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- anonymous

Please take a look at the electricity problem on my website: http://www.romakoulikov.com/dc-electricity-review/
Thank you!

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- anonymous

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- anonymous

You wrote "We know that the resistances of I3 and I4 are equal. The voltage will be split in two, and so the currents should be equal as well." How do you know that? This information is not given. Yes, the resistances are equal, but how do you know that the voltage is split in two? You can't just assume that the currents through both resistors will be equal. All you know is that the total current in both resistors will be 170 mA. The current could be split in different ways.

- radar

You however, do know, Kirchoffs law which states that the sum of current entering a point equal the sums of current leaving a point. In this case the point I am referencing is the point at the junction of the four resistors. It is shown that 170 ma is entering the junction (I1 + I2) thus it is know 170 ma is leaving the point. It is shown that 100ma I3 goes through one path. So it is obvious that I4 is 70 ma.
!1+I2+I3+I4=0
70+100-100-70=)

- anonymous

radar i agree with U in use of KCL for the solution...that i3 + i4 should be 170 mA, but what do U know that i4 is 70 mA?

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- radar

I know that because of the amount and directions of I1, I2, and I3 which were given in the problem.

- radar

It is aribtrary, but one must be consistent. I usually give values of positive to currents entering a pont, and negative values leaving a point.

- anonymous

I'm sure that the current of I1 can flows in i3 and i4
so, let i4 has 2 possible answer 70 and 100 mA

- radar

It very well may but you are given the values in the other three resistors, use them wisely.

- anonymous

the value of each resisitor is 100 ohm

- radar

If you wish to say that I4 is 100 ma, are you agreeing that I2 is 100 ma as stated in the problem.

- anonymous

what do U think?

- radar

If I4 is 100 ma, then I2 can not be 100 ma as shown.

- anonymous

why?

- radar

Then the sums of all the currents are not zero at the junction of the four resistors, Kirchoff told me so. lol

- radar

I'm sticking with choice b. 70 ma. Check the answer. Good luck with your review. I don't know why they say their is insufficient info They have all three values of currents with one branch not specified, 4 equal value resistors.

- anonymous

yeah thats reasonable

- anonymous

Gentlemen, thank you for your input! Radar, you write that it is shown that I3 = 100mA. How do you we know that?
So far, what PhysMath wrote seems reasonable... Basically that my assumption that the current splits in half is incorrect. Although if the wires have absolutely the same properties, I don't see why this would be a faulty assumption. Thank you everyone!

- anonymous

Correction, physmath is a lady. And a student of my alma mater Stuyvesant High school at that. Represent!

- radar

I must admit I made a mistake. I thought the problem stated the value of I3 as 100 ma, after reviewing the diagram I see that I3 is not given. Also there was a typo in my reply stating that the sum was not zero at the junction of the four resistors please disregard the word "not".
I am now agreeing with the solution that there is insufficient information to determine the current.

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