What is the equivalent resistance when diagonal resistors are present?

  • Thread starter Roary
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In summary, Roary is trying to find the R(eq) for a circuit with six resistors and a CD resistor. The diagonal 6 ohm resistor is not in parallel with the other resistors, and the 2.4 ohm resistor is not in parallel with the other resistors either.
  • #1
Roary
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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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  • #2
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.
 
  • #3
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?)?
 
  • #4
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.
 
  • #5
6 ohm + 9 ohm = 3.6 ohms
Then the rest is series?
3.6ohms + (3)6 ohms + 2.4 ohms = 24 ohms?
 
  • #6
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?
 
  • #7
I believe you have it.
 
  • #8
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?
 
  • #9
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)
 

What is equivalent resistance?

Equivalent resistance refers to the overall resistance of a circuit or combination of resistors. It is the single resistance value that can replace multiple resistors while maintaining the same current flow.

How do you calculate equivalent resistance?

The formula for calculating equivalent resistance depends on the type of circuit. For series circuits, it is the sum of all individual resistances. For parallel circuits, it is the reciprocal of the sum of the reciprocals of each resistance. For more complex circuits, you can use Kirchhoff's laws or the delta-star transformation method.

Why is equivalent resistance important?

Equivalent resistance is important because it allows us to simplify complex circuits and analyze them more easily. It also helps us understand how different resistors affect the overall resistance of a circuit.

Can equivalent resistance be negative?

No, equivalent resistance can never be negative. Resistance is a physical property of materials and it cannot have a negative value. If you encounter a negative value while calculating equivalent resistance, it means there is an error in your calculation.

What happens to equivalent resistance when resistors are added in parallel?

When resistors are added in parallel, the equivalent resistance decreases. This is because adding more paths for current to flow decreases the total resistance of the circuit. In other words, parallel resistors provide an easier path for current to flow, resulting in a lower overall resistance.

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