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Kirchoff's Laws Series-Parallel Question

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Ryan Thornwell

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So here is the circuit, I am trying to figure current and volt drop through resistors.


I've got the circuit down to this by combining parallel and series. Also worked out total current.


I cannot think where to start with calculating the volt drops,

I have this table



Would I start by going back to the original circuit and first calculate E = I * R5//R6_R4? then do the same for each branch in the circuit?
 

MikeMl

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Doesn't the total current (0.543106A) flow in R1?

If you know the current in R1, then you should be able to calculate the voltage at the node where R1, R2, and R3 connect. Similarly, you can work your way to the right progressively for the remaining nodes. I would label them...

Your calculation of the magnitude of the current from the voltage source is correct, but I do not see where you got the values you put in the table.

Also, if you ever hope to get a real job in the Electronics Industry, then the current in R1 flows from left to right...
 
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Ryan Thornwell

New Member
If you know the current in R1, then you should be able to calculate the voltage at the node where R1, R2, and R3 connect. Similarly, you can work your way to the right progressively for the remaining nodes. I would label them...

Your calculation of the current from the voltage source is correct, but you screwed-up when filling in the table.

Doesn't the total current (0.543106A) flow in R1?

Also, if you ever hope to get a real job in the Electronics Industry, then the current in R1 flows from left to right...
So this is the first node right?



Also fixed the table and current direction


 

MikeMl

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Note that your equivalent circuit is not correct:

kir1.gif


Study this:

kir.gif

Do you understand why I put the current source where I did, and how that defines V(b)?
 
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Gasboss775

Member
Also, if you ever hope to get a real job in the Electronics Industry, then the current in R1 flows from left to right...
Speaking of this, there is another electronics site on the internet that insists on using the current flowing from negative to positive convention. I thought that issue had been settled years ago.
 

MikeMl

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Speaking of this, there is another electronics site on the internet that insists on using the current flowing from negative to positive convention. I thought that issue had been settled years ago.
One of the reasons that I don't post there anymore...

Science, Education, and Industry use Conventional current which defines current as 1 Coulomb of positive charge flowing past a point in the circuit every second.

I agree with Wiki's definition of the convention:

concur.gif
 

MikeMl

Well-Known Member
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My equivalent circuit is wrong because it gives the wrong current right?
Correct.

You put the current source there because there the current will flow through the node?
To say it a bit better: The 0.543A current source forces the current in R1 (and everywhere else in the circuit) to be the same as you already correctly calculated...

Note that you can arrive at V(b) by subtracting (I*R1) from 5.000V, as I tried to show in post #4. I only put the current source there so the sim could solve for V(b) without knowing the equivalent of R23456.
 

Ryan Thornwell

New Member
Correct.


To say it a bit better: The 0.543A current source forces the current in R1 (and everywhere else in the circuit) to be the same as you already correctly calculated...

Note that you can arrive at V(b) by subtracting (I*R1) from 5.000V, as I tried to show in post #4. I only put the current source there so the sim could solve for V(b) without knowing the equivalent of R23456.
Okay so my equivalent resistance should be 3.486?

Could you send the file for the sim?
 

MikeMl

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You should be able to do this with pencil and paper....

Here is my effort:
 

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MikeMl

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Here is the verification of the reduction steps:

e.g., compare V(b), V(b1) and V(b2)

kir.gif
 

Ryan Thornwell

New Member
I definitely went for the longer approach o_O You've helped my understanding a lot thanks

As for the currents of the remaining resistors, they're given right?
 

MikeMl

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Most Helpful Member
...How does I (R4) for instance = 0.239?

If its just I = V/R, then I (R4) = 2.298/3.684 = 0.624
I(R4)=V(c)/R4 = 1.19958/5 = 0.239916, which agrees with the tabulated value of I(R4). The current into node C via R3 equals the sum of the currents that flow to ground via R4 and R5+R6, or I(R3) = I(R4) + I(R5). We can find the current through the R5 branch the same way. I(R5) = V(c)/(R5+R6) = 1.19958/(9+5) = 0.0856842, which matches the tabulated values for I(R5) and I(R6). The sum of I(R4) and I(R5) = 0.32560, which matches the tabulated value of I(R3).

KCL works for every node. Ohms Law works for every resistor...
 

Ratchit

Well-Known Member
So here is the circuit, I am trying to figure current and volt drop through resistors.


I've got the circuit down to this by combining parallel and series. Also worked out total current.


I cannot think where to start with calculating the volt drops,

I have this table



Would I start by going back to the original circuit and first calculate E = I * R5//R6_R4? then do the same for each branch in the circuit?
There are a zillion ways to do this problem. Some of the ways are by loop, node, or branch analysis. Another way is the current divider law. For instance, the current present in R4 is I(R3) * (R5 + R6)/(R5+R6+R4). Since you already calculated the series-parallel values up the chain, they should be available for you to use.

Ratch
 

Ratchit

Well-Known Member
Speaking of this, there is another electronics site on the internet that insists on using the current flowing from negative to positive convention. I thought that issue had been settled years ago.
Why not name it? "All about Circuits", right? Yes, they are a rather quirky site, aren't they? They seem to think that current direction should be dependent on the polarity of the charge carriers. That means they have to think one way when they calculate electron flow in wires, and double-think reverse when they calculate hole current in semiconductors. It is much better to use the mathematical convention of assuming charge carriers are positive and charge flow is positive when it comes from the positive terminal of the voltage source. Then, if necessary to find the real direction of the current carriers, just reverse direction if the charge carriers are negative. Also, ammeters and semiconductors are marked using the mathematical convention. Too bad LTSpice is backwards.

Ratch
 

MikeMl

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Ratchit

Well-Known Member
http://forum.allaboutcircuits.com/t...nt-always-measured-negative-in-ltspice.40854/

http://www.electro-tech-online.com/threads/ltspice-current-probe.109196/
Look at the attached file in post #3. It shows both voltage sources as having a negative current.

Finally, look at the attached file documentation. In figure "a" it shows the current in the opposite direction to the mathematical convention. The descriptive paragraph in the middle of the page also describes the current direction as being opposite to the mathematical convention.



Ratch
 

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