In this video, I will show you how to solve a complex series parallel circuit to find out a current through a resistor which is inside that complex structure. See in the circuit I have to find the source current and the current through 80 ohm resistor for this circuit. See 80 ohm is this resistor and I. Have to find out the current flowing through this 80 .
Resistor. I have to also find out the source current that is out from this source. Okay .With this kind of structure the circuit is looking complicated. Therefore ,at first I will simplify this circuit by using so-called point method of simplifying circuit.
Lets say this is our source terminal and this is our a terminal and this is our B terminal. This will be our B terminal this will be our B terminal and this will be our B terminal. Our point will continue unless we encounter a resistor. See, I cannot go beyond this beyond this or beyond this because I.
Will encounter resistors. Okay Lets say, this is our Point C so this will be our Point C, let's say this is our point D, this will be our point D. Therefore, we have to have four points. Lets say this is our a point, this is our C point, this is our D point and this is our B.
Point .So this will be our a ,this will be C point, this will be our D point and this will be our B Point .See if I want to go from A to C I will encounter a resistor of 5 ,so this is my 5 . Resistor. After that if I want to go from C to b if I want to go from C to B. I will encounter this 15 resistor.
So if I continue my journey from C to B. You see that I will encounter a resistor of 15, so look here from this point to this point by this way I can go from C to B. So from this C to this B, I. Will encounter another resistor 60 .
Let us say this is my another resistor of 60 , so far I have drawn this 5 , 15 and this 60 . Now if I want to go from C to D - C to D I will encounter a resistor of 12 Ohm and from this C. To B, I will encounter a resistor of 6 . So if I elongate my C point from here to here you see - if this or C point from C to D, I will encounter this 12 ohm resistor.
So now, let me draw that - this will be our 12 ohm resistor and from this C to this D, I will I will encounter another resistor of 6. Let's say, this is our D point this is 12 and this is 6. So I will cross this 6 and this 12. Now from these D to this B, I will encounter a resistor of 20 .
See if I elongate this, you will see from D to B, I will encounter a resistor of 20 ,so I will cross out this. From this D to this B, I will encounter another resistor of 80 . Now from this A to B, I have to connect a voltage source of 40 volt. Therefore if I.
Draw it, that circuit will be our simplified circuit. Now see this circuit and this circuit are same but this is a more simplified structure. If I show you the neat sketch of the circuit , new circuit will look like this. So this will be our simplified circuit.
In this circuit I will have to calculate that total source current, lets say I will denote that current with i1. See up to this point, this i1 will not divide .When this I 1 gets into this point, it will have to portions - 1 will go in this direction and another will go in this direction. Lets say the current that is flowing through this path will be our I 2 and the current that will be flowing through this path will be our i3. See up to this point, this I3 will remain intact s,o here this i3 will divide into two portions but at this point we will get this i3 therefore this I the current that will be flowing through this point to this point after this point will be our i3.
Now see this i3 will divide into two portions - one will go through this 80 and another portion will go through this 20 resistor. The current that will be flowing through this 80 is i-4. In this circuit I will have to calculate the total current i1 and the current that is flowing through our 80 ohm resistor i-4. Now at first I have to calculate the source current i1 , so if I want to calculate the source current i1, I have to calculate the total resistance with respect to this terminal and this terminal and that resistance will be our total resistance of the circuit.
So, to calculate the total resistance R T with respect to that two terminals, at first I. Have to start my simplification from this right side. Now, you will see with respect to these two terminals, this 80 . And this 20 ohm resistor will be in parallel and their equivalent parallel resistance I can calculate by using this formula 80 into 20 divided by 80 plus 20 which will be equal to 16 that means I.
Can replace these two resistances with 16 . Single resistance. See with respect to this terminal and this terminal this 6 and this 12 resistors will be in parallel. I can also reduce these 6 and this 12.
With a single resistor and I. Will calculate their equivalent resistance using this formula 6 into 12 divided by 6 plus 12 .You know that when we have total two resistances connected in parallel, let's say r1 and r2 I can calculate the resistor between these two terminal lets say R T by using this formula r1 into r2 divided by r1 plus r2 so 6 into 12 divided by 6 plus 12 will give us a value of 4 ohm that means I. Will replace these two resistors with 4 and these two resistors with 16 . Now you will see with respect to this terminal and this terminal, this 15 .
And this 60 ohm resistors are connected in parallel, so I can replace them with a single resistor and I will calculate that equivalent resistance between these two terminal with this formula 15 into 60 divided by 15 plus 60 which will be equal to 12 . Now you'll see if I. Replace these two with 4 and these two with 16 ohm - this 4 and this 16 ohm resistors will be connected in series that means I can replace the resistors between this terminal and this terminal that means I can replace this branch with a resistor of 20. I can replace the resistors in this branch with a resistor of 12 , therefore, our new circuit will look like this I'm calculating our equivalent to each respect to this terminal and this terminal here you will see with respect to this terminal and this terminal this 12 ohm and this 20 ohm resistors are connected in parallel I can replace this 12 and these 20 with a single resistor and I will calculate that resistor using this formula 20 into 12 divided by 12 plus 20 which will give us a value of 7.5 Ohm and that resistor will be connected in series with this 5 ohm.
Therefore in this circuit I will get a total RT equal to 7 point 5 plus 5 which will be equal to 12.5 So our total resistor RT equal to 12.5 . Here our supply voltage is 40 volt therefore our total current I 1 will be equal to total supply voltage will be 40 volt and total resistance will be 12.5 , Therefore I. Will get i1 equal to 3.2 Ampere and this 3 point 2 ampere will be our total current now if I want to calculate the current that is flowing through this branch that means the branch between this point to this point if I want to calculate the current through this branch or if I want to calculate the current through this branch I have to reduce these all these resistors into a single resistor and this branch resistor also into a single resistor. And previously I have shown you that these two resistors will be in parallel and they are equivalent parallel resistor will be 16 , these two resistors will be in parallel their equivalent resistor will be 4 .
Therefore, this branch will have a resistor equal to 16 + 4 = 20 . As 15 and 60 are connected in parallel, their equivalent parallel resistor will be 12 ohm that means I will represent this branch with 12 resistor and this branch with 20 resistor . So our i1 is equal to 3.2 A and here I have replaced this branch resistor with 20 ohm and this branch resistor with 12 ohm. Now if I want to calculate the i3, I have to apply our special current divider rule what is that? If I want to calculate the current through this 20, I will take the resistor of opposite branch to a 12 * 3.2 A/ ( 12+ 20) that is entering at this terminal that will be 3 point 2 ampere therefore our i3 will be equal to 1.2 A.
Now look at the circuit here our i1 = 3.2A, i3 = 1.2A. Therefore our i2 should be equal to 2 ampere. Now, I will calculate this i4 that will be flowing through this 80. Resistor .If the current that is entering at this point is equal to 1.2 Ampere, if I want to calculate the current through this 80 ohm resistor, I.
Will take the resistance of opposite branch. So, i4 will be equal to i4 = (20*1.2) / (80+20) = 0.24 A. Therefore, I will get i4 the current that is flowing through this 80 ohm resistor will be equal to 0.2 4 Ampere. Now, to validate my circuit analysis approach I have set the circuit into multi-sim simulator now I will run the simulation and show you the total current will be equal to 3 point 2 ampere and the current that will be flowing power 80 ohm resistor will be equal to 0.24 A or 240 mA.
See this multimeter is showing the reading of 3.2 Ampere and this is our source current I have deduced that. After that this current that is flowing through this 80. Equal to 240 milliampere which is equal to 0.24A ampere okay that means my circuit analysis is completely okay that's it thank you.
Labels:
CURRENT
Thanks for reading Current and Voltage in Complex Series Parallel Circuit - 2 (W subtitles). Please share...!
0 Comment for "Current and Voltage in Complex Series Parallel Circuit - 2 (W subtitles)"