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Kirchhoff’s circuit laws
2 the most important laws in electrotechnics

Analog-Digital-Electronic-Components“Nothing dies in nature”. This proverb reflects very well two the most important laws in electrotechnics : Current and Voltage Kirchhoff’s laws which are fundamental in study of electrotechnics. Regardless you are designer or electronic technician, well understanding how  current behaves in closed circuit is prerequisite if you are going to fix or design electronic appliances. 





1st Kirchhoff’s law – The Current Law says :

“At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node”

What does it mean ? Let’s check application in simple electrical circuit :

1skirch

According to KCL (Kirchhoff Curent Law)

 

I1 + I2 + I3 + I4 = 0  

 

How is it possible that added values are worth 0 ? Because some charges will have positive potential the others will have negative potential, in other words some charges will get into the node , some will leave the node.

so in above case

I1 + I2  =  I3 + I4   

Let’s substitute real value  I1 = 5A  ,   I2 = 7A  ,  I3 = 4A    

what is the value of I= ?  of course 8A

 


 

Let’s check this in real life !

My power source has 120 volts, I want to energize 4 bulbs (100W, 60W, 60W, 40W) with different  resistance. The question is : 

How many Ampers my electrical circuit will require to works properly?

1stblub

I will use the formula : Itotal =  I1 +  I2  +  I3  + I4

But before I do that I need to know the value of I1 I2 I3 I4, I am going to calculate it using 

Ohm’s law formula  I = U/R 

I= 120V/144Ω = 0.83A

I= 120V/240Ω = 0.5A

I= 120V/240Ω = 0.5A

I= 120V/360Ω = 0.33A

Itotal = 0.83A + 0.5A + 0.5A + 0.33A = 2.16A

So, in order to energize 4 bulbs i my electrical circuit  I have to provide 2.16 amps






2nd Kirchhoff’s law – The Voltage Law says :

 

 “The sum of the electromotive forces in any closed loop is equivalent to the sum of the potential drops in that loop “

In the other words :

V+ V+ V3 – Vn… = 0

Let me draw your attention at this electrical circuit:

2ndkirch2

We can see two circuits ORANGE and BROWN, let’s count voltage drops for these paths

ORANGE    4V + 1V + 5V – 10V = 0

BROWN      4V – 3V + 9V  – 10V = 0

 


 

Let’s check this in real life !

I want to energize 4 bulbs connected in series each bulb should be powered by 2,2V source of energy. The question is:

How many Volts do I have to provide to switch them all on ?

bulbsvolts

2,2V + 2,2V + 2,2V + 2,2V – Vtotal =  0

Vtotal = 8,8V

So, in order to switch on 4 bulbs we have to provide 8,8 Volts.


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