**Ohm's law**is one of the most fundamental principles in the field of electrical engineering and physics. This law explains the relationship between voltage, current, and resistance in a circuit. Ohm's law was first discovered and formulated by the German physicist

**Georg Simon Ohm**in the early 19th century. Since then this law has been widely used in various fields.

Understanding Ohm's Law is important for those interested in pursuing a career in the field of electrical engineering or physics. In this article, we will discuss what Ohm's law is, its significance, and how this law can be applied in practical situations. We will also cover various formulas used to calculate voltage, current and resistance values in circuits.

**What is Ohm's Law?**

**Statement:**Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, but the temperature and other physical conditions of the conductor remain constant.

In other words, the greater the voltage applied across a circuit, the greater the current that flows through it. So the mathematical expression of this law is written as

*V = IR*

Where

*I*is the current flowing through the conductor,*V*is the voltage across the conductor, and*R*is the resistance of the conductor.**The hydraulic analogy of Ohm's Law**

The hydraulic analogy of Ohm's Law is a way of visualizing the relationship between voltage, current, and resistance in an electrical circuit using water flowing through pipes.

In this analogy, voltage is represented by the pressure of water, the current is represented by the rate of flow of water, and resistance is represented by the constriction produced by the diameter of the pipes.

Just like how the rate of flow of water increases when the pressure is increased, the current in an electrical circuit increases when the voltage is increased.

Similarly, just as how the rate of flow of water decreases when the diameter of the pipes is decreased, the current in an electrical circuit decreases when the resistance is increased.

Using this analogy, we can clearly understand how Ohm's Law works in an electrical circuit. Just as how the rate of flow of water can be calculated using the pressure and the diameter of the pipes.

Similarly, the current flowing through the electrical circuit can also be calculated using the voltage and the resistance.

**Formulas to calculate Voltage, Current, and Resistance**

Below are some simple formulas, which we can use to calculate voltage, current and resistance values in a circuit.

*1. V = IR*

This is the most commonly used formula for Ohm's Law. It states that the voltage (V) across a conductor is directly proportional to the current (I) flowing through the conductor and the resistance (R) of the conductor.

*2. I = V/R*

This formula is a rearrangement of the first formula and states that the current (I) flowing through a conductor is directly proportional to the voltage (V) applied across the conductor and inversely proportional to the resistance (R) of the conductor.

*3. R = V/I*

This formula is another rearrangement of the first formula and states that the resistance (R) of a conductor is directly proportional to the voltage (V) applied across the conductor and inversely proportional to the current (I) flowing through the conductor.

*4. P = IV*

This formula is used to calculate the power (P) dissipated in a circuit. It states that the power (P) dissipated is equal to the product of the current (I) flowing through the circuit and the voltage (V) applied across the circuit.

*5. P = I²R*

This formula is another way to calculate the power (P) dissipated in a circuit. It states that the power (P) dissipated is equal to the product of the square of the current (I) flowing through the circuit and the resistance (R) of the circuit.

*6. P = V²/R*

This formula is yet another way to calculate the power (P) dissipated in a circuit. It states that the power (P) dissipated is equal to the product of the square of the voltage (V) applied across the circuit and the inverse of the resistance (R) of the circuit.

**Limitations of Ohm's Law**

Although Ohm's law is widely applicable and useful, it has some limitations which are as follows:

**1. Non-ohmic materials:**Ohm's law applies only to materials that exhibit a linear relationship between voltage and current, such as metals, which are called ohmic. Non-ohmic materials such as diodes, transistors, and capacitors do not obey Ohm's law.

**2. Temperature:**The temperature of a conductor can also affect its resistance and therefore, the current flowing through it. As a result, Ohm's Law is only applicable to materials whose temperature remains constant.

**3. Frequency:**At high frequencies, the resistance of a conductor can change, and Ohm's Law may no longer be applicable.

**4. Power:**Ohm's Law only applies to resistive loads, which dissipate power as heat. It is not applicable to reactive loads, such as capacitors and inductors, which store and release energy.

**Ohm’s Law Solved Problems**

Now let's take some numerical to understand Ohms law better -

**Q1: A circuit has a resistance of 5 ohms and a voltage of 10 volts. What is the current flowing through the circuit?**

**Solution:**

Using Ohm's Law, we can calculate the current flowing through the circuit as follows:

*I = V/R*

*I = 10 V / 5 Ω*

*I = 2 A*

So the current flowing through the circuit is 2 amps.

**Q2: A circuit has a resistance of 100 ohms and a current of 0.5 amps. What is the voltage across the circuit?**

**Solution:**

Using Ohm's Law, we can calculate the voltage across the circuit as follows:

*V = IR*

*V = 0.5 A*

*×*

*100 Ω*

*V = 50 V*

So the voltage across the circuit is 50 volts.

**Q3: A resistor has a resistance of 30 ohms and is connected to a battery with a voltage of 12 volts. What is the power being dissipated in the resistor?**

**Solution:**

Using Ohm's Law and the formula for power, P = VI, we can calculate the power being dissipated in the resistor as follows:

*I = V/R*

*I = 12 V / 30 Ω*

*I = 0.4 A*

*Now, P = VI*

*P = 12 V*

*×*

*0.4 A*

*P = 4.8 W*

So the power being dissipated in the resistor is 4.8 watts.

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