# 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

i_{1 } + i_{2 } + i_{3 } +……..+ i_{N } = 0

## EXAMPLE

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

i_{A} + i_{B} + (-i_{C}) + (-i_{D}) = 0

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

( -i_{A })+ ( -i_{B })+ ( i_{C }) + ( i_{D }) = 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:

i_{A } + i_{B } = i_{C } + i_{D}

## PRACTICE PROBLEMS

### PRACTICE PROBLEMS NO: 1

**Q: For the circuit that is given below, compute the current through resistor R _{3 } if it is known that the voltage source supplies a current of 3A.**

#### Identify the goal of the problem

The current through resistor **R _{3}**, labeled as

**i**on the circuit diagram.

#### Collect the known information

This current flows from the top node of **R**_{3, } 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 **R**_{1}, we may write a KCL equation at the top node of resistors **R**_{2} and **R**_{3}

#### Construct an appropriate set of equation

**Summing** the currents flowing into the **node**:

i_{R1} – 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 **i _{R1}** .

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 i _{x} = 3A and the 18 V source delivers 8 A current, What is the value of R_{A.}**

**Branches = 5** (As there are 5 elements in the circuit)

**Nodes = 3**

we have given,

I_{vs1} = 8A

I_{vs} = 3A

To find **R _{A}** = ?

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

8 – I_{RA } – 3 + 13 = 0**I _{RA } = 18A**

Now by using **ohm’s law**,

V=IR

R = I/V

R= 18/18**R = 1 Ω**

I hope these **practice problems** will help you to **understand well** about **KCL Current Law**.

I have also **explain** **Kirchhoff’s Voltage Law**. Click on the click below to **read more** about **Kirchhoff’s Voltage Law (KVL)**.

Click below to give quizzes !