What is the equivalent resistance when diagonal resistors are present?

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The discussion centers on calculating the equivalent resistance (R(eq)) in a circuit with diagonal resistors. The diagonal 6 ohm resistor is identified as being in parallel with a 9 ohm resistor, while a 2.4 ohm resistor is in series with this parallel combination. The participants clarify the arrangement of the resistors and confirm that the total resistance can be calculated as 15 ohms after considering the series and parallel configurations. Additionally, they discuss methods for finding current and potential differences across each resistor using Ohm's law. The conversation highlights the complexities of visualizing circuit diagrams and the importance of correctly identifying resistor arrangements for accurate calculations.
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To begin with, I need to find the R(eq) of the following (beg my endeavor with paint). What I don't seem to get at all is what to do with the diagonal 6 ohm resistor. I don't see the grouping at all..
http://www.azuretek.com/forums/phys.gif
 

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I think you are letting the diagonal position of it throw you off.
That R is simply in parallel with the 9 ohm between points d and f.

Try redrawing the diagram, but with the 6 ohm R running vertically on your paper, and kick the 9 ohm to the right and turn it vertical as well. Just be sure to keep all your connections and labeled points in the correct relation. I think that will aid your visualization of the problem.

Let us know how this works for you.
 
So... the de 6 ohm is parallel to the 9 ohm, but the 2.4 ohm isn't parallel to those two (b/c of the cd resistor?)?
 
Draw your picture as above, you will see that the 2.4 ohm res. is in series with the 6 and 9 ohm parallel pair.
 
6 ohm + 9 ohm = 3.6 ohms
Then the rest is series?
3.6ohms + (3)6 ohms + 2.4 ohms = 24 ohms?
 
Originally posted by Roary
6 ohm + 9 ohm = 3.6 ohms
Then the rest is series?
3.6ohms + (3)6 ohms + 2.4 ohms = 24 ohms?
Wait, or is it, the 6ohm and 9 ohm are parallel.
The 2.4 ohm is series to the (6 & 9).
The 6 ohm in the middle is parallel to the combined (2.4 ohm series to (6 &9)).
Then all series? To get 15 ohms?
 
I believe you have it.
 
Atleast I got that part , so I can traverse around the circuit
a. I need to find I1, I2, .. I5
This seems to be a bit of a problem for me. I know I = V/R
So the absolute total would be 1 amp for each way, but where next?

b. Find potential difference across each resistor.
So V=IR
For the 6 Ohm on the a-c, that would be 6V? (same for the bottom b-d) Then I use the remaining 3V and break into the 2nd half, correct?

c. Find power dissipated by each resistor.
P=IV, Use the V's figured from b. to get the wattage for each?
 
So I1 = 1 amp is confirmed,
someone said I2 & I3 = .5 amps each. (I4 & I4 = .25 amps each?)

Now if those are what they are, I'm all confused on the current to use for part C & D. Anyone with any insight? (Someone got 2.5v from the a to c 6ohm resistor and the bottom resistor too, from 15v/6ohm)
 
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