How to find current direction when analysing a cricuit using KVL, KCL

[vord_ler]

Homework Statement

Good afternoon, I am having a problem to understand how to set up the current directions when analyzing a circuit using KVL and KCL. I know that it's probably solvable using other more advanced techniques, such as NVM (node-voltage method) or mesh-current method, but it's not yet covered in my classes and in the book, so I want to understand the algorithm on how to find the current directions properly in order to apply the KVL and KCL. I also would like to get some hints on how to apply the KVL and KCL to the given circuit, after defining the current directions. I already tried to solve the problem by setting up the currents going out of the node, but the final answer is usually not correct.

Relevant Equations

V = i*R

 

[BvU]

Hello @vord_ler ,

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Please read the PF homework rules and guidelines: you'll need to post your best effort before we are allowed to help

In the mean time, you can simplify the drawing considerably and work out the voltages over some of the resistors/resistor combinations and some currents as well

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[vord_ler]

Hello @vord_ler ,

$\qquad$!​

Please read the PF homework rules and guidelines: you'll need to post your best effort before we are allowed to help

In the mean time, you can simplify the drawing considerably and work out the voltages over some of the resistors/resistor combinations and some currents as well

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Thank you for your answer. I would like to know at leats how to set up the currents direction properly, so I can continue analyzing the circuit. Could you please provide some algorithm maybe that can help me to set the direction of currents, or any resources that contain that info. Appreciate your response.

 

[BvU]

set up the currents direction properly

There is no prescribed way to do that. Just pick a direction; the math will come up with a sign.

I had fun redrawing your circuit, so I might as well post it. See if you agree and fill in the dots

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[vord_ler]

There is no prescribed way to do that. Just pick a direction; the math will come up with a sign.

I had fun redrawing your circuit, so I might as well post it. See if you agree and fill in the dots View attachment 333953

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Oh finally I got the correct answer, thank you! Actually, there was no problem with the current directions (as you said, it doesn't matter) but the problem was with the voltage signs across resistors, I messed them up. This time, I just set the 5 ohm resistor from + - (voltage drop) and other resistors accordingly. I am so sorry to take your time by such a silly question and thank you for directing me towards the solution!

 

[vord_ler]

There is no prescribed way to do that. Just pick a direction; the math will come up with a sign.

I had fun redrawing your circuit, so I might as well post it. See if you agree and fill in the dots View attachment 333953

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I am sorry to distrupt you again, but can you please tell, why the solution to find Vg (the solution I provided belong) is incorrect? I compared my answer to the answer given in the book, and the answer in the book is 190V (the solution in the book is quite different from mine and doesn't really make sense to me)

P.S. the second attachment is the book solution

 

[alan123hk]

the solution in the book is quite different from mine and doesn't really make sense to me

Please double check your Kirchhoff loop rule equation, it seems problematic to me.

 

[BvU]

I don't understand what you do in loop 1: I agree that V_B - V_D = 160 V (labels in post #4) but to me i₁ is still an unknown

[edit] my bad, overlooked i₁ = -6 A. @alan123hk points you to the sign correction.

(and I read 100 - (-6)*(15) -- here's why you should use $\LaTeX$ -- so I looked at the lower left resistor; but even then the current there is 4 not 6 A --- I should have read more thoroughly instead of hurrying this useless post $\$

 

[vord_ler]

I don't understand what you do in loop 1: I agree that V_B - V_D = 160 V (labels in post #4) but to me i₁ is still an unknown

[edit] my bad, overlooked i₁ = -6 A. @alan123hk points you to the sign correction.

(and I read 100 - (-6)*(15) -- here's why you should use $\LaTeX$ -- so I looked at the lower left resistor; but even then the current there is 4 not 6 A --- I should have read more thoroughly instead of hurrying this useless post $\$

I am sorry, but I still cannot understand where is the mistake in my solution. Can you please provide more details or maybe send your own solution using loop rule here?

 

[vord_ler]

Please double check your Kirchhoff loop rule equation, it seems problematic to me.

Can you please provide more details on what is exactly problematic in my solution, please?

 

[BvU]

In your loop 1 you write $$100-(-6)*(5)-v_g +(-4)*(15)$$which is KVL starting from point D in the picture in post #4. But, going from point B to point D the voltage over the 15 $\Omega$ is + 60 V
as you correctly indicate in your picture !

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[alan123hk]

Can you please provide more details on what is exactly problematic in my solution, please?

#11 already pointed out the problem.

You write the voltage change across the 5 ohm resistor as $-(-6*5)~$ , then why don’t you write the voltage change across the 15 ohm resistor as $~+(-4*15)~$ ?

 

[vord_ler]

#11 already pointed out the problem.

You write the voltage change across the 5 ohm resistor as $-(-6*5)~$ , then why don’t you write the voltage change across the 15 ohm resistor as $~+(-4*15)~$ ?

I wrote exactly like that and got: 100 + 30 - 60 = 70 V, while the answer in the book is 190 V.

 

[vord_ler]

In your loop 1 you write $$100-(-6)*(5)-v_g +(-4)*(15)$$which is KVL starting from point D in the picture in post #4. But, going from point B to point D the voltage over the 15 $\Omega$ is + 60 V
as you correctly indicate in your picture !

View attachment 333984​

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But why it's + 60V? I traverse in clockwise direction and there is a voltage increase across the 15 ohm resistor, but the direction of the 4 A current is counterclockwise, which means we finally get (-4) * 15 = -60 V, why don't we?

 

[BvU]

I wrote exactly like that and got: 100 + 30 - 60 = 70 V, while the answer in the book is 190 V.

And it should have been 100 + 30 + 60
the voltage goes UP when you go across the resistor. The highest voltage is where the current enters the resistor.

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[vord_ler]

And it should have been 100 + 30 + 60
the voltage goes UP when you go across the resistor. The highest voltage is where the current enters the resistor.

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I see now. Still it's strange for me why the 4 A current doesn't change its sign, since its original direction(with + sign) was counterclockwise, while we are traversing the loop in the clockwise direction.

 

[BvU]

Does it look better when presented like this ?

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[vord_ler]

Does it look better when presented like this ?

View attachment 333996

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Thank you for you representation, but why there is 30 V increase across 5 ohm resistor, if we are going from the higher to lower voltage?

 

[BvU]

You don't mean Ohm's law

I think I get it (te source of your confusion): your loop 1 is going clockwise, starting at point C in the picture in post #4 (not point D as I mistakenly wrote in #11 ).

So you start with 100 V,

Then subtract the voltage over the 5 $\Omega$ which is (-6 A)*(5 $\Omega$)
you subtract -30 V because you defined the voltage as + on the left, - on the right
meaning you effectively add 30 V and at point E you are at +130 V

Now down by v_g to get at point D . so 100 + 30 - v_g

Last step: from D to C

and here you go from right to left in your clockwise loop, so you add (4A)*(15 $\Omega$) is +60 V

i.e. we are not

if we are going from the higher to lower voltage

but in the direction in which we go through the loop

Clearer ?

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[vord_ler]

You don't mean Ohm's law

I think I get it (te source of your confusion): your loop 1 is going clockwise, starting at point C in the picture in post #4 (not point D as I mistakenly wrote in #11 ).

So you start with 100 V,

Then subtract the voltage over the 5 $\Omega$ which is (-6 A)*(5 $\Omega$)
you subtract -30 V because you defined the voltage as + on the left, - on the right
meaning you effectively add 30 V and at point E you are at +130 V

View attachment 333997

Now down by v_g to get at point D . so 100 + 30 - v_g

Last step: from D to C

View attachment 333998
and here you go from right to left in your clockwise loop, so you add (4A)*(15 $\Omega$) is +60 V

i.e. we are not

but in the direction in which we go through the loop

Clearer ?

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Yes, It's clear that we should add 60V because we are going from - to + across 15 ohm resistor, but the fact that the 4 A current is flowing in the opposite direction of our traverse but we still DO NOT change its sign to -4A (as we usually do in order to specify that the current flows in opposite direction relative to its original position) still unclear for me. (I am so sorry to take your time on such simple things, but I just want to make everything clear now so I will not have problems with it any further)

 

[BvU]

Yes, It's clear that we should add 60V because we are going from - to + across 15 ohm resistor, but the fact that the 4 A current is flowing in the opposite direction of our traverse but we still DO NOT change its sign to -4A (as we usually do in order to specify that the current flows in opposite direction relative to its original position) still unclear for me.

We don't change the sign of the current because the current is flowing in the direction of the arrow (under 4A).
The sign of the voltage is defined by the + to the left and the - to the right of the 15 $\Omega$.

And Ohm's law says $V = I\,R$ with the sign/direction as shown here

I am so sorry to take your time on such simple things, but I just want to make everything clear now so I will not have problems with it any further

No need to apologize. Best thing to do now is try another exercise on the same subject.

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[vord_ler]

We don't change the sign of the current because the current is flowing in the direction of the arrow (under 4A).
The sign of the voltage is defined by the + to the left and the - to the right of the 15 $\Omega$.

And Ohm's law says $V = I\,R$ with the sign/direction as shown here

View attachment 334000

No need to apologize. Best thing to do now is try another exercise on the same subject.

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Now everything is clear, thank you! Let me finally clarify it: when the direction and value of the current is defined, and we are applying KVL, we omit the direction and take the given value of the current.