# Calculate the Ratio of Ib/Ia in an Electrical Circuit

• SuperHero
In summary: Now I see, thanks.Ahha ... thus the wink. Now I see, thanks.In summary, the given circuit consists of a single battery V and several resistors with values of R and 2R. Using Kirchhoff's laws, the ratio of the two currents Ib/Ia can be calculated as (10R^2 + 6)/(2R^2 + 1). However, due to the symmetry of the circuit, a simpler solution can be found by considering the last three resistors at the right end, resulting in a current ratio of 1/2.
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

SuperHero said:

## 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

ehild said:
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?

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?

gneill said:
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.

phinds said:
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.

#### Attachments

• Fig1.gif
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gneill said:
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.

phinds said:
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.

gneill said:
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.

## 1. What is the purpose of calculating the ratio of Ib/Ia in an electrical circuit?

The ratio of Ib/Ia, also known as the input bias current ratio, is used to measure the level of accuracy and stability of an electrical circuit. It is an important parameter in determining the performance of amplifiers, filters, and other electronic components.

## 2. How is the input bias current ratio calculated?

The input bias current ratio is calculated by dividing the input bias current for the inverting input (Ib) by the input bias current for the non-inverting input (Ia). This can be done by measuring the currents at each input using a multimeter or by using mathematical equations.

## 3. What factors can affect the input bias current ratio in a circuit?

There are several factors that can affect the input bias current ratio in a circuit, including temperature, component tolerances, and aging of components. Any changes in these factors can cause the ratio to deviate from its initial value, which can impact the circuit's performance.

## 4. Why is it important to consider the input bias current ratio in circuit design?

The input bias current ratio is an important factor to consider in circuit design because it can affect the accuracy, stability, and linearity of the circuit. It can also impact the overall performance and reliability of the circuit, especially in precision applications.

## 5. How can the input bias current ratio be minimized in a circuit?

To minimize the input bias current ratio, precision components with low input bias currents should be used. Additionally, careful circuit design and layout techniques can also help reduce the effects of temperature and component tolerances on the ratio. Regular maintenance and calibration can also help ensure the ratio remains within acceptable levels.

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