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KIRCHHOFF’S CURRENT LAW EXAMPLE PROBLEMS WITH SOLUTIONS

Kirchhoff’s Current Law (KCL) is a fundamental principle in electrical engineering that states that the total current entering a junction in a circuit must equal the total current leaving the junction. This law is essential for analyzing complex electrical circuits and is used to solve a variety of problems in circuit analysis. In this article, we will explore some example problems that demonstrate the application of Kirchhoff’s Current Law and provide solutions to help you better understand this concept.

Problem 1: Simple Circuit

Consider the following circuit:

Apply Kirchhoff’s Current Law at the junction point A:

Let I1 be the current flowing through the 6-ohm resistor.
Let I2 be the current flowing through the 4-ohm resistor.

According to Kirchhoff’s Current Law:

I1 = I2

Now, apply Ohm’s Law to find the values of I1 and I2:

V = I1 * 6 ohms

V = I2 * 4 ohms

Substitute the values of V, I1, and I2 to solve for the currents:

6V = 4V

2V = 0

V = 0

Therefore, the current flowing through both resistors is 0A.

Problem 2: Parallel Circuit

Consider the following parallel circuit:

Apply Kirchhoff’s Current Law at the junction point A:

Let I1 be the current flowing through the 10-ohm resistor.
Let I2 be the current flowing through the 5-ohm resistor.

According to Kirchhoff’s Current Law:

I1 + I2 = I

Apply Ohm’s Law to find the values of I1 and I2:

V = I1 * 10 ohms

V = I2 * 5 ohms

Substitute the values of V, I1, and I2 to solve for the currents:

10V = 5V

5V = 0

V = 0

Therefore, the total current I flowing through the circuit is 0A.

Problem 3: Complex Circuit

Consider the following complex circuit:

Apply Kirchhoff’s Current Law at the junction points A, B, and C:

Let I1 be the current flowing through the 8-ohm resistor.
Let I2 be the current flowing through the 6-ohm resistor.
Let I3 be the current flowing through the 4-ohm resistor.

According to Kirchhoff’s Current Law:

I1 = I2 + I3

Apply Ohm’s Law to find the values of I1, I2, and I3:

V = I1 * 8 ohms

V = I2 * 6 ohms

V = I3 * 4 ohms

Substitute the values of V, I1, I2, and I3 to solve for the currents:

8V = 6V + 4V

8V = 10V

-2V = 0

V = 0

Therefore, the current flowing through the resistors in the circuit is 0A.

Summary

In conclusion, Kirchhoff’s Current Law is a powerful tool for analyzing electrical circuits and solving complex problems. By applying this law at junction points in a circuit, you can determine the currents flowing through different branches of the circuit. Understanding Kirchhoff’s Current Law is essential for electrical engineers and technicians working with circuits.