Kirchhoff laws (KCL)
There are two Kirchhoff laws. These two laws are somewhat comprehensive than Ohm's law. And also these laws can be used in more complicated networks.
Kirchhoff's first law (Current law):
- The algebraic sum of current to a node in a electric circuit is equal to zero.
or
- In any network of conductors the algebraic sum of the currents meeting at a point (or junction) is zero.
This simply says that the total current leaving the junction is equal to the total current entering the junction.
See figure 7.1
We get the current leaving the junction as (-)ve and current entering the junction as (+)ve
Using KCL1 , we can write,
(- I1 ) + (-I2) + I3 + I4 + (-I5) = 0
Or
I1 + I2 + I5 = I3 + I4
Mathematically,
∑ i = 0 ..... at a junction .
See figure 7.2 . This is the practical way of using KCL1 .
When you have to analyse this type of circuit, don't get confused. Although there are many voltage sources available it is very easy to analyse by using KCL .
First of all you must understand that you don't need to mark the current that draws from each and every battery. And don't get confused with the current flowing direction. Just forget those things and choose one source. Then mark the current draw from that source. You don't need to worry about the actual current flowing direction, just mark it in any direction.
If the assumed direction of current is not the actual direction, then on solving the equation, this current will be found to have a minus sign. If the answer is positive, then the assumed direction is the same as actual direction.
If the assumed direction of current is not the actual direction, then on solving the equation, this current will be found to have a minus sign. If the answer is positive, then the assumed direction is the same as actual direction.
In figure 7.2 I have selected the E1 Source in AB branch and marked the current I . Then that current goes to point C. Here we apply the KCL1 . In the figure I have already applied the KCL1. you can see that I1 current goes through the CD branch and I - I1 current goes through the FE branch. So the sum of the algebraic current to point C is zero.
Then again at point D the current I1 join I - I1 current. and current I flows to the E1 source again.
Kirchhoff's Second law (Voltage law) :
- The algebraic sum of the product of current and resistance in any closed mesh (or loop, or path) in a network is equal to the algebraic sum of the e.m.f s in that path .
Mathematically,
∑ IR = ∑ e.m.f ... for a closed loop
NOTE : Direction of "IR" and "e.m.f" is very important.
In Figure 7.3 it is mentioned how to decide the direction of e.m.f s and IR product
Following Example will show you how to apply the second KCL to a network.
In Figure 7.3 it is mentioned how to decide the direction of e.m.f s and IR product
- Direction of IR product is decide by the direction of current.
Following Example will show you how to apply the second KCL to a network.
Example :
(Remember that the direction is very important)
Consider the network shown in figure 7.2 again.
We have already applied the first KCL in the diagram.
Applying Second KCL
To apply Second law, first we have to choose a loop.
Applying KCL to ABCD loop
(Remember that the direction is very important)
∑ IR = ∑ e.m.f
I *R1 + I1* R2 = E1 - E2 ----------------------------- (1)
Applying KCL to DCFE loop
- I1*R2 + (I - I1)* R3 + (I - I1)* R4 = E2 - E3 ------------------------------(2)
- It should be noted that the direction of E2 is in the opposite direction we choose (ABCD) and therefore E2 has the negative sign in our equation.
Applying KCL to DCFE loop
- I1*R2 + (I - I1)* R3 + (I - I1)* R4 = E2 - E3 ------------------------------(2)
- Note that the Current through R3 flows from C to D and therefore it has the negative sign in our equation.
If you have any problems regarding this post please leave your problem as a comment. I will reply as soon as possible.
Pabindu lakshitha
B.Sc (Engineering undergraduate)
Why isn't the current due to e2 or e3 taken in kcl2?
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