# KIRCHHOFF'S CURRENT LAW

In this post, I will try to teach you perfectly about the Kirchhoff’s Current Law. The things that we are going to discuss here are given below.

• INTRODUCTION
• STATEMENT
• EXAMPLE
• PRACTICE PROBLEMS ## INTRODUCTION

We will now consider the law named for Gustav Robert Kirchhoff (two h’s and two f’s), a German university professor who was born about the time when Ohm was doing his experimental work. This axiomatic law is known as Kirchhoff’s Current Law (abbreviated KCL).

## STATEMENT

“The algebraic sum of all the currents entering any node is equal to zero”

This law represents a mathematical statement of the fact that charge cannot accumulate at a node. A node is not a circuit element, and it certainly cannot store, destroy, or generate charge. Hence, the currents must sum to zero. A hydraulic analogy is sometimes useful here:

for example, consider three water pipes joined in the shape of a Y. We define three “currents” as flowing into each of the three pipes. If we insist that water is always flowing, then obviously we cannot have three positive water currents, or the pipes would burst. This is a result of our defining currents independent of the direction that water is actually flowing.

Therefore, the value of either one or two of the currents as defined must be negative.

A compact expression for Kirchhoff’s current law is which is just a shorthand statement for

i1 + i2 + i3 +……..+ iN = 0

## EXAMPLE Consider the node shown in the figure below. The algebraic sum of the four currents entering the node must be zero:

iA + iB + (-iC) + (-iD) = 0

It is evident that the law could be equally well applied to the algebraic sum of the currents leaving the node:

( -iA )+ ( -iB )+ ( iC ) + ( iD ) = 0

We might also wish to equate the sum of the currents having reference arrows directed into the node to the sum of those directed out of the node:

iA + iB = iC + iD

## PRACTICE PROBLEMS

### PRACTICE PROBLEMS NO: 1

Q: For the circuit that is given below, compute the current through resistor R if it is known that the voltage source supplies a current of 3A. #### Identify the goal of the problem

The current through resistor R3, labeled as i on the circuit diagram.

#### Collect the known information

This current flows from the top node of R3 which is connected to three other branches. The current flowing into the node from each branch will add to form the current i.

#### Devise a plan

If we label the current through R1, we may write a KCL equation at the top node of resistors R2 and R3 #### Construct an appropriate set of equation

Summing the currents flowing into the node:

iR1 – 2 – i + 5 = 0

The currents flowing into this node are shown in the expanded diagram below. #### Determine if additional information is required

We see that we have one equation but two unknowns, which means we need to obtain an additional equation. At this point, the fact that we know the 10V source is supplying 3A comes in handy: KCL shows us that this is also the current iR1 .

Now after substituting, we find that

i = 3 – 2 + 5 = 6 A

It is always worth the effort to recheck our work. Also, we can attempt to evaluate whether at least the magnitude of the solution is reasonable. In this case, we have two sources —– one supplies 5 A, and the other supplies 3 A. There are no other sources, independent or dependent. Thus, we would not expect to find any current in the circuit in excess of 8 A.

### PRACTICE PROBLEMS NO: 2

Q) Count the number of branches and nodes in the circuit given below. If ix = 3A and the 18 V source delivers 8 A current, What is the value of  RA. Branches = 5 (As there are 5 elements in the circuit)

Nodes = 3

we have given,

Ivs1 = 8A
Ivs = 3A

To find RA = ?

by using KCL, writing KCL equation for the above circuit:

8 – IRA  – 3 + 13 = 0
IRA  = 18A

Now by using ohm’s law,

V=IR
R = I/V
R= 18/18
R = 1 Ω