How Do You Solve for x in an Electrical Circuit with Iin/I=10?

[ilyas415]

1. Hi. The question I am trying to work out is Question 6 part (v). I'm stuck for a solution to the question. I know that it involves the use of the forumula for resistance and possibly is a potential dividers questions. I don't know quite where to start and I don't understand the following phrase in the question:

"Find x so that I_in/I=10"​

I know I need to find the length of x, but I'm stuck as to what they mean when they say that it must be done so that I_in/I=10.

2. Homework Equations : - Resistance = $$\rho$$(L/A)[/b]

3. The Attempt at a Solution : I worked out the resistance of the wire AB = 159 ohms

Thank you guys for your help. Sorry for any poor grammar!

 

[jegues]

1. Hi. The question I am trying to work out is Question 6 part (v). I'm stuck for a solution to the question. I know that it involves the use of the forumula for resistance and possibly is a potential dividers questions. I don't know quite where to start and I don't understand the following phrase in the question:

"Find x so that I_in/I=10"​

I know I need to find the length of x, but I'm stuck as to what they mean when they say that it must be done so that I_in/I=10.

2. Homework Equations : - Resistance = $$\rho$$(L/A)[/b]

3. The Attempt at a Solution : I worked out the resistance of the wire AB = 159 ohms

Thank you guys for your help. Sorry for any poor grammar!

I'm not 100% sure if my approach is correct, but here's my attempt at the problem.

Consider to two pieces of the wire from A to P and P to B as 2 separate resistors(R1 & R2), then simply write out you're desired currents, and solve for the distance X using a current divider.

The length of the portion of wire from B to P is x meters long, then the portion from P to A must be (1-x) meters long.

I hope this helps, again I'm not 100% sure whether this is the correct approach or not.

Good luck!

 

[vk6kro]

I agree with the resistance of the wire.

So, if the resistance of the wire across the input is R then the rest of the wire must have a resistance of 159 - R.

So the total in the right hand loop is (159 - R) + 100.

But this must have a resistance of 9 times the resistance of R to split the current in the required ratio.

So, you can say 9 * R = 259 - R.

You can work out R and the distance x from that. I think this is the same approach as Jegues used.