Can someone help me? for a. i got 87.14 ohms
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You need to show your work. Without it, the most we can tell you is whether your answer is right or wrong. If it's wrong, we can't help you find where it is wrong or what you need to do differently because we can't see what you did period.Can someone help me? for a. i got 87.14 ohms
You need to show your work. Without it, the most we can tell you is whether your answer is right or wrong. If it's wrong, we can't help you find where it is wrong or what you need to do differently because we can't see what you did period.
The first step is to redraw the circuit with the connections as described.
Then the second step should be to put bounds on the possible resistance values. Without even redrawing the circuit and just picking the most obvious path between Node A and Node x shows that the MAXIMUM resistance between those two nodes, even without making any of the described connections, is 60 Ω. With the connections there is a path that is only 40 Ω.
You also know that the absolute minimum resistance is 10 Ω and can actually bound it at 15 Ω very easily.
So before you begin you know the answer has to be between 15 Ω and 40 Ω.
The first problem can be done by inspection (particularly once it is redrawn) without using deltawye transforms. But if you want (or are required) to use them, then the following article might help.
http://www.allaboutcircuits.com/technicalarticles/deltawyetransformsbehindthescenes/
Consider just the red path between A and x. What is the resistance of that path? Since any other paths are going to be, at least in part, in parallel with this path, you know that total resistance from A to x has to be less than the resistance of this path.is 87 too high? our instructor gave us a clue that we have to transform it twice to thrice. I'll just have to redraw it like you said and think of other ways.
Nope.i got 28.3333
Nope.
Look at the post I just made and see what the upper bound is by considering the blue and the yellow (described in words) paths. That bound is less than your result, so we know that it is wrong.
Draw the circuit with the connections made. Since you have several points that are all connected together, use that as your common node along the bottom. Then draw the circuit from x to that common node (which includes A). You may need to sketch it a couple of times to get it clean. But once you have this this diagram you can use a simple symmetry argument to effectively remove one of the resistors and turn it into a simple seriesparallel network.
here is it. sorry i replied late.We are NOT mind readers!
How can I tell you what might be wrong with your redrawn circuit or your work when you won't show either your redrawn circuit or your work?!
here is it. sorry i replied late.
I can't really absorb what you're trying to say. Can you give me a more simpler explanation or a figure sir?Hi,
For one thing you can use the concept of a virtual equal node. We might call it an "equanode". It's like a virtual ground only instead of being a virtual zero volts it is some other virtual voltage. Because the second node is equal in voltage to the first node, they can be connected with a short and that reduces the complexity because then some resistances appear in parallel which can then be reduced to one single resistor instead of two.
Note this is for calculation purposes only.
First, let's redraw the circuit with the stated connections:thanks! i thought i had the wrong answer since my classmates also have different answers.
sure? how?
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