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Understanding Physics Topics is essential for solving complex problems in many fields, including engineering and medicine.

## Ohm’s Law Explanation and Verification

Ohm’s law gives a relationship between current and potential difference. According to Ohm’s law : At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. If I is the current flowing through a conductor and V is the potential difference (or voltage) across its ends, then according to Ohm’s law :

I ∝ V (At constant temp.)

This can also be written as : V ∝ I

or V = R × I

where R is a constant called “resistance” of the conductor. The value of this constant depends on the nature, length, area of cross-section and temperature of the conductor. The above equation can also be written as :

\(\frac{V}{I}\) = R ……………… (1)

where V = Potential difference

I = Current

and R = Resistance (which is a constant)

The above equation is a mathematical expression of Ohm’s law. Equation (1) can be written in words as follows :

\(\frac{\text { Potential difference }}{\text { Current }}\) = constant (called resistance)

We find that the ratio of potential difference applied between the ends of a conductor and the current flowing through it is a constant quantity called resistance.

We have just seen that : \(\frac{V}{I}\) = R

or \(\frac{V}{R}\) = I

So, Current, I = \(\frac{V}{R}\)

It is obvious from this relation that:

- the current is directly proportional to potential difference, and
- the current is inversely proportional to resistance.

Since the current is directly proportional to the potential difference applied across the ends of a conductor, it means that if the potential difference across the ends of a conductor is doubled, the current flowing through it also gets doubled, and if the potential difference is halved, the current also gets halved.

On the other hand, the current is inversely proportional to resistance. So, if the resistance is doubled, the current gets halved, and if the resistance is halved, the current gets doubled. Thus, the strength of an electric current in a given conductor depends on two factors :

- potential difference across the ends of the conductor, and
- resistance of the conductor.

We will now discuss the electrical resistance of a conductor in detail.

### Resistance of a Conductor

The electric current is a flow of electrons through a conductor. When the electrons move from one part of the conductor to the other part, they collide with other electrons and with the atoms and ions present in the body of the conductor.

Due to these collisions, there is some obstruction or opposition to the flow of electron current through the conductor. The property of a conductor due to which it opposes the flow of current through it is called resistance. The resistance of a conductor is numerically equal to the ratio of potential difference across its ends to the current flowing through it. That is :

Resistance = \(\frac{\text { Potential difference }}{\text { Current }}\)

or R = \(\frac{V}{I}\)

The resistance of a conductor depends on length, thickness, nature of material and temperature, of the conductor. A long wire (or conductor) has more resistance and a short wire has less resistance. Again, a thick wire has less resistance whereas a thin wire has more resistance. Rise in temperature of a wire (or conductor) increases its resistance.

The SI unit of resistance is ohm which is denoted by the symbol omega, Q. The unit of resistance ohm, can be defined by using Ohm’s law as described below.

According to Ohm’s law :

\(\frac{\text { Potential difference }}{\text { Current }}\) = Resistance (A constant)

That is, \(\frac{V}{I}\) = R

So, Resistance, R = \(\frac{V}{I}\)

Now, if the potential difference V is 1 volt and the current I is 1 ampere, then resistance R in the above equation becomes 1 ohm.

That is, 1 ohm = \(\frac{1 \text { volt }}{1 \text { ampere }}\)

This gives us the following definition for ohm : 1 ohm is the resistance of a conductor such that when a potential difference of 1 volt is applied to its ends, a current of 1 ampere flows through it. We can find out the resistance of a conductor by using Ohm’s law equation \(\frac{V}{I}\) = R. This will become more clear from the following examples.

**Example Problem 1.**

Potential difference between two points of a wire carrying 2 ampere current is 0.1 volt, Calculate the resistance between these points.

**Solution:**

From Ohm’s law we have :

\(\frac{\text { Potential difference }}{\text { Current }}\) = Resistance

or \(\frac{V}{I}\) = R

Here, Potential difference, V = 0.1 volt

Current, I = 2 amperes

And, Resistance, R = ? (To be calculated)

Putting these values in the above formula, we get:

\(\frac{0.1}{2}\) = R

0.05 = R

or Resistance, R = 0.05 ohm (or 0.05 Ω)

**Example Problem 2.**

A simple electric circuit has a 24 V battery and a resistor of 60 ohms. What will be the current in the circuit ? The resistance of the connecting wires is negligible.

**Solution:**

In this case :

Potential difference, V = 24 volts

Resistance, R = 60 ohms

And, Current, I = ? (To be calculated)

Now, putting these values in the Ohm’s law equation :

\(\frac{V}{I}\) = R

we get, \(\frac{24}{I}\) = 60

So, 60 I = 24

And, I = \(\frac{24}{60}\) ampere

I = 0.4 ampere (or 0.4 A)

Thus, the current flowing in the circuit is 0.4 ampere.

**Example Problem 3.**

An electric iron draws a current of 3.4 A from the 220 V supply line. What current will this electric iron draw when connected to 110 V supply line ?

**Solution:**

First of all we will calculate the resistance of electric iron. Now, in the first case, the electric iron draws a current of 3.4 A from 220 V supply line. So,

Potential difference, V = 220 V

Current, I = 3.4 A

And, Resistance, R = ? (To be calculated)

Now, \(\frac{V}{I}\) = R

So, \(\frac{220}{3.4}\) = R

Resistance, R = 64.7 Ω

Thus, the resistance of electric iron is 64.7 ohms. This resistance will now be used to find out the current drawn when the electric iron is connected to 110 V supply line. So,

\(\frac{V}{I}\) = R

\(\frac{110}{I}\) = 64.7

Current, I = \(\frac{110}{64.7}\)

= 1.7 A

Thus, the electric iron will draw a current of 1.7 amperes from 110 volt supply line.

### Graph Between V and I

If a graph is drawn between the potential difference readings (V) and the corresponding current values (I), the graph is found to be a straight line passing through the origin (see Figure). A straight line graph can be obtained only if the two quantities are directly proportional to one another.

Since the ‘current- potential difference’ graph is a straight line, we conclude that current is directly proportional to the potential difference. It is clear from the graph OA that as the potential difference V increases, the current I also increases, but the ratio \(\frac{V}{I}\) remains constant. This constant is called resistance of the conductor. We will now solve one problem based on the graph between V and I.

Example Problem.

The values of current I flowing through a coil for the corresponding values of the potential difference V across the coil are shown below :

Plot a graph between V and I and calculate the resistance of the coil.

Solution:

We take a graph paper and mark the potential difference (V) values of 1, 2, 3, 4, 5, 6 and 7 on the Y-axis. The current (I) values of 0.1, 0.2, 0.3 and 0.4 are marked on the y-axis (see below Figure ).

- On plotting the first reading of 0.85 V on x-axis and 0.05 A on the y-axis, we get the point A on the graph paper (see Figure 16)
- On plotting the second reading of 1.70 V on the x-axis and 0.10 A on the y-axis, we get a second point B on the graph paper.
- On plotting the third reading of 3.5 V on the x-axis and 0.20 A on the y-axis, we get a third point C on the graph paper.
- On plotting the fourth reading of 5.0 V on x-axis and 0.30 A on the y-axis, we get a fourth point D on the graph paper.
- And on plotting the fifth reading of 6.8 V on x-axis and 0.4 A on the y-axis, we get a fifth point E on the graph paper.

Let us now join all the five points A, B, C, D and E. In this way we get a straight-line graph between V and I. This straight-line graph shows that current (I) is directly proportional to the potential difference (V). And this conclusion proves Ohm’s law.

Let us calculate the resistance now. If we look at the above graph, we find that at point E, potential difference (V) is 6.8 volts whereas the current (I) is 0.4 amperes. Now, we know that :

Resistance, R = \(\frac{V}{I}\)

So, R = \(\frac{6.8}{0.4}\)

R = 17 ohms

Thus, the resistance is of 17 ohms.

### Experiment to Verify Ohm’s Law

If we can show that for a given conductor, say a piece of resistance wire (such as a nichrome wire), the ratio \(\frac{\text { Potential difference }}{\text { Current }}\) is constant, then Ohm’s law will get verified. Alternatively, we can draw a graph between the potential difference (V) and current (I), and if this graph is a straight line, even then Ohm’s law gets verified. Let us see how this is done in the laboratory.

Suppose we have a piece of resistance wire R (which is the conductor here) (below Figure), and we want to verify Ohm’s law for it, that is, we want to show that the conductor R obeys Ohm’s law. For this purpose we take a battery (B), a switch (S), a rheostat (Rh), an ammeter (A), a voltmeter (V) and some connecting wires. Using all these and the conductor R we make a circuit as shown in below Figure.

To start the experiment, the circuit is completed by pressing the switch S. On pressing the switch, a current starts flowing in the whole circuit including the conductor R. This current is shown by the ammeter. The rheostat Rh is initially so adjusted that a small current passes through the circuit. The ammeter reading is now noted. This reading gives us the current I flowing through the conductor R.

The voltmeter reading is also noted which will give the potential difference V across the ends of the conductor. This gives us the first set of V and I readings. The current in the circuit is now increased step by step, by changing the position of the sliding contact C of the rheostat. The current values and the corresponding potential difference values are noted in all the cases. The ratio or \(\frac{\text { Potential difference }}{\text { Current }}\) or \(\frac{V}{I}\) is calculated for all the readings.

It is found that the ratio T has constant value for all the observations. Since the ratio of potential difference and current, \(\frac{V}{I}\) is constant, Ohm’s law gets verified because this shows that the current is directly proportional to potential difference. The constant ratio \(\frac{V}{I}\) gives us the resistance R of the conductor.

So, this Ohm’s law experiment can also be used to determine the resistance of a conductor. If a graph is drawn between potential difference readings and corresponding current readings, we will get a straight line graph showing that current is directly proportional to potential difference. This also verifies Ohm’s law.

### Good Conductors, Resistors and Insulators

On the basis of their electrical resistance, all the substances can be divided into three groups : Good conductors, Resistors and Insulators. Those substances which have very low electrical resistance are called good conductors. A good conductor allows the electricity to flow through it easily. Silver metal is the best conductor of electricity. Copper and aluminium metals are also good conductors.

Electric wires are made of copper or aluminium because they have very low electrical resistance (see Figures). Those substances which have comparatively high electrical resistance, are called resistors. The alloys like nichrome, manganin and constantan (or eureka), all have quite high resistances, so they are called resistors. Resistors are used to make those electrical devices where high resistance is required (see Figures).

A resistor reduces the current in a circuit. Those substances which have infinitely high electrical resistance are called insulators. An insulator does not allow electricity to flow through it. Rubber is an excellent insulator. Electricians wear rubber handgloves while working with electricity because rubber is an insulator and protects them from electric shocks (see Figures). Wood is also a good insulator.