Series and Parallel Circuit Help

[Missvux3]

Homework Statement

helllo! i am having a lot of trouble on this circuit. Each resistor is 2ohms and the total voltage is 9V. I'm having trouble trying to find the current for each resistor and the voltage for each resistor. Thank you so much for your time! Here is the url to the image that i drew. :D

http://i570.photobucket.com/albums/ss145/Missvux3/Physics.jpg

Homework Equations

The Attempt at a Solution

So far what I solved is that the total resistance is 0.6ohms and the total voltage is 15A. I don't know how to do the rest.

 

[Cyosis]

You mean the total current is 15 A. This seems to be correct. What do you know about the potential differences across parallel resistors?

 

[Missvux3]

The voltage is the same as the total voltage for each parallel resistor and the total current is the same for each resistor in a series.

I'm not really sure how to solve it though :(

For the resistors in the series, since the current is the same as the total current, would the current be 15 as well? or would it be 5A for each because there are 3 resistors?

Thank you very much.

 

[Cyosis]

The voltage is the same as the total voltage for each parallel resistor and the total current is the same for each resistor in a series.
Thank you very much.

This is correct. So you know that the potential difference across the single resistors in the three branches to the right is 9 V (the same as the source). Using V=IR now should give you the current through those 3 resistors. You have a total of 15 A that is split between four branches so if you can calculate three branches easily you should be able to find the fourth one as well.

If you're still stuck just ask again.

 

[Missvux3]

Thank you so much. =]
When i tried solving it, for the resistors to the right I got 4.5A for each of them and for the resistors in the series, i got 2.5V for each and 3.75A for each. Is this correct?

 

[Cyosis]

The 4.5A is correct, the other values are not. If you have a total of 15A and it splits over 4 branches you get. I=I1+I2+I3+I4, you already know I2, I3, I4 and I, so what is I1? If it would be 3.75A as you say we would get 3.75+3*4.5=17.25 A, but we only have 15A to divide among the branches! Something must be wrong.

 

[Missvux3]

thank you so much for your time :] i really do appreciate you for helping me =D for the resistors in the series i got 0.5A for the current and 1V for the voltage for each of the resistors. :D is this correct? thank you so much again =D

 

[Cyosis]

How did you get the 0.5 A value for the resistors in series? Did you divide the total current (1.5A) through that branch in three? You are not allowed to do this for resistors in series. Imagine it being a pipe instead that gets narrower at three places (resistors). If you put water in one side of the pipe you expect everything you put into it on one end to go out at the other end right? Same for electrical current. Current does not get used up! So the current passing every resistor in the first branch is?

 

[Missvux3]

oh isee xD so would it be 1.5A for the current for each of the resistors in the series and 4.5V for the voltage? :D

 

[Cyosis]

The current is correct now, but how did you calculate the voltage? V=I R=1.5*2=?

 

[Missvux3]

ohhhh ooopss ! omgoshh very bad mistake =[ sorry about that ! D: would it be 3V? :D

 

[Cyosis]

Yep that's correct. Note that it should be, because the total potential difference across resistors in series should be equal to the sum of the potential differences across each resistor.

 

[Missvux3]

thank you so much for your time and help :] i really appreciate it! =DDDDDD

 

[Cyosis]

You're welcome.