Contents

__Ohm’s Law Definition__

__Ohm’s Law Definition__

Georg Simon ohm a Germen physicist developed a relationship between current in conductor, voltage across the ends of a conductor and resistance of conductor, which is called ohm’s Law. It states current in a conductor is directly proportional to the potential difference between the terminals of the conductor, provided all physical conditions and temperature remain constant.

__Explanation of Ohm’s Law__

__Explanation of Ohm’s Law__

Consider a conductor as shown in figure. Let “v” is the potential difference across the terminals of the conductor. Then current (I) in the conductor according to ohm’s law is \\ I \propto V\\I = \frac{V}{R}\;\;\;\;\;\;\;(1)

Where 1/R is constant of proportionality, “R” is called Resistance, having SI unit ohm (Ω).\\The equation (1) can rewritten as V = IR or R = \frac{V}{I}

__Analogy for ohm’s Law__

__Analogy for ohm’s Law__

Ohm’s law is basically a relationship between voltage, current and resistance. This can be better understand by water-pipe analogy.

In this analogy the water pressure is equivalent to voltage, amount of water flowing is representing amount of current and size of pipe is representing the resistance. More water (current) will flow through the pipe when more pressure (Voltage) is applied and the bigger the pipe, (lower the resistance).

__Experimental Verification of ohm’s Law__

__Experimental Verification of ohm’s Law__

To Verify the ohm’s law in a laboratory, the apparatus needed are

- Resistor
- Ammeter
- Voltmeter
- Variable voltage source
- switch

Connect the apparatus as shown in figure

__Procedure__

- initially keep the switch in open position, set the variable voltage source at particular value and note down the value from voltmeter.
- Close the switch and note reading from ammeter which gives magnitude of current
- Now open switch again and change the value of voltage and note down the value.
- Close the switch again and note reading from ammeter.
- Repeat the steps from 1 to 4 for different voltages.
- After taking readings of currents at different voltages. Find the ratio of V/I, you will come to notice that it is the same for each case.
- Plot a graph of the current against the potential difference, it will be a straight line. This shows that the current is proportional to the potential difference.

## Observation Table

S.No | Voltage (V) | Current (I) | Resistance (R=V/I) |
---|---|---|---|

1 | |||

2 | |||

3 |

__Applicability of ohm’s Law__

__Applicability of ohm’s Law__

Ohm’s law is an empirical law which is found true for maximum experiments but not for all. Ohm’s law is applicable to only linear device where the volt-ampere characteristics are linear.

Ohm’s law can also be applied where volt – amper characteristics are not linear (e.g. incandescent lamp) provided the ohm’s law is expressed in differential form.\\di=\frac{dV}{R}\\Ohm’s law is not applicable to non-linear device such as diodes, transistors, Vacuum tubes etc.

## Calculating Electrical Power Using Ohm’s Law

The rate at which energy is converted from the electrical energy to some other form of energy like mechanical energy, heat, magnetic fields or energy stored in electric fields, is known as electric power. The unit of power is the watt, denoted by “W”. The electrical power can be calculated using the Ohm’s law.

P = VI

P = I^{2}R\;\;\;\;\;\;\;\;\;\;(\because V = IR)

## Ohm’s Law solved problems

Q1: Find the current I through a resistor of resistance R = 2 Ω if the voltage across the resistor is 6 V.\\Solution: Given data\\Voltage (V) = 6V\\Resistance (R) = 2 Ω \\Find Current (I)

According to ohm’s Law

V = IR\\\;\\I=\frac{V}{R}\\\;\\I= \frac{6}{2}=3A

Q2: The current passing through a resistor in a circuit is 0.01 A when the voltage across the same resistor is 5 V. What current passes through this resistor when the voltage across it is 7.5 V?

Given Data

Voltage ( V_{1} ) = 5V and Current (I_{1}) = 0.01A

Voltage ( V_{2} ) = 7.5 V.

Find current ( I_{2} )

First we will find the Resistance of Resistor.

According to ohm’s Law

V = IR

R = V_{1}/I_{1} = 5/0.01 = 500 Ω

Now

V_{2} = I_{2} R

I_{2} = V_{2} / R = 7.5/500 = 0.015A

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