Calculate the Ratio of Ib/Ia in an Electrical Circuit

[SuperHero]

Homework Statement

Part of an elecrial circuit for the olympic score board consisted of a single battery V and several resistors having on of two values. R or 2R. They are hooked up in the confriguration as shown. Calculate the ratio of the two currents shown Ib/Ia
http://s1302.beta.photobucket.com/user/Rameel17/media/gggg_zps5b6383dc.jpg.html#/user/Rameel17/media/gggg_zps5b6383dc.jpg.html?&_suid=135753198581409194345584751531

The Attempt at a Solution

So for Ib i did

Rt = R + 1/2R
Rt = 2R^2 /2R+ 1/2R
Rt = 2R^2 + 1 /2R

Ib = V/2R^2 + 1 /2R
Ib = V/4R^3 + 2R

for Ia
Rp = 1/2R + 1/2R + 1/2R +1/2R +1/2R +1/2R
Rp = 6/2R

Rs = R + R + R + R + R
Rs = 5R

Rt = 5R + 6/2R
Rt = 10R^2 + 6 / 2R

Ia = V / 10R^2 + 6/2R
Ia= V/ 20R^3 + 12R

Ib/Ia= V/4R^3 + 2R / V/ 20R^3 + 12R
= 20R^3 + 12R/4R^3 + 2R
= 2R (10R^2 + 6) / 2R (2R^2 + 1)
= (10R^2 + 6) /(2R^2 + 1)

i don't know what to do next

 

[ehild]

Homework Statement

Part of an elecrial circuit for the olympic score board consisted of a single battery V and several resistors having on of two values. R or 2R. They are hooked up in the confriguration as shown. Calculate the ratio of the two currents shown Ib/Ia
http://s1302.beta.photobucket.com/user/Rameel17/media/gggg_zps5b6383dc.jpg.html#/user/Rameel17/media/gggg_zps5b6383dc.jpg.html?&_suid=135753198581409194345584751531

The Attempt at a Solution

So for Ib i did

Rt = R + 1/2R
Rt = 2R^2 /2R+ 1/2R
Rt = 2R^2 + 1 /2R
...

What is Rt? Your last equation is dimensionally incorrect.

ehild

 

[SuperHero]

What is Rt? Your last equation is dimensionally incorrect.

ehild

oh sorry i re-did this this way now:
Ib= V/R
Rtot = R + 2R
Rtot = 3R
Ib = V/3R

for Ia = V/R
Rseries = 5R
Rparallel = 6/2R
Rtot = 10R^2 + 6 / 2R
Ia = V/20R^3 + 12R

right?

 

[gneill]

It is unclear what resistors you're combining and how. It appears that whatever approach you're taking it's not producing the results you want; the units in your expressions don't match (they are dimensionally incorrect as ehild mentioned).

Looking at the diagram there are very few candidates for parallel or series connected components for direct simplification purposes. So you'll either have to make a brute-force kirchhoff law attack (mesh or nodal) or find a clever approach employing symmetry if you can spot it

Hint: consider just the last three resistors at the right end. What equivalent resistance do they 'present' to the rest of the circuit?

 

[phinds]

Looking at the diagram there are very few candidates for parallel or series connected components ...

Uh ... I'm wondering if we are looking at the same circuit? The value of Ib is truly trivial and Ia isn't much harder

EDIT: and I mean just sort of by looking at it ... there is NO sophisticated analysis needed.

 

[gneill]

Uh ... I'm wondering if we are looking at the same circuit? The value of Ib is truly trivial and Ia isn't much harder

EDIT: and I mean just sort of by looking at it ... there is NO sophisticated analysis needed.

I'm looking at this circuit:

The symmetry of the "repeated cells" allows for a quick solution for the current ratio, otherwise it could be a bit of a slog.

 

[phinds]

I'm looking at this circuit:

The symmetry of the "repeated cells" allows for a quick solution for the current ratio, otherwise it could be a bit of a slog.

Exactly. That's why I was puzzled by your original statement.

 

[gneill]

Exactly. That's why I was puzzled by your original statement.

I was looking at the OP's attempts and it didn't appear that he'd spotted the short cut, which prompted my post; I wanted to find out more about his approach, and see if he could spot the short cut.

 

[phinds]

I was looking at the OP's attempts and it didn't appear that he'd spotted the short cut, which prompted my post; I wanted to find out more about his approach, and see if he could spot the short cut.

Ahha ... thus the wink.