pop000 Posts: 352, Reputation: 6 Full Member #1 Nov 15, 2011, 12:35 PM
Find the electrical potential difference
Given electrical network (in the picture).

I asked to find the electrical potential difference between A and B. UAB
I tried and not find the answer.
I really need help.

Thanks.
 ma0641 Posts: 15,681, Reputation: 1012 Uber Member #2 Nov 15, 2011, 12:43 PM
Where is the picture? I don't see one.
 pop000 Posts: 352, Reputation: 6 Full Member #3 Nov 15, 2011, 12:47 PM
Oh sorry.

Here is it
Attached Images

 ebaines Posts: 12,132, Reputation: 1307 Expert #4 Nov 15, 2011, 01:12 PM
Start by finding the equivalent resistance that can replace the 4-ohm, 3-ohm, and two 6-ohm resistors. What do you get for that resistor? That equaivalent resistor splits the battery voltage in conjunction with the 3 ohm reistor on the left. Post back with what you get for an answer.
 pop000 Posts: 352, Reputation: 6 Full Member #5 Nov 15, 2011, 01:52 PM
well I find the equivalent resistance that can replace the 6 ohm and the 3 ohm (they are parallel) so I get:
6*3/(6+3)=2ohm.

and I get a new electrical network (in the picture).

my big problem is how to keep from this point, I know there is a law that helps in this kind of problem (I not sure how is call in english), maybe is call a Converting triple star.

thanks.
Attached Images

 ebaines Posts: 12,132, Reputation: 1307 Expert #6 Nov 15, 2011, 02:02 PM
Remember that resistors in parallel don't add - you must use the formula:

$
R_t = \frac 1 { \frac 1 {R_a} + \frac 1 {R_b}}
$

Hence the 3-ohm and 6-ohm resistors in parallel are equivalent to:

$
R_t= \frac 1 { \frac 1 6 + \frac 1 3} = \frac 1 { (\frac 3 6)} = 2$
ohms.

Next step is that this 2-ohm resistor plus the 4-ohm in series adds to 6 ohms, which is in parallel with the remaining 6-ohm resistor.

Can you take it from here?
 pop000 Posts: 352, Reputation: 6 Full Member #7 Nov 15, 2011, 02:17 PM
yes I get 6 ohms which is in parallel with the remaining 6-ohm resistor so is equal to 6*6(12)=3ohms
and this equivalent resistance is series to the last 3 ohms resistor and I get RT=6 ohms

then I do E=6V/RT so I get 6/6=1v Uab=1v
do I correct ?
 ebaines Posts: 12,132, Reputation: 1307 Expert #8 Nov 15, 2011, 02:28 PM
You have RT = 6 ohms which is correct. But voltage divided by resistance is current, not voltage. From Ohm's Law:

V=IR,
6 volts = I * 6 ohms,
I = 1 amp.

Now you can use use Ohm's Law again to find the voltage drop across the 3-ohm resistor that spans points a and b:

V = IR = 1 amp x 3 ohms = ? Volts
 pop000 Posts: 352, Reputation: 6 Full Member #9 Nov 15, 2011, 02:39 PM
Oh yes you correct.
But I have to ask why u took the 3-ohm resistor voltage that spans points a and b:
And not the 6 ohms resistor that also spans points a and b ?

 ebaines Posts: 12,132, Reputation: 1307 Expert #10 Nov 15, 2011, 03:01 PM
Yes - I made an error - sorry! The progression of how we got from your original circuit diagram to two 3-ohm resistors in series is shown in the figure below. The current around the circuit is 1 amp. But only half of that goes through each of the 6 ohm resistor equivalents. So the voltage drop across ab should be V = IR = 1/2 amp x 2 ohms = 1 volt.
Attached Images

 pop000 Posts: 352, Reputation: 6 Full Member #11 Nov 15, 2011, 03:06 PM
wow you made is clear now :) now I understand it.
so as u told beofoe V = IR = 1 amp x 3 ohms = 3 volts

so this is the final answer Uab=3v ?

 pop000 Posts: 352, Reputation: 6 Full Member #12 Nov 15, 2011, 03:07 PM
 ebaines Posts: 12,132, Reputation: 1307 Expert #13 Nov 15, 2011, 03:34 PM
Originally Posted by pop000
wow you made is clear now :) now i understand it.
so as u told beofoe V = IR = 1 amp x 3 ohms = 3 volts

so this is the final answer Uab=3v ?
After I made my previous post I realized I had an error, so was in the midst of correcting it when you posted. Sorry for the confusion.
 pop000 Posts: 352, Reputation: 6 Full Member #14 Nov 15, 2011, 03:41 PM
that OK :) .so you can tell me if the final answer is Uab=3v?

 ebaines Posts: 12,132, Reputation: 1307 Expert #15 Nov 16, 2011, 09:18 AM
Originally Posted by pop000
that ok :) .so you can tell me if the final answer is Uab=3v?
No - 1/2 amp of current flowing through a 2 ohm resistor yields? Voltage drop across that resistor.
 pop000 Posts: 352, Reputation: 6 Full Member #16 Nov 17, 2011, 01:01 AM
Ok now I got it.

Thank you very much.

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