The basic laws of currents governing the electrical networks derived by Gustov Robert Kirchhoff are Kirchhoff’s laws.

This law is the foundation for the systematic analysis of electrical circuits. Kirchhoff’s law defines the relationship between the current and the voltage that must be satisfied within the circuit. In the analysis of the complex network to find the equivalent resistance and the current flowing through various branches, this law is very helpful.

The two laws are; the first law which is Kirchhoff’s Current Law (KCL) and the second law which Kirchhoff’s Voltage Law (KVL).

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## Kirchhoff’s First Law or Kirchhoff’s Current Law (KCL) or Point Law

This law states that in any network the algebraic sum of the currents meeting at a point or junction is equal to zero. In other words, the sum of the currents incoming at any point or junction is equal to the sum of the current leaving the point or the junction.

Consider the figure below. Figure: Current Flowing Through a Node

Here, for the currents I1, I2, I3, I4, I5, and I6, taking the incoming current at the junction ‘O’ as positive and the current leaving the junction ‘O’ as negative. We have, Where, I1, I4, and I5 are incoming currents and are I2, I3, and I6 outgoing currents.

KCL is based on the conservation of the charge as the charge entering a node or junction must leave the node or junction. This is because the charge can neither be created nor destroyed.

## Kirchhoff’s Second Law or Kirchhoff’s Voltage Law (KVL) or Mesh Law

This law states that the sum of the EMFs acting along a closed circuit or mesh is equal to the algebraic sum of the products of the current and the resistances or impedances of each part of the closed circuit.

Consider the circuit shown below. Figure: Electrical Circuit

According to the KVL, the KVL equation for the mesh AFCBA is The KVL equation for the mesh FEDCF is The KVL equation for the mesh AFEDCBA is KVL is based on the conservation of energy as the voltage is the energy or the work done per unit charge. The total energy loss or gain in a system is equal to zero. Similarly, the algebraic sum of the voltages in a closed circuit must also be equal to zero.

## Application of Kirchhoff’s Law

For the application of Kirchhoff’s law in the circuit, firstly we apply the KCL to show the current distribution in the various branches. Secondly, we apply the KVL to each mesh and finally form an algebraic equation for further analysis.

To apply the KCL, we can take the algebraic sign of the currents to be positive for the incoming current and negative for the outgoing current from the node.

For KVL this can be confusing and note the following points:

1. The voltage drop in resistances for the current flowing in the clockwise direction is taken as a positive drop. Such as in the KVL illustration above for mesh AFCBA, the voltage drops in the resistances R1, R2 and R5 are taken positively as the current flowing through them is in a clockwise direction. 1. The voltage drop in resistances for the current flowing in the counter-clockwise direction or anti-clockwise direction is taken as a negative drop. Such as in the KVL illustration above for mesh FEDCF, the voltage drops in the resistances R1, R2, R3, and R4 are taken negatively as the current flowing through them is in a counter-clockwise direction. 1. For the source emf; the source emf which is causing the current to flow in a clockwise direction in a mesh or closed circuit is taken to be positive. For the source emf causing the current to flow in a counter-clockwise direction in a mesh or closed circuit is taken to be negative. Such as in the above KVL illustration, the source emf E1 causes current to flow in a clockwise direction and E1 is taken positive whereas, the source emf E2 causes the currents to flow in the counter-clockwise direction and is taken negatively.

## Example Problems

### Example-1:

Here in the given circuit, we will find the current through the 2 Ω resistor. Figure: Example 1

At first, the KCL has applied in the circuit above and the current distribution is shown below Figure: Circuit Showing Branch Currents

Secondly, we apply KVL

Appling KVL in mesh ABEFA Then applying KVL in mesh BCDEB Further solving equation (i) and (ii) Finally, the current flowing in the 2 Ω resistor is ### Example-2:

Here in the given circuit, we will find the current through the 5 Ω resistor. Figure: Example 2

At first, the KCL has applied in the circuit above and the current distribution is shown below Figure: Circuit Showing Branch Currents

Now we apply KVL

Appling KVL in loop ABFGA Now, the current flowing in the 5 Ω resistor is ENGINEERING NOTES ONLINE: RECIPROCITY THEOREM

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