Basic Electronics

# Kirchhoff’s Laws, Kirchhoff’s Voltage Law & Kirchhoff’s Current Law

Table of Contents

## Introduction:

Kirchhoff’s Laws: these laws are use to find the current in different branches of an electrical network which may be easily solved by ohms law, Kirchhoff’s laws are two in number

1. Kirchhoff’s first law or Kirchhoff’s current law “KCL”
2. Kirchhoff’s loop law or Kirchhoff’s voltage law or Kirchhoff’s mesh law “KVL”.

Kirchhoff’s laws are applicable to both Ac and Dc.

1. ## Kirchhoff’s current Law

Kirchhoff’s current Law this law states in any electrical network that

“The algebraic sum of current meeting at a junction or node or point is zero”

Or

In other words “the sum of currents flowing towards the junction is equal to the sum of the currents flowing away from the junction”.

Let we consider the arrangement of 4-wires connected together at point “0” as shown in fig below.

The arrows indicate the direction of flow of current. The currents I1 and I4 are coming towards the junction and current I2 and I3 are going out from junction we assume (suppose ) +ive sign for incoming current and –ive sign for outgoing current. Now according to the Kirchhoff’s Current Law “KCL”

∑I=0

Now from fig

I1+I4+(-I2)+(-I3)=0

I1+I4-I2-I3=0

I1+I4=I2+I3

Or incoming currents = outgoing currents

### Problem1:

Determine the currents I3 and I4 for the giving circuit using Kirchhoff’s Current Law “KCL”.

#### Solution:

At point a by Kirchhoff’s

∑ Iincoming = ∑ Ioutgoing

I1+I2=I3

2+3=I3

= I3 = 5

At point “b” by Kirchhoff’s law

∑ Iincoming = ∑Ioutgoing

I3 + I5 =I4

5 +1=I4

I4=6amp

### Problem2:

Find I1, I3, I4, and I5 in the network given

#### Solution:

At point a

I= I1+I2

5=I1+4

5-4=I1

I1=1amp

As R1 and R3 are connected in series and we know that in series same current will pass through each resistor so

I3=1amp

And we know that R2 and R4 are connected in series so,  current will be same therefore

I4=4amp

As I5 = I4 + I3

I5 =4+1

I5=5amp

### Problem3:

Find the magnitude and direction of I3 and I4

I6 and I7 for the network using Kirchhoff’s Current Law KCL.

#### Solution:

Incoming current I1= 10amp

Outgoing current I7=10amp

At point b we see that

I2 = I4 +I5

I2 = I4 +8

4= I4

As at point d we see that

I7 = I5 + I6

10 =8 +I6

10-8 = I6

I6 =2amp

As at c point we see

I6 + I3 = I4

2 +I3 = 4

I3 = 2amp

1. ## Kirchhoff’s Voltage Law:

This law states that “the algebraic sum of all the voltage taken in a specified direction taken around the closed loop is zero”.

Mathematically

∑v=0

### Closed loop:

A closed loop or a closed path is any continuous path that reaches a point in one direction and return to the same point from another direction without leaving the circuit.

Consider the circuit

Apply Kirchhoff’s Voltage Law KVL to the circuit we have

∑v=0

-(xR1) + (-v1)- R2(-z)+ v2 –R3(-y)

-xR1-v1+R2z+v2+R3y=0

For conformity the clock wise direction will be closed throughout all application of Kirchhoff’s Voltage Law KVL also anti clock wise direction can be used and both will gives us the same result.

### Problem1:

Write the voltage equation for the given circuit

-v1-v2+v=0

V=v1+v2

∑vrise =∑vdrop

### Problem2:

Find the voltage v1 for the network shown

-v1-v22+1=0

Putting values

-v1 -4.2-9+16=0

V1= -4.2-9-16

V1=16-13.2

V1=2.8vout