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physics problem

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Old Apr 25, 2005 | 05:28 PM
  #1  
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alright, this page is due tomorrow and i pretty lost. i figured out the first two, but im stuck on this one.
it reads:
a.) Find the equivalent resistance (4.43)
b) If I=1.0 A in the 5 ohm resistor, find the potential difference between points a and b (46.5)

i was given the answers, i just dont understand how to get there. i also dont under stand which resistors are in series and which are in parallel. any help would be great
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Old Apr 25, 2005 | 06:08 PM
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yo dude gimmie a min ill hook you up
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Old Apr 25, 2005 | 06:12 PM
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alright man, thanx
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Old Apr 25, 2005 | 06:13 PM
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first off add the 4, 5, 3 ohm resistors... at teh end cause they are in series so it would be 12 ohms
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Old Apr 25, 2005 | 06:15 PM
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then the 6 and (new 12) are in parallel so use 1/Req= 1/R1 + 1/R2
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Old Apr 25, 2005 | 06:16 PM
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should help
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Old Apr 25, 2005 | 06:19 PM
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once you get the two parallel resistors added the resulting one is in series with the 1 ohm resistor so add them
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Old Apr 25, 2005 | 06:20 PM
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You need to simplify the circuit, and find the equivalent resistance of the entire circuit.

It helps if you look at it like this:




Start from the bottom, work up.

You have 4 ohm, 5 ohm, 3 ohm in series, so that equals one 12 ohm resistor. That 12 ohm resistor is in parallel with the 6 ohm resistor. I assume you know how to do parallel and series equivalent resistances...so the 12 ohm in parallel with the 6 ohm gives one equivalent resistor of 4 ohms. This 4 ohm resistor is in series with a 1 ohm, giving a 5 ohm equivalent. This 5 ohm is in parallel with a 2 ohm, which gives a 1.43 ohm equivalent. This is in series with a 3 ohm, which gives 4.43 ohms.

As for part b:

You know the current through the 5 ohm resistor is 1A, so you know the voltage difference through that section of the 3 resistors using ohms law....that voltage difference is 12V. The voltage difference across the 6 ohm resistor in parallel with that portion must be the same, so you can then determine the current through it also using ohms law, 2A. Add those 2 currents together, 3A, and you have the current going through the 1 ohm resistor. You then know this voltage difference also using ohms law, to be 3V across the 1 ohm resistor. Now you know the total voltage difference for that side of the circuit to be 15V. This must be the same across the 2 ohm resistor, so you know the current through that resistor to be 7.5A. Now you know the current going through the 3 ohm resistor on top to be 7.5A + 3A = 10.5A...and thus the voltage across it is 31.5V. Add this 31.5V to the bottom portion's 15V, and you get 46.5V.

If you need more help, my AIM screen name is Purespeedintegra.
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Old Apr 25, 2005 | 06:22 PM
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so now the 2 ohm and resulting resitor (comprised of 1,4,6,5, and 3 ohm) are in parallel so use the equation to solve them... and once you get that resistor it is in series with teh first 3 ohm

hope that helps
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