Kirchhoff's Law

Ever wondered how electrical circuits follow specific rules to function efficiently? Whether designing power grids or working with simple electronic devices, understanding electrical circuits is crucial. This is where Kirchhoff’s Laws come into play.

These fundamental laws help engineers and students analyze and solve complex electrical circuits by determining current and voltage relationships. But what exactly are Kirchhoff’s Laws, and why are they essential in circuit theory? This article delves into Kirchhoff's Laws, their uses, and step-by-step procedures for solving circuit problems.

What are Kirchhoff’s Laws?

Kirchhoff's Laws, derived by Gustav Kirchhoff in 1845, are two basic rules applied in the analysis of a circuit:

• Kirchhoff’s Current Law (KCL) – Alternatively referred to as the Node Rule, it states that the current entering a junction (node) in an electrical circuit is equal to the current leaving the junction.
• Kirchhoff's Voltage Law (KVL) – Also referred to as the Loop Rule, it states that the sum of the entire voltage rises and drops through a closed path in a circuit equals zero.

The laws rest upon the theory of conservation of charge (KCL) and conservation of energy (KVL).

Kirchhoff’s Current Law (KCL)

The algebraic sum of currents entering and leaving at any node of a circuit is zero.

Mathematically: ∑Iin = ∑Iout

Example: Suppose that three currents I1, I2, and I3 flow into a node and two currents I4 and I5 flow out, then:

I1 + I2 + I3 = I4 + I5

Application: Applied in solving complex circuits having many branches.

Kirchhoff’s Voltage Law (KVL)

The algebraic sum of voltages in a closed loop must be zero.

Mathematically: ∑V = 0

Example: In a loop with a battery (V) and resistors (R1, R2), the sum of voltage drops (Ohm’s Law: V = IR) must equal the supplied voltage.

V − IR1 − IR2 = 0

Application: Used to determine unknown voltages and currents in looped circuits.

Examples of Kirchhoff’s Laws

Example 1: At a circuit node, three currents enter: 5A, 3A, 2A, and two currents exit: I1, 6A. Find I1.

Solution: 5A + 3A + 2A = I1 + 6A

10A = I1 + 6A

I1 = 4A

Example 2: A circuit loop has a 12V battery, two resistors 3Ω and 5Ω, and an unknown current I. Find I.

Solutions: Using KVL: 12 V − 3Ω(I) − 5Ω(I) = 0

12 V = 8Ω(I)

I = 1.5 A