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.

Where is the picture? I don't see one.

Oh sorry.

Here is it

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.

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.

Remember that resistors in parallel don't add - you must use the formula:



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

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?

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 ?

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

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 ?

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.

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 ?

*made it

Quote:

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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 ?

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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.

that OK :) .so you can tell me if the final answer is Uab=3v?

thanks so much for your help and for your time.

Quote:

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Originally Posted by pop000

that ok :) .so you can tell me if the final answer is Uab=3v?

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No - 1/2 amp of current flowing through a 2 ohm resistor yields? Voltage drop across that resistor.

Ok now I got it.

Thank you very much.