Hi,

Have been sitting with this question for some time now, and about to got nuts. I know how to do it when its with volt as input, but this time it is amps. I have been calculating back and forth and i gives me 6 amps (which is the number to start with).
Hoping someone can help me out

 

Normal person would apply either Node-Voltage or Mesh-Current Method.
Why aren't you?

 

Hi,

How you solve this depends partly on what you know already.
For example, do you know voltage division or current division using resistors?

If so, combine resistors from right to left until you are left with just two resistors, then calculate the voltage at the junction, then work from left to right to calculate the other quantities you need.

If you dont know how to do any of this you'll have to say so.

 

I have calculated resistors in parallel, so R23 ended up being 150ohm, and R56 being 100. From there i used knowledge about resistors in series, and now i have two resistors. R123 being 350 ohm, and R456 being 300 ohm. From there on im not quite sure what to do?

 

I have calculated resistors in parallel, so R23 ended up being 150ohm, and R56 being 100. From there i used knowledge about resistors in series, and now i have two resistors. R123 being 350 ohm, and R456 being 300 ohm. From there on im not quite sure what to do?

R456 is correct. R4 and R56 are in series.

But R23 and R456 ARE NOT in series. They are in parallel. You will want to redo this part.

 

I see.

So i use resistors in parallel and R23 in parallel with R456 gives me a new resistance of 100 ohm. Now they are in series. Then i can multiply, giving me an resistance of 300 ohm for the whole circuit. What now then?
So calculate the current, I, i use the formula I =V/R, but because i don't know the V, i can calculate my V from amps, and watt. Therefore i calculate my Power, W using P = I^2 * R = 10.800. Then calculating V by using I and V gives me 1800 V. Now i use the formula I = 1800/300 = 6 amp, which is the number i now from the start?

Am i totally wrong here?

 

I see.

So i use resistors in parallel and R23 in parallel with R456 gives me a new resistance of 100 ohm. Now they are in series. Then i can multiply, giving me an resistance of 300 ohm for the whole circuit. What now then?
So calculate the current, I, i use the formula I =V/R, but because i don't know the V, i can calculate my V from amps, and watt. Therefore i calculate my Power, W using P = I^2 * R = 10.800. Then calculating V by using I and V gives me 1800 V. Now i use the formula I = 1800/300 = 6 amp, which is the number i now from the start?

Am i totally wrong here?

Resistors in series ADD. R1=300 Ohm, R23456=100 Ohm. R=R1+R23456=400 Ohm

Since you don't know that resistors in series ADD together. It seems to me that you R456 is wrong.

 

Some people redraw the circuit with each step so they can clearly follow their own thinking.

 

Some people redraw the circuit with each step so they can clearly follow their own thinking.

That's advanced college way of doing things. I am pretty sure OP wants to acquire this advanced skill through painful experience.

 

Im sorry, i was waaay to quick on the keyboard. Of course they add together, and gives me a total of 400 ohm resistance for the whole circuit.

Then i calculate the power, w, by P = I^2*R, which gives me total power of the circuit being 14.400 W.

From there i calculate my 6 I, amps to volt since i can then use the formula.

Gives me a voltage of 2400.

Then i use the formula I = V/R to calculate the current. 2400/400 = 6 amp.

Am i right here?

Btw. i am redrawing on papir by my side, i just don't have the possibility to upload, so therefore explaining and making it clearer for myself.

 

Don't see where you trying to get at? I'm all new where so don't blame for not having the formulas on my spine, but in the end, we got the same numbers, so no big deal. Not trying to be rough.

So will you please try helping from here and on? I got the numbers, i just need to know the way go with this

I get the result of 6 amps through the resistor, but from the assignment description i already knew that?

 

Btw. i am redrawing on papir by my side, i just don't have the possibility to upload, so therefore explaining and making it clearer for myself.

Use Paint (or other drawing software that comes with virtually all operating systems). Use boxes for the resistors.

You said in your original post that you know how to do it if it was a voltage source and not a current source. Well, this is a linear circuit which means that if you double the voltage you double the current. So solve the circuit using a 1 V source and see what current you get for the source. Let's say that it comes out to be 1.5 A. So you need four times the current to get to your 6 A, so your current source is putting out 4 V. That's not the 'best' way to do it, but it is 'a' way to do it. Something to keep in mind on exams should you be faced with a similar situation.

 

Can you maybe elaborate what? Is there a specific way to do it with a current source? Because it is not something i have heard of so far?
I know it when it is a voltage source, by using I = V/R.

When i do it your way, it gives: I = 1/400 = 0,0025 Amps

What can i say from that? It is that part of current through the resistor that confuses me, because i know it is 6 Amps. from the start, and by calculating it also gives me 6 Amps. and that is current, I....

 

Where did the 1V come from?

Another technique is to create a table ... so you can fill in the answers as you develop them ... and double check them using KVL and KCL.

 

No specific place, i was just following WBahn's example, but wanted him to elaborate the amps i got, and what i could tell from that

 

Where did the 1V come from?

Another technique is to create a table ... so you can fill in the answers as you develop them ... and double check them using KVL and KCL.

It's simply the easiest value to work with. You can pick whatever value you like, since at the end you are going to scale it to the correct value.

 

ok

 

Can you maybe elaborate what? Is there a specific way to do it with a current source? Because it is not something i have heard of so far?
I know it when it is a voltage source, by using I = V/R.

When i do it your way, it gives: I = 1/400 = 0,0025 Amps

What can i say from that? It is that part of current through the resistor that confuses me, because i know it is 6 Amps. from the start, and by calculating it also gives me 6 Amps. and that is current, I....

Okay, so with a 1V source you would have a current of I = 1V/400Ω = 2.5 mA.

What would the voltage need to be in order to make I = 6A?

 

I would need to multiply by 2.400, so its putting out 2.400 V?

It all adds up to the 6 amps which is already given from the start, so the current flowing through the resistor is just the 6 amps, now i have just showed it by calculation? Gesh, it's some rather complicated questions.. or at least to get the correct info

 

I would need to multiply by 2.400, so its putting out 2.400 V?

We are going to have some confusion because you use the , as the radix and . as the thousands separator whereas as most of us do the opposite. We'll just have to work through it.

But, yes, your current source is putting out 2400 V.

It all adds up to the 6 amps which is already given from the start, so the current flowing through the resistor is just the 6 amps, now i have just showed it by calculation? Gesh, it's some rather complicated questions.. or at least to get the correct info

There are lots of ways to solve this problem. If you reduce the circuit to a single resistor you have 400 Ω and you know that 6 A is flowing through it. Thus you know that the applied voltage is 2400 V. What is the voltage drop across R1? That will let you get the voltage across R2 and R3, from which you can calculate the currents in those two resistors. After subtracting those currents from the 6A, you have the current in R4, from which you can calculate the drop across R4 to get the voltage across R5 and R6, from which you can calculate the currents in those two resistors. Since those resistors are equal, you also know that the currents splits and so has to be half of the current in R4, so you have a built in check. None of those computations involves math that you shouldn't be able to do in your head. They key is to be systematic and walk from one end of the circuit to the other.