential difference at the resistor. Look at the figure below.

We can write for the relationship between potential differences and the electromotive force in the above circuit:

emf = ΔVbattery + ΔVAB + ΔVBC + ΔVCD

If we write the total resistance of the circuit by Rtot, we obtain (applying Ohm's Law):

emf=I ∙ r + I ∙ RAB + I ∙ R + I ∙ RCD

Given that RABCD = Rwire, we obtain

emf = I ∙ r + I ∙ R + I ∙ Rw

or

emf = I ∙ (r + R + Rw)

The above equation gives Ohm's Law for the whole circuit. We can write it in a shorter way as

emf = I ∙ Rtot

### Example 4

A 20 m long copper wire of 4 mm2 thickness (ρ = 1.69 × 10-8 Ω ∙ m) is used to build an electrical circuit that contains a 24 V battery and a resistor. The battery has a 0.6 Ω internal resistance. What is the resistance of the resistor if the current flowing through the circuit is equal to 12A? Use the figure shown in the theory as a reference.

### Solution 4

Clues:

L = 20 m
A = 4 mm2 = 4 × 10-6 m2
emf = 24 V
r = 0.6 Ω
I = 12 A
R = ?

First, we find the resistance of wire. We have

Rw = ρ ∙ L/A
= (1.69 × 10-8 Ω ∙ m) ∙ (20m)/4 × 10-6 m2 )
= 8.45 × 10-2 Ω
= 0.0845 Ω

Applying Ohm's Law for the whole circuit, we obtain

emf = I ∙ (r + R + Rw)
r + R + Rw = Rtot = emf/I
= 24 V/12 A
= 2 Ω

Thus,

R = Rtot - r - Rw
= 2 Ω - 0.6 Ω - 0.0845 Ω
= 1.3155 Ω

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