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[SOLVED] Calculating current in paraller/series circuit ??

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osa1313

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Hi i have been stuck on this circuit for a long time and it seems impossible how current labelled I4 on the circuit below is 4mA.
Maybe somebody is kind enough to explain this to me, i have searched 5 books and none of them do an example on more than 4 resistors connected in this way.


The image, you can enlarge it by clicking on it.
5555555555555555555.png


Thank You for any help as this will help me pass my GCSE Electronics without being able to attend to the class.
 

Hi i got this as a test from my teacher over the holiday and by using Circuit Wizard it show i1 = 4mA a i want to know exactly what i have to do to get the correct answer 4mA.

And lets say the battery has unlimited current, that's what my teacher said think that it's being power by a power supply.
 

Hi i have been stuck on this circuit for a long time and it seems impossible how current labelled I4 on the circuit below is 4mA.
Click to expand...
I agree with Brad that it's literally impossible.

i have searched 5 books and none of them do an example on more than 4 resistors connected in this way.
Click to expand...
Rather than scanning various books for an exact solution, you should try to read about the general method to solve similar problems, as it's e.g. sketched in the DC network analysis chapter of this book https://www.allaboutcircuits.com/vol_1/chpt_10/index.html

The present circuit is actually a simple one, you don't need a pocket calculator or pencil and paper to "see" the results intuitively. In a few words, it's about evaluating the equivalent circuit resistance from right to left (by calculating alternatingly resistor series and parallel circuits) until you know the total resistance connected to the battery. Than calculate the total current and how it's shared by partial resistances from left to right.
 

To analyze the above circuit, one would first find the equivalent of R2 and R3 in parallel, then add R1 in series to arrive at a total resistance. Then, taking the voltage of battery B1 with that total circuit resistance, the total current could be calculated through the use of Ohm's Law (I=E/R), then that current figure used to calculate voltage drops in the circuit. All in all, a fairly simple procedure.

00206.png
Click to expand...

This is what i am able to do but with the circuit i have give on the photo i just don't know which resistors are connected in series and which in parallel, giving me this information would help a lot.
 

If you replace R3 by two series connected resistors in the post #5 circuit (e.g. name it R5 and R6), how does the calculation change? If you can solve the slightly modified problem, you have already halve of the original problem. Than take the new equivalenr resistance and apply the same schem again. As said, simply proceed from rigth to left.
 

(look circuit on post #1)

First i add up R5 + R6 = R56 = 2k

Then i calculate the parallel resistance of R456 = (2k * 2k)/(2k + 2k) = 1k

And this is where i get confused i need to calculate the overall resistance of this circuit to calculate I1 but i don't know if i do as follows

R4 + R3 = 4k then R234 = (4k * 3k)/(4+3) = 1.7k

then

R456 + R234 = 2.7k + R1 = 3.7k

You see yourself i get very confused because i have no idea is it R5, R6 and R4 that are connected in parallel or is it R2, R3 and R4.

- - - Updated - - -

And what role does R3 play is it in parallel with R2 and R4 ?
 

Then i calculate the parallel resistance of R456 = (2k * 2k)/(2k + 2k) = 1k

And this is where i get confused i need to calculate the overall resistance of this circuit to calculate I1 but i don't know if i do as follows

R4 + R3 = 4k then R234 = (4k * 3k)/(4+3) = 1.7k
Click to expand...
Not quite right. R3 is series connected (resistance added) to the previously calculated "R456". So you get R3456 = 3k.

Then parallel circuit with R2. R23456 = 1.5k

Then series circuit of R1. R123456 = 2.5k

The previously calculated partila resistances will help you to calculate current sharing.
 
Thank you very much i understand this now, but one last question on one website it said that "Two components are connected in series if they are connected to each other at exactly one point and no other component is connected to that point." so according to this information we divide the circuit into loops so we take R4, R5 and R6 as loop 3 which is lets say one big resistor that is connected to R3, in series.

And from my point of view in this circuit we can make 6 loops is that correct ?

5555555555555555555.png
 

There are 3 independent loops, the other loops can be created by adding or substrating others. We usually choose loops 1,2,3. The number three will represented by 3x3 matrix in the general numerical network solution (if we don't reduce it sequentially to an equivalent resistor).
 
I think i am going to loose all my hair by 0:00 :)

To get the voltage drop across resistor R1 i do I * R so 4mA * 1k = 4V but when i do it in circuit wizard it shows 6V, so how is that ?

Do i need to subtract the answer 4V from the total supply voltage 10V for the first resistor ?
 

Hi Osa1313,

There are a couple of ways to solve this problem without simulations. One is the brute force method where you take into account the 3 closed loops and apply Kirchoffs law for them. you also have 2 nodes to work out 2 more current relations. So you have 5 equations and 5 variables, hence you can figure it out. Its just a long drawn out process, thats all.

Another way to do it is to use superposition to solve this circuit.

Hope it helps,
K

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Another way to do it is to follow FvM's method to find I1. Once you have done that calculate the voltage across R2 (V - 4mA*R1). divide this by R2 to give you I2. From Kirchoffs law you can find I3. So you have I1 and I3 now so you can find the Voltage across R4 (V - I1*R1-I3*R3), then divide that by R4 to get I4 . then you can find I5 similarly

Hope it helps
K
 

I wish i had the ability to attend to electronics class so the teacher could explain me these easy methods.

To calculate U1 the voltage drop on resistor R1 do we do as follows:

Get the R123456 resistance which is 2.5k and then R123456 - R1 = 1.5k
4mA * 1.5k = 6V ??? The 4mA is the current I1

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One thing i don't understand i built a similar circuit but used different value resistors and 15V

I was able to get the voltage drop across R1 using this formula V = I * R but on this circuit that i gave in post #1 the outcome is very different compared to what you get by the equation.
 

Hey osa1313,

If you want voltage drop across R1 you should do as follows;

I1*R1 = 4mA * 1k = 4V ;

or you could also do 10 - 6V (the voltage you calculated is the drop across R2 which is in parallel with the rest of the circuit)

Hope it helps

Regards
K
 

karthick1987 said:
Hey osa1313,

If you want voltage drop across R1 you should do as follows;

I1*R1 = 4mA * 1k = 4V ;

or you could also do 10 - 6V (the voltage you calculated is the drop across R2 which is in parallel with the rest of the circuit)

Hope it helps

Regards
K
Click to expand...


I have built the circuit and voltage drop across resistor R1 is 6V.
Across R2 6V.
But i still don't know why ?
 

the solution is this the total current i1 is 4mA

i2 is 2mA
i3 is 2mA
i4 is 1mA
i5 is 1mA

ask me in pm if you want to know how i solved

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AMSA84 said:
Hello guys,

I tried to solve this problem, but I don't know if it's correct. Could someone see it? Maybe will help osa1313.

View attachment 78385

Best regards.
Click to expand...

you have done a great job but we can do it in voltage method that would have been much easier
 

Thank you.

But, now I am a little bite confused. You said in your post that the i1 is 4mA, but my calculations gave me 1,818mA. We can say, approximately 2mA. Right?

It looks like it is giving me half of the values.
 

your total circuit current is 4mA. In parellel circuit current will divide and voltage is constant.
Initially i1=4mA, then i2 = i3 = 2mA, after that i4 = i5 =1mA.
According to KCL i1=i2+i3; i3=i4+i5.
For these calculations you can gothrough simulation softwares like Proteus.
 
AMSA84 said:
Thank you.

But, now I am a little bite confused. You said in your post that the i1 is 4mA, but my calculations gave me 1,818mA. We can say, approximately 2mA. Right?

It looks like it is giving me half of the values.
Click to expand...

there is a prob of error sir i checked it out
 

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