Electrical Engineering: Basic Laws (19 of 31) The Bridge Network
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- Опубликовано: 29 сен 2024
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In this video I will find the 6-equations and 6-unknowns of a 5-resistor bridge network.
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Mr. Van Biezen has left out one important fact. The equations he came up with are correct as far as he has gone. Mr. Van Biezen has left out one very important fact. We need one more KVL equation that is a loop that includes the voltage source. Then you can solve for the currents in the resistors. If you express the currents I4 and I5 as functions of I1, I2 and I3 then you have three equations in three unknowns.
You are correct. If the wrong combination of loops (equations) are picked, a solution cannot be found. To appreciate that, it is useful for the student to experiment with that so they can see how that works.
yeah - it's unsolvable otherwise
sir , instead of using kirchoff's junction rule , can we use current division rule to relate the branching current variables ?
Best basic explanation I've seen of this, after a lot of searching! Thanks very much. :)
REALLY ENJOY YOUR HARD WORK
Thank you. Glad you do. 🙂
Can I instead solve the problem by using three loop currents? The two you wrote and a third that is in the mesh between the battery and the bridge?
All equations have nothing to do with 10 V of source. How come?
The number of equations for Kirchoff's Current Law you can writer to solve a circuit equals the number of nodes minus 1. This circuit has 4 nodes, but you have written 4 KCL equations (the first 4). So aren't you still one equation short of being able to solve the circuit?
You need an equation for each of the unknown currents. (you can use Kirchhoff's current equation to eliminate one of the unknowns).
One of the biggest problems in understanding these problems is that many examples have the different I sections in different sides of the diamond and it gets very confusing. So when a book problem says that r1 is always I1? r2 to I2 and so forth and so forth?
That is a good point. (But different authors are going to use different combinations).
@@MichelvanBiezen now in wheatstone bridge circuits does current always travel and split at the very first top node? why do books say there are loops of different kinds? again very confusing.
Current always splits in some ratio at every circuit junction. More current will flow through the branch with less resistance. Note that bridge circuits are complicated and it is better to start with some examples of simpler circuits before attemping to use Kirchhoff's rules to solve a bridge circuit.
Dear Sir,
Thank you very much for all your videos. I need your notes in ppt .
total current is 1.74 AMPS
VERY GREAT EXPLANATION SIR, i really love to watch your videos, be HAPPY sir.
thnak you for the lecture how can i simplify this circuit like you did in the previous vedio
You're welcome!
🤗🤗
Glad you liked it. 🙂
I got I= 2.1 and Req=4.74 and I1=1.4 and I2=0.7, I3=0.93, I4=0.4667, I5=1.63
is is right??
Plz comment
no
yeah, that isn't correct - did you include another equation, factoring in the voltage source? Otherwise, no solutions are possible
Answer is wrong
thank you for saving my life
That may be a bit dramatic, but we are glad that we were able to help.
HI,
in case of R1 > R2 the I3 would have flown in opposite direction, isn't it?
thanks!
It also depends on the value of R4 and R5. Try a few combinations and see what happens. The best way to learn.
How do you know which side I3 will flow into
We assume that some current (we don't know how much) flows through each of the branches. Then we label those currents for each branch. So the current flowing through R3 is now labeled as I3. Then we use one of the circuit evaluation techniques to determine what the value of that current is. (as illustrated in the video)
@@MichelvanBiezen What I am assuming is that due to Ohm's Law, more current will take the lower resistance path, which is 5 ohms as opposed to 10, so the higher current node won over the lower current node (the one connected to 10 ohms) and became our assumed current direction. Could be wrong.
what to do if the source is located inside the diamond.
we can interchange souce and dimond resistance
result will be same
Is the total resistance 16.10 ohms using d to y formula ?
This is most helpful as I could understand further
How do you know which way the current in the center flows?
It is usually possible to determine by looking at the size of the resistors. The smallest resistors carry the greatest current. If you can't tell then you can just guess and if you guess wrong, you'll get a negative answer for the chosen current.
@Michel Van Biezen, Sir what would be the answers?? I get
I1 = 1.15 Amps
I2 = 0.59 Amps
I3 = 0.09 Amps
I4 = 1.06 Amps
I5 = 0.68 Amps
Total current = 1.74 Amps
Am I correct??
@@MichelvanBiezen @Michel Van Biezen, Sir what would be the answers?? I get
I1 = 1.15 Amps
I2 = 0.59 Amps
I3 = 0.09 Amps
I4 = 1.06 Amps
I5 = 0.68 Amps
Total current = 1.74 Amps
Am I correct??
@@cyrilrelatado3874 I obtained the same total current (1.74 Amps) when working the problem with two different methods (mesh analysis & delta-y conversion).
@@cyrilrelatado3874 yes
Sir I am from India and I like you explaing us very good
Thank you and welcome to the channel!
if you use a matrix you cant use equation 4 because its the same as equation 1 and the determinant will be zero won't it? Plus as rickmcn1986 said, you do need an equation that includes the system's emf.
I tried to get the final solution with this six ecuations with six unknowleges. Using an app in my mobile (Gauss-Jordan) there are infinite solutions. The determinant of coeficient matrix is zero. So, at least one ecuation is a linear combination of other.
These equations will not yield a result in amperes. The best you can do (as far as I can see) is end up with a series of relationships between currents.
For those who want the solution, we need to include another equation that factors in the voltage source: 10 - 5(I_1) - 4(I_4) = 0, a consequence of Kirchhoff's Voltage Law.
If we solved the following equations simultaneously:
-> (I_T) = (I_1) + (I_2)
-> (I_1) = (I_3) + (I_4)
-> (I_5) = (I_2) + (I_3)
-> 5(I_1) - 10(I_2) + 2(I_3) = 0
-> 4(I_4) - 2(I_3) - 6(I_5) = 0
all are derived in the video (NOTE: I am not including: (I_T) = (I_4) + (I_5))
and -> 10 - 5(I_1) - 4(I_4) = 0
We obtain the following answers:
I_T = 1.74216A
I_1 = 1.14983A
I_2 = 0.59233A
I_3 = 0.08711A
I_4 = 1.06272A
I_5 = 0.67944A
Note, these values also satisfy the equation we excluded: (I_T) = (I_4) + (I_5):
1.06272A + 0.67944A = 1.74216A = I_T
Thanks for the input.
In 10-5(i_1)-4(i_4)=0
Why did you only use 2 resistors if it includes the whole circuit?
Im just confused
my formal understanding of math is somewhat "arrested" but conceptually I reckon that the purpose of the six equations is to calculate the unique current value of each branch (I1, I2, I3, etc ...) that would satisfy each and every one of them, hence arriving at the solution. Correct?
Yes. (That would be a big task, but it can be done).