# How do I get the Current to be closer to 510uA instead of 1mA?

• Sevarin
In summary, Sevarin is asking for help with a homework problem. He has correctly analyzed the resistors but has not been able to find a solution for the rest of the circuit.
Sevarin
Homework Statement
Calculate V7 and Is, I5 in the network below. Show your work.
Relevant Equations
Ix=(Rt/Rx)*I , (R1*R2)/(R1+R2)=parallel resistors. I=V/R, V=I*R, R=I/V, Vx = Rx*(E/Rt)
Homework Statement: Calculate V7 and Is, I5 in the network below. Show your work.
Homework Equations: Ix=(Rt/Rx)*I , (R1*R2)/(R1+R2)=parallel resistors. I=V/R, V=I*R, R=I/V, Vx = Rx*(E/Rt)

Hi there everyone Sevarin here and I am asking for some help on this question i have been stuck with for a while. It's to do with adding up resistors in series and parallel then taking the whole Rt (Resistor Total) and finding out the current. Knowlage needed is ohm's law and current divider rule.

https://cdn.discordapp.com/attachments/597514035017285643/639618783488639006/unknown.pngSimulator picture

So these are the pictures I have currently. I have ran it through my simulator and i am supposed to get around 510.703uA but as you can see I am getting 1mA granted yes its close... but not close enough since it needs to be 0.5mA to be closer or at least near to that. Have I made a mistake any where or got lost along the way? Please let me know.

Thank you so much for your time and help.

I think that triangle on the right was thrown into confuse you. You still do not have it quite right. Try labeling each vertex with A, B, C, etc. Then redraw that part of the circuit with all right angles and horizontal and vertical lines with the vertices labeled. Make sure each vertex is connected to the same vertices in each drawing the same resistors are is between vertices A and B, Band C, etc. in each drawing. If you do that carefully, you will be able to see which ones are in parallel and which are in series.

Sevarin
tnich said:
I think that triangle on the right was thrown into confuse you. You still do not have it quite right. Try labeling each vertex with A, B, C, etc. Then redraw that part of the circuit with all right angles and horizontal and vertical lines with the vertices labeled. Make sure each vertex is connected to the same vertices in each drawing the same resistors are is between vertices A and B, Band C, etc. in each drawing. If you do that carefully, you will be able to see which ones are in parallel and which are in series.
forgive me if i am wrong but did i not redraw it on the other page? i found that the combination between those ones was 8.9kOhms instead of the 30k that i thought it was. unless I am still missing something?

I think i see what's going on the pages arent going any bigger so here's a link redesigned diagram on back

first page

Sevarin said:
forgive me if i am wrong but did i not redraw it on the other page? i found that the combination between those ones was 8.9kOhms instead of the 30k that i thought it was. unless I am still missing something?

I think i see what's going on the pages arent going any bigger so here's a link redesigned diagram on back

first page
Yes, I saw that diagram and it is still not right. Try the vertex labeling trick that I mentioned in post #2 and see if you can make things match up.

Hello @Sevarin .

I see that you are new to PF, so here are a few comments regarding posting in the PF Homework Help forums. First of all: Familiarize yourself with the guidelines given in the "sticky" thread found in this, the "Engineering and Comp Sci Homework Help" forum. As such, much of the information you have relied on your images to provide to any potential helper should have been typed in as text. The details of the problem, including the circuit diagram are virtually impossible to read.

Due to you being new to PF, I will fill in some of the info which I was able to gather (with considerable effort) from your images. This is a one time service.

The modified (as per the info in the image) resistance values are given in the following table.

 R1 R2 R3 R4 R5 R6 R7 R8 R9 3.9kΩ 8.2kΩ 12kΩ 22kΩ 12kΩ 12kΩ 10kΩ 3.3kΩ 5.6kΩ
.
The battery voltage is changed to 12 volts.

In the following enlarged image, I have crudely labelled the resistors, R1, R2, ... , R9. (I have also made one small modification to the circuit.)

Regarding your attempt to solve the problem:
You appear to have analyzed the left hand side of the circuit, R1 through R5 correctly.

The solution for R6 through R9 is incorrect.

In the above image, does modifying the circuit as I have done alter anything?

Last edited:
jim mcnamara
Sevarin said:
...

So these are the pictures I have currently. I have ran it through my simulator and i am supposed to get around 510.703uA but as you can see I am getting 1mA granted yes its close... but not close enough since it needs to be 0.5mA to be closer or at least near to that. Have I made a mistake any where or got lost along the way? Please let me know.

Thank you so much for your time and help.
After struggling to figure out what you were doing and what you were asked to find and what various fuzz balls in the images could possibly be, I actually think that I know what you did although, there's no way to know why you did some of it as well as figuring out why you failed to do some other things.

You did state that you were to find IS, I5, and V7. You should have stated what each of those is. I suppose it's clear enough that I5 is the current through R5 and V7 is the potential drop across R7. I'm guessing that I is the current through the battery, but we shouldn't be expected to guess the meaning of any of these quantities.

You have stated that the simulator gives 0.510703 mA, but that you keep calculating a value of 1 mA. Which current are you trying to calculate, and which current is the simulator calculating? IS or I5 ?

As @tnich and I have pointed out, you have errors in your analysis of what combinations are in parallel versus series. However, you do have the correct result for the equivalent resistance (you call it R12345) of the branch comprised of , R1, R2, R3, R4, and R5. From that I came up with the current through R5 being 0.5106 mA. (I got this using your value, 23.5kΩ, for R12345.)

berkeman

## 1. How does the current flow through a circuit?

Current flows through a circuit from the positive terminal of a voltage source, through the components, and back to the negative terminal of the source.

## 2. What is the relationship between current and resistance?

According to Ohm's Law, the relationship between current and resistance is inversely proportional. This means that as resistance increases, current decreases, and vice versa.

## 3. How can I decrease the current in my circuit?

To decrease the current in a circuit, you can increase the resistance by adding resistors or using components with higher resistance values.

## 4. What is the difference between 510uA and 1mA?

The difference between 510uA (microamperes) and 1mA (milliamperes) is the magnitude of current. 1mA is equal to 1000uA, so 510uA is a smaller amount of current.

## 5. How do I measure and adjust the current in a circuit?

To measure current, you will need a multimeter. To adjust the current, you can modify the circuit by adding or changing components to achieve the desired current value. You can also use a variable resistor to fine-tune the current.

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