hi, how do i calculate this resistance?

Thread Starter

DreamFM

Joined Aug 5, 2012
7


the image is here,

thanks alot to anyone who can explain to me..

Always seem to have problems towards complex circuit.. :(
 

mlog

Joined Feb 11, 2012
276
Thank you for posting the sketch here.

First, you need to redraw the figure as you attempt to solve it. It may require redrawing it more than once. Second, when you redraw it, make the diagonal 6Ω as vertical.

Third, do you see any series or parallel combinations that might simplify the network? I see several. Redraw the network as you combine the combinations.
 

Thread Starter

DreamFM

Joined Aug 5, 2012
7
Thank you for posting the sketch here.

First, you need to redraw the figure as you attempt to solve it. It may require redrawing it more than once. Second, when you redraw it, make the diagonal 6Ω as vertical.

Third, do you see any series or parallel combinations that might simplify the network? I see several. Redraw the network as you combine the combinations.
Hi thanks alot for the reply.

I tried all my means to draw and do some research on complicated circuit resistance solving too..

But as i progress, i attained weird circuit. For example, in a mesh, so far i have encountered mostly 2 resistances, but in this particular circuit, as i redrew the diagramme, i have gotten some 3 paralled resistance with different current flowing through each individual element in a single mesh.

That confused me and impedes the progress.

I have been trying really hard for past few hours.. would it be possible that you gave some hints regarding it ?
 

mlog

Joined Feb 11, 2012
276
Forget about meshes for now. Did you redraw it the way I suggested? First, straighten up the slanting 6Ω resistor (make it vertical). (Don't be afraid to extend lines and rotate components as long as you maintain the connections.)

Now that you have redrawn the network with the 6Ω as vertical. What other resistor is the 6Ω in parallel with? (There is actually more than 1 other resistor in parallel with the 6Ω, but only 1 of them is obvious.) Combine the parallel resistors. Do you know the equation for combining 2 resistors in parallel?

Next, do you see any series combinations? What about the 2Ω resistors? Are they in series with each other? If so, combine them into 1 resistor.

Redraw after combining these. It will greatly simplify the network. Then you probably see more combinations that will help you get to the answer.
 

WBahn

Joined Mar 31, 2012
30,077
It can also help to label the nodes at each resistor. For instance, call the top node A and but an 'A' at the end of each resistor connected to that node. Then do the same for each of the remaining nodes. How many total nodes do you have? Now look for all of the resistors that are connected between the same two nodes; these are in parallel with each other. Replace them with their equalivent resistances. Also look for internal nodes that are connected to exactly two resistors (the terminal nodes don't count since they can also be connected to things external to this circuit)' those resistors are in series. Replace them with their requiivalent resistances. You now have a simplified circuit and can repeat this process since to find additional parallel/series combinations. Eventually, in this case, you will get to a single resistor. Hint, the total resistance is equal in value to one of the resistors in the original circuit.
 
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