How to solve resistance or current in a circular circuit?

A wire is bent in the form of circle whose resistance is 40 Ω. A and B are two points on the wire dividing it into a quadrant.Find the value of I 1 and I 2. The diagram is here: http://www.flickr.com/photos/78780315@N0… How to solve this?And how to solve this type of sum,when the circuit is circular?Plz help

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Sorry the link for the diagram is this one:

http://www.flickr.com/photos/78780315@N06/8412264087/in/photostream

The figure of the wire and of the equivalent circuit is attached below.

Arriving at point A the current `I_0 =1A` splits into two currents `I_1` and `I_2` that are inversely proportional to the resistance of the two portions of the wire (or equivalent to the lengths of the two portions of the wire).

`R_1 =rho*L_1/S =rho/S *(R*pi/2)`

`R_2 =rho*L_2/S =rho/S*(R*(3*pi)/2)`

From Kirchhoff laws we have

`I_1+I_2 =I0`

`I_1*R1 =I_2*R_2`

From the second equation we get

`I_1 =I_2*R_2/R_1 = 3*I_2`

`3*I_2 +I_2 =I_0 rArr I_2 =I_0/4 and I_1 =(3/4)*I_0`

Numerical values are

`I_1 =3/4 A =0.75 A and I_2 =1/4 A =0.25 A`

If you want to find also the resistances you have

`R_2 =3R1` and `R_1+R_2 =40 Omega`

`4R_1 =40 Omega rArr R_1 =10 Ohm and R_2 =30 Ohm`

**Answer: The currents in the wire are `I_1 =0.75 A` and `I_2 =0.25A` **

**Sources:**

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