**Ohm's Law:** Resistance is one of the foundational principles of Ohm’s law and can be found in virtually any device used to conduct electricity. GS Ohm was a German physicist who conducted some experiments on electricity. His discovery of the relationship between current, voltage, and resistance is the fundamental law of current flow, and the formula that connects these three measurements is named in his honour.

## Ohm's Law Explanation

A foundation law regarding the flow of currents was discovered by G.S. Ohm before the discovery long before the physical mechanism responsible for the flow of currents in the year 1828. Let us suppose, a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor.

Then Ohm’s law states that,

**V ∝ I**

or, **V = R I**

Where,

**R is the proportionality constant (called Resistance)**

**The SI unit of resistance is ohm and is denoted by the symbol Ω.**

**So, According to Ohm's law, " The voltage of the conductor is directly proportional to the current flowing through that conductor.**

Three Quantities Related to Ohm's Law | |||

Quantity | Symbol | Unit | Unit Symbol |

Current | I | Ampere | A |

Voltage | E or V | Volt | V |

Resistance | R | Ohm | Ω |

## Ohm’s Law: Relationship Between Voltage, Resistance, and Current

In the early 19th century, German physicist Ohm discovered that the current passing through a conductor is directly proportional to the voltage across the conductor. His discovery came to be known as Ohm’s Law, which describes the relationship between voltage (V), current (I) and resistance (R):

**V = I R, ( For calculating Voltage)**

**I = V/R, (For calculating Current)**

**R= V/I (For calculating Resistance)**

## Ohm's Law Formula Wheel

The Ohm’s law formula wheel helps to show the calculations that are required to solve for voltage, current, resistance, as well as power. Power is the rate at which work is done. To use the formula wheel, it is required to know two variables to solve for the third. For instance, if you know the current and resistance in a circuit, you can calculate the power produced by squaring the current and multiplying that by the resistance: I² x R.

## Experimental Verification of Ohm's Law

Suppose A conductor of resistance (R) ohms is connected in series with an ammeter (A), rheostat (variable resistance) and a battery (B)B through key K(K). A voltmeter (V) is connected across the ends of the resistance which helps to measure the potential difference across it.

Then a key (K) is pressed and readings of the voltmeter and ammeter are recorded. After reverting, the key (K) the setting of the rheostat is changed, Key(K) is again pressed and voltmeter and ammeter readings are recorded. Thus, for various settings of a rheostat, the reading of the voltmeter and ammeter are noted. From the monitoring, it can be seen that the ratio of voltmeter reading and ammeter reading always remains the same every time i.e.

V1/I1 = V2/I1 = V3/I3 = V4/I4 = ……… = Constant = R( Resistance of the Conductor)

## Application for Ohm's Law

The application for Ohm's law is-

- In fuses
- To know the power consumption.
- To control the speed of fans.
- For deciding the size of resistors

## Limitations of Ohm's Law

Some of the limitations of Ohm's law are-

- Ohm’s law is not applicable to Unilateral electrical elements such as diodes and transistors as the following of current is only in one direction only.
- Ohm's law is applicable only to Ohmic Conductors not to non-ohmic conductors.

## Solved Examples

**Example 1: If the resistance of an electric coil is 60 Ω and a current of 3.2 A flows through the resistance. Find the voltage between two points.**

**Solution: **According to the question, Resistance = 60 Ω

Current = 3.2 A

So in order to calculate the voltage we can use the formula, V = IR

**V = I × R****V = 3.2 A × 60 Ω = 192 V**

So the Voltage between the two points in the electric coil is 192 V**.**

**Example 2: An EMF source of 10.0 V is connected to a purely resistive electrical appliance (a light bulb). An electric current of 5.0 A flows through the bulb. Consider the conducting wires to be resistance-free. Calculate the resistance of the appliance.**

**Solution: **According to the question, the Voltage of the source = 10V

Current = 5.0 A

So in order to calculate the resistance we can use the formula, R= V/I

**R = V/I****R = 10 V/5 A **

**R= 2Ω**

**Example3: If the resistance of an electric iron is 150 Ω and the current flowing is 5.2 A through the resistance. Find the voltage between two points.**

**Solution: **According to the question, Resistance is 150 Ω,

The current flowing is 5.2 A

So in order to calculate the voltage we can use the formula, V = IR

V = 5.2 x 150

**V= 780Ω**

**Example 4: Find the current I through a resistor of resistance R = 9Ω if the voltage across the resistor is 18 V.**

**Solution: **According to the question, Resistance is = 9Ω

and the Voltage across the resistor is 18V

So in order to calculate the voltage we can use the formula, **I = V/R**

I = 18/9

**I = 2A**

**Example5: In the circuit below resistors R1 and R2 are in series and have resistances of 10Ω and 15 Ω, respectively. The voltage across resistor R1 is 5 V. Calculate the current passing through resistor R2 and the voltage across resistor R2.**

**Solution: **According to the question, R1 = 10Ω

R2 = 15 Ω

Voltage across R1 = 5V

So let us first calculate the current(I) across the R1.

So, I = V/R

I = 5/10

I1 = 0.5 A

**We got the current for R1, that is, 0.5 A**

So, when two resistors are connected in series then the current flowing through the R1 will be the same across R2.

**I1 = I2 = 0.5A**

Let us calculate the resistance across the resistor

V2 = I2R2

V2 = 0.5 X 15

**V2= 7.5 volt **

Related Links- | |

Boyles's Law | Newton's Law of Motion |

Charles Law | Newton's First Law of Motion |

Newton's Second Law of Motion | Newton's Third Law of Motion |