-
What would be the value of Rx if the current flowing throuth it is 1.5A
Posted by ZeTeX on 30 Apr, 2017 11:56 -
Hi,
my friend need to solve this circuit and I'm not sure how to do it on paper with simple mesh current theorem, I still didn't study other theorems.
they ask you to find the resistance of Rx, from LTspice putting a current source I find the resistance to be 8.3333ohms.
what's the easiest way to solve such thing? maybe model Rx as a current source?
EDIT: Solved! quite easy! -
Hi,
my friend need to solve this circuit and I'm not sure how to do it on paper with simple mesh current theorem, I still didn't study them.
they ask you to find the resistance of Rx, from LTspice putting a current source I find the resistance to be 8.3333ohms.
what's the easiest way to solve such thing? maybe model Rx as a current source?
Not 10 ohms?
Sent from my iPhone using Tapatalk -
what's the easiest way to solve such thing?
You calculate the replacement source which is (V1 + V2) / 2 (because R1 = R3) and R1 || R3
Giving a 35V source with 15 Ohms.
For 1.5 A you will need a total resistance of 35V/1.5A = 23.33 Ohms
leaving 8.33 Ohms for R2.
With best regards
Andreas
-
what's the easiest way to solve such thing?
You calculate the replacement source which is (V1 + V2) / 2 (because R1 = R3) and R1 || R3
Giving a 35V source with 15 Ohms.
For 1.5 A you will need a total resistance of 35V/1.5A = 23.33 Ohms
leaving 8.33 Ohms for R2.
With best regards
Andreas
I don't think that's correct.
Kirchoff's current law says that the sum of the currents through the two resistors is 1.5A
So
20/(30 + Rx) + 50/(30 + Rx) = 1.5
Re-arranging gives Rx = 16.66 ohms
Edit: I think
-
Probably the canonical way they want you to solve it, Thevenin/Norton equivalents:
- Recognize the sources are VSRC + R Thevenin circuits.
- Transform them to Norton equivalents.
- 20V + 30R --> 2/3 A || 30R
- 50V + 30R --> 5/3 A || 30R
- Now the entire circuit is parallel. Three resistors in parallel, and two current sources in parallel.
- The unknown resistor is drawing 3/2 A (given). The node is supplied with 2/3 + 5/3 = 7/3 A. That leaves 5/6 A total flowing into the two parallel 30R resistors.
- 5/6A * (30R || 30R) = 5/6A * (15R) = 12.5V
- Rx = 12.5V / 1.5A = 8 1/3 R.
(Using fractions just because it's better than writing out decimal thirds.)
Or you could do nodal analysis, or superposition, or whatever.
However, the special-case observation notes that:
- Two Thevenin sources are in parallel, and have equal resistances.
- The equivalent of these is the average of the voltages, and half the resistance. (More generally, any number of parallel Thevenin sources has an equivalent where the voltage is the weighted average, and the resistance is all resistors in parallel. It is up to the student to prove this...)
- Halfway between 20 and 50V is 35V. Half of 30R is 15R.
- 1.5A into 15R is 22.5V, subtract that from 35V (= 12.5V).
- 12.5V / 1.5A = 8 1/3 ohm.
Tim -
Andreas and Tim are correct.
Kirchoff's current law says that the sum of the currents through the two resistors is 1.5A
So
20/(30 + Rx) + 50/(30 + Rx) = 1.5
Re-arranging gives Rx = 16.66 ohms
Edit: I think
Wrong, the current through the resistors is not 20/(30 + Rx) or 50/(30 + Rx) respectively. You can't just totally disregard the current flowing through the common resistor from the other branch.
If you want to apply superposition, then you have to set the other voltage source to zero, i.e., a short circuit, and you have to contend with the part of the current going through the other resistor, which is a whole can of worms.
Andreas' approach is much simpler, and also correct. -
Andreas and Tim are correct.
OK - I thought I had probably missed something, should have taken a bit more time to think it throughEdit: I think
Wrong, the current through the resistors is not 20/(30 + Rx) or 50/(30 + Rx) respectively. You can't just totally disregard the current flowing through the common resistor from the other branch.
If you want to apply superposition, then you have to set the other voltage source to zero, i.e., a short circuit, and you have to contend with the part of the current going through the other resistor, which is a whole can of worms.
Andreas' approach is much simpler, and also correct.
-
Yes, that's how I'd do it.what's the easiest way to solve such thing?
You calculate the replacement source which is (V1 + V2) / 2 (because R1 = R3) and R1 || R3
Giving a 35V source with 15 Ohms.
For 1.5 A you will need a total resistance of 35V/1.5A = 23.33 Ohms
leaving 8.33 Ohms for R2.
With best regards
Andreas
Attached is the .asc file for completeness.
-
I so much prefer Nodal Analysis for these kinds of problems...
Define a node VA at the junction of the 3 resistors
Assume current is flowing into the node from V1 and that current is (V1 - VA) / R1
Assume current is flowing into the node from V2 and that current is (V2 - VA) / R3
Assume current is flowing out of the node VA and that current is VA / Rx and is given as 1.5A
So, you wind up with 2 equations and 2 unknowns (VA and RX)Code: [Select]eq1 : 0 = (V1-VA)/R1 +(V2-VA)/R3 - (VA/Rx);
eq2 : 1.5 = (VA/Rx);
And, yes, RX = 8.333 Ohms
To solve the problem using Mesh Analysis, you wind up with 3 equations and 3 unknowns (I1,I2,RX)Code: [Select]eq1 : 0 = V1 - (I1*R1) - RX*(I1+I2);
eq2 : 0 = V2 - (I2*R2) - RX*(I1+I2);
eq3 : 1.5 = I1+I2;
For the mesh, I assume the current flow is from ground up through each source. The left loop is clockwise, the right loop is counterclockwise.
Attached is wxMaxima solution for both Nodal and Mesh. I am trying to maximize my use of wxMaxima, hence the reply.
-
I too prefer a simple nodal analysis like rstofer did.
The trouble with "clever tricks" like Thevenin/Norton equivalents, parallel circuits and special cases is that you have to remember how to apply them correctly, which is a lot to remember for a beginning student. Many chances to make mistakes and get it wrong.
If you do a simple nodal analysis like a current balance around the middle node, it is all first principles. All you have to do is apply Ohm's law correctly and not make any mistakes in your algebra. Not only is it less to remember, but it also helps to reinforce fundamental principles like Kirchhoff's laws, things which need to be deeply embedded in the mind until they become second nature. -
Each to their own I suppose. I've not done nodal analysis since college and that was just to pass the exam. I'd often mess up the algebra somewhere, then have to redo it.
One of the advantages of using these clever tricks, is they require a greater understanding of the circuit, than just following a procedure and doing the algebra, even if it is more difficult to remember. -
Nodal analysis wasn't emphasized in college, it seems we spent a great deal more time with mesh analysis. In general, nodal analysis will wind up with fewer equations (probably just one fewer) but if you plan to solve the matrix by hand, a 2x2 is a lot easier than a 3x3 and if we get to 4x4, all bets are off.
Nodal analysis is simpler (in my view) for this kind of circuit. We just define the two currents coming into the node and the one current (given) leaving the node.
In any event, op amps are better understood with nodal analysis. Summing the currents at the (-) input is the classic way to solve these feedback equations.
-
I so much prefer Nodal Analysis for these kinds of problems...
Thanks! quite a simple problem I see, don't know why I struggled.
Define a node VA at the junction of the 3 resistors
Assume current is flowing into the node from V1 and that current is (V1 - VA) / R1
Assume current is flowing into the node from V2 and that current is (V2 - VA) / R3
Assume current is flowing out of the node VA and that current is VA / Rx and is given as 1.5A
So, you wind up with 2 equations and 2 unknowns (VA and RX)Code: [Select]eq1 : 0 = (V1-VA)/R1 +(V2-VA)/R3 - (VA/Rx);
eq2 : 1.5 = (VA/Rx);
And, yes, RX = 8.333 Ohms
To solve the problem using Mesh Analysis, you wind up with 3 equations and 3 unknowns (I1,I2,RX)Code: [Select]eq1 : 0 = V1 - (I1*R1) - RX*(I1+I2);
eq2 : 0 = V2 - (I2*R2) - RX*(I1+I2);
eq3 : 1.5 = I1+I2;
For the mesh, I assume the current flow is from ground up through each source. The left loop is clockwise, the right loop is counterclockwise.
Attached is wxMaxima solution for both Nodal and Mesh. I am trying to maximize my use of wxMaxima, hence the reply.
-
Thanks! quite a simple problem I see, don't know why I struggled.
It's a process, a simple yet methodical writing of the equations. It may be worthwhile to take the equations I wrote for each of the approaches along with the schematic and see where I got them.
Nodal: One node VA with two currents coming in and one current leaving and that current is given. This is absolutely simple since the total current is given.
Mesh: Two loops where the only complication is the center resistor and the polarity of the voltage drop across the resistor from each loop current. In this case (most cases if you do it right), the currents add so the same voltage term [RX(I1+I2)] winds up in both equations and the factor (I1+I2) is given.
There are many videos on the Internet but there is an entire semester long video course at Digilent Inc
https://learn.digilentinc.com/classroom/realanalog/
-
Thanks, I completely understand how you did that, it makes sense, but I wasn't thinking hard enough because this question was a little bit outside the box for mesh analysis, usually the resistance is given and they want you to find the current, but now that I know how you did that I know how to solve such questions.Thanks! quite a simple problem I see, don't know why I struggled.
It's a process, a simple yet methodical writing of the equations. It may be worthwhile to take the equations I wrote for each of the approaches along with the schematic and see where I got them.
Nodal: One node VA with two currents coming in and one current leaving and that current is given. This is absolutely simple since the total current is given.
Mesh: Two loops where the only complication is the center resistor and the polarity of the voltage drop across the resistor from each loop current. In this case (most cases if you do it right), the currents add so the same voltage term [RX(I1+I2)] winds up in both equations and the factor (I1+I2) is given.
There are many videos on the Internet but there is an entire semester long video course at Digilent Inc
https://learn.digilentinc.com/classroom/realanalog/
digilent looks interesting, thanks! -
Thanks, I completely understand how you did that, it makes sense, but I wasn't thinking hard enough because this question was a little bit outside the box for mesh analysis, usually the resistance is given and they want you to find the current, but now that I know how you did that I know how to solve such questions.
digilent looks interesting, thanks!
My first attempt was Mesh Analysis and it wasn't a blazing success for the same reason you stated. I didn't know the resistance. How could I solve the thing? That's not what I remember from school...
So, I completed the Nodal Analysis first while spending a little more time thinking about that pesky resistor.
Eventually, I realized that the combined current was given and the voltage drop was common to both loops. Then I got it!
I would like to say the solution just flowed to the keyboard but it's not true. I had to spend a few minutes thinking about the Mesh Analysis. The Nodal Analysis was simple.
-
I built that circuit and it didn't do anything (beside get really hot).
Is this how you would solve the problem of having an infinitely large grid or array of resistors and trying to find the current/volts of one in the middle?
What would happen if you made a cube instead of a flat sheet? -
Is this how you would solve the problem of having an infinitely large grid or array of resistors and trying to find the current/volts of one in the middle?
Given that the answer to that problem involves pi, it is self-evident that that particular problem is non-trivial and involves some sort of calculus, with perhaps a healthy dose of trigonometric substitution. But sure, the same basic nodal analysis laws are still the basis of the solution.
If you have a cube, then yeah the maths is probably a little bit harder but it's the same principle. -
It might be hard to do Nodal Analysis on a cube or sphere but Mesh Analysis would work The matrix would be large andd a hand solurion might not be practical. That's why we have computers.
-
Morning all.
SPOILER ALERT.
READ NO FURTHER IF YOU WANT TO WORK IT OUT.
I have had the odd miscalculation with LTSPICE and am still just learning to use it so I may be wrong.
I've just used LTSPICE to calculate the value of R2.
I started with the 8.333
and went from there.
It came up with 2.5
The voltage at the node where all 3 resistors meet is 5V.
Maths is my weak point and 30 years ago I couldn't get Norton / Thevenin when I did a coarse at a local TAFE. Technical And Further Education.
Please let me know if this is incorrect.
Bill
-
Post the .asc file. I'm pretty confident the resistor should be 8.333 ohms.
What are the various currents? -
rstofer
I just found your reply.
Unfortunately I did'nt save the file and I am about to go to work.
5 minutes before I have to go.
I think someone else with LTSPICE would be able to recreate the file.
sorry.
BILL. -
It has to be 8.333 ohms. The maths doesn't lie, no matter whether you use nodal or mesh analysis or simplification to a Thevenin equivalent source. I've got absolutely no idea how you could manage to get 5V at the junction unless you goofed and calculated for 2A through R2 instead of 1.5A.
You can easily solve it in LTspice by using a current source in place of R2 (or a behavioural resistor that passes 1.5A), then calculating the resistance from Ohm's law. Sim attached.
Being able to work Thevenin equivalents in your head is a useful skill when dealing with real world circuits because a potential divider feeding a load is an incredibly common circuit configuration.
-
I tried it with LTSpice using the 8.333 Ohms calculated earlier.
-
Yes, that's how I'd do it.what's the easiest way to solve such thing?
You calculate the replacement source which is (V1 + V2) / 2 (because R1 = R3) and R1 || R3
Giving a 35V source with 15 Ohms.
For 1.5 A you will need a total resistance of 35V/1.5A = 23.33 Ohms
leaving 8.33 Ohms for R2.
With best regards
Andreas
Attached is the .asc file for completeness.
Yep