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
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
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`
Sorry the link for the diagram is this one: