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Electronics current calculation

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  • #2
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Can you let me know if this is correct? Seems to make sense to me.
Almost.
You're reasoning is flawless! :smile:

However, at the end you have two resistors, and you accounted for only one of the two.
 
  • #3
Femme_physics
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Hmm. I'm looking back and I'm asking myself, what did I do?!? I must've spent too much time in the sun yesterday (I actually did!)

Okay, it really should be

RT=R1+R2,3 = 38.75

NOW I do

I1=10/38.75

I'm so stupid for not getting it right initially. I mean, seriously, I really gotta double check myself more often. (*self-annoyance*)
 
  • #4
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Hmm. I'm looking back and I'm asking myself, what did I do?!? I must've spent too much time in the sun yesterday (I actually did!)

Okay, it really should be

RT=R1+R2,3 = 38.75

NOW I do

I1=10/38.75

I'm so stupid for not getting it right initially. I mean, seriously, I really gotta double check myself more often. (*self-annoyance*)
Yep! :smile:

Oh dear, you're really getting good at electronics! :wink:
 
  • #5
Femme_physics
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Haha. Thanks. You're too nice :) Wait, this isn't over! I figured to ease you in as I ease myself in. Now I have to find I2

http://img268.imageshack.us/img268/454/fori2.jpg [Broken]

(do you like my notation analogy to moment calculations? ;) but the way, is it okay to do it?)
 
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  • #6
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Haha. Thanks. You're too nice :) Wait, this isn't over! I figured to ease you in as I ease myself in. Now I have to find I2

(do you like my notation analogy to moment calculations? ;) but the way, is it okay to do it?)
*attention wandering*

Oh hey, there's something else here! ;)

No this is not quite right. :(
You seem to be mixing up the current law and the voltage law.
[EDIT] Hold on, you're not mixing it up! Revising post ... See next reply. [/EDIT]

As for the notation, yes, you can use this type of notation, but you should use a different symbol.


On the wikipedia page here: http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws
you can see that they write it as:

[tex]\sum_{k=1}^{n} V_k = 0[/tex]


I would suggest writing it as:

[tex]\sum V = 0[/tex]
 
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  • #7
gneill
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It's a series circuit, so the current should be the same for each component; I2 = I1.
Also, where did the 12V come from? I thought you'd decided that the net source voltage was 10V?
 
  • #8
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I didn't expect to see the drawing of the loop being isolated from the rest of the circuit.
I guess people never do that.

Usually, the entire circuit is always drawn, possibly with one or more loop-symbols in it (a circle ending in an arrow).

There are a few other things wrong though.
You have the wrong voltage and you didn't properly substitute the values for I1 and R1.
 
  • #9
Femme_physics
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No this is not quite right. :(
You seem to be mixing up the current law and the voltage law
Voltage law states that the e directed sum of the electrical potential differences (voltage) around any closed circuit is zero. Voltage is IR. The current that flows through R1 is I1, the current that flows through R2 is I2

It's a series circuit, so the current should be the same for each component; I2 = I1.
But I know that it's only a section of the circuit, the real circuit has I1 splitting to I2 and I3. Am I suppose to just ignore that fact?


Also, where did the 12V come from? I thought you'd decided that the net source voltage was 10V?
*smacks forehead* My bad. Should be 10V.


I would suggest writing it as:

Duly noted :) Thanks.
 
  • #10
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Voltage law states that the e directed sum of the electrical potential differences (voltage) around any closed circuit is zero. Voltage is IR. The current that flows through R1 is I1, the current that flows through R2 is I2

But I know that it's only a section of the circuit, the real circuit has I1 splitting to I2 and I3. Am I suppose to just ignore that fact?
No you're right. Please look at my post in between.

Duly noted :) Thanks.
Pity you dewed spell it correctly this time. ;)
 
  • #11
gneill
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Ah. I see. So R3 is still there in spirit!

In that case, your I1 from before still applies to R1. Find the voltage drop across R1 due to I1, and along with the 10V source voltage, find the net voltage "presented" to R2. The current through R2 will then follow.
 
  • #12
Femme_physics
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I didn't expect to see the drawing of the loop being isolated from the rest of the circuit.
I guess people never do that.

Usually, the entire circuit is always drawn, possibly with one or more loop-symbols in it (a circle ending in an arrow).

There are a few other things wrong though.
You have the wrong voltage and you didn't properly substitute the values for I1 and R1.
Now you and gneill are contradicting each other?

Edit: Nevermind, he wrote a new reply. This is going fast lol


I didn't expect to see the drawing of the loop being isolated from the rest of the circuit.
I guess people never do that.

Usually, the entire circuit is always drawn, possibly with one or more loop-symbols in it (a circle ending in an arrow).
As you please :)

http://img855.imageshack.us/img855/1184/doner.jpg [Broken]


But I'm still calculating the first loop algebraically the same way I did with the "section drawing", right?

You have the wrong voltage and you didn't properly substitute the values for I1 and R1.
Yes, noted with the wrong voltage, as far as I1 and R1, oh you're right, I used my earlier wrong calculation of I1. *self mantra* Double check double check double check....

Pity you dewed spell it correctly this time. ;)
^^
 
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  • #13
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Now you and gneill are contradicting each other?

Edit: Nevermind, he wrote a new reply. This is going fast lol
Right! gneill was also thrown off by the isolated part of the circuit!
So effectively he confirmed my observation! :wink:

But I'm still calculating the first loop algebraically the same way I did with the "section drawing", right?
Yep! :smile:

And your new drawing is looking good.
Although... what is the function of the large circle around the entire circuit?

I meant that you should mark only the part of the circuit that you're using in the voltage law.

Yes, noted with the wrong voltage, as far as I1 and R1, oh you're right, I used my earlier wrong calculation of I1. *self mantra* Double check double check double check....
:)
 
  • #14
Femme_physics
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Right! gneill was also thrown off by the isolated part of the circuit!
So effectively he confirmed my observation
I was bestowed with great confusion powers at birth... :-)

Although... what is the function of the large circle around the entire circuit?

I meant that you should mark only the part of the circuit that you're using in the voltage law.
I see.

http://img859.imageshack.us/img859/9899/therei2.jpg [Broken]

Is there another way to calculate I2?
 
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  • #15
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I was bestowed with great confusion powers at birth... :-)
You have indeed been confusing me. But I like it! :smile:

I see.

Is there another way to calculate I2?
Sure.
You could draw another loop between R2 and R3 and do the voltage law there.
Then you would combine that with the current law at the crossroad above R2 and R3.

This would a great fullfillment of your self-mantra (*double-check* *triple-check* *double-check*). :wink:
 
  • #16
Femme_physics
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You have indeed been confusing me. But I like it!
You've been bestowed with great "confusion-untangling" powers! Or rather, unconfusion powers! :)

Sure.
You could draw another loop between R2 and R3 and do the voltage law there.
Then you would combine that with the current law at the crossroad above R2 and R3.
Yes, right, sorry; I meant if there is another way other than using Kirchhoff laws. I am aware of this 2 equations 2 unknowns thing I could've done (indeed I wanted to spare myself some mathy work so I choose the easier loop^^)

This would a great fullfillment of your self-mantra (*double-check* *triple-check* *double-check*).
LOL indeed

I think I would make THAT as my new mantra (with the "triple check" added!)
 
  • #17
gneill
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You have I1 already, so you know what the voltage drop across R1 must be, right?

The voltage at the top of R2 and R3 should be the voltage of the source minus the voltage drop on R1. So that gives you the voltage across both R2 and R3. The currents for both R2 and R3 then follow from Ohm's Law for those components.
 
  • #18
Femme_physics
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You have I1 already, so you know what the voltage drop across R1 must be, right?
That's interesting. This is the first time I recall hearing the term "voltage drop", but it makes perfect sense and frankly fills a hole in my mind about why circuit laws behave the way they do.

So the voltage drop is
0.258 x 20 = 5.16

Hmm.

The voltage at the top of R2 and R3 should be the voltage of the source minus the voltage drop on R1. So that gives you the voltage across both R2 and R3. The currents for both R2 and R3 then follow from Ohm's Law for those components.
Hmm!

So, 10-5.16 = 4.84V

Now I use ohm's law.

IR2 = 4.84/30
IR3 = 4.84/50

Incredible!!!! Really, really, awesome :D

Thank you sooooooooooo much gneil!

And you're also more than incredible ILS!!!!
 
  • #19
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May i help you to solved the problem with another approach (only use V = I R rather than Kirchoof rule).

Like this: with R23 = 18.75 you can find voltage here V23 = I1 . R23 = 4.84 V

then back to your original circuit diagram.

I2 that flow through R2 is [tex]I_2 = \frac{V_{23}}{R_2}[/tex] = about 0.16 A

then [tex]I_3 = \frac{V_{23}}{R_3}[/tex] = 0.097 A or you find it by I3 = I1 - I2

if you have solved this problem using Kirchoof rule, please check if the result is same as above.
I think it should.

Note: in paralel, there is same voltage diff current.
 
  • #20
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That's interesting. This is the first time I recall hearing the term "voltage drop", but it makes perfect sense and frankly fills a hole in my mind about why circuit laws behave the way they do.

So the voltage drop is
0.258 x 20 = 5.16

So, 10-5.16 = 4.84V

Now I use ohm's law.

IR2 = 4.84/30
IR3 = 4.84/50

Incredible!!!! Really, really, awesome :D

Thank you sooooooooooo much gneil!

And you're also more than incredible ILS!!!!
I agree that this feels like a totally different way of looking at it, and it is good to confirm it this way.
But just for fun, could you compare the numbers you have here with the numbers you had when using the voltage law?
You should find that you made exactly the same calculations! :smile:
 
  • #21
Femme_physics
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May i help you to solved the problem with another approach (only use V = I R rather than Kirchoof rule).

Like this: with R23 = 18.75 you can find voltage here V23 = I1 . R23 = 4.84 V

then back to your original circuit diagram.

I2 that flow through R2 is [tex]I_2 = \frac{V_{23}}{R_2}[/tex] = about 0.16 A

then [tex]I_3 = \frac{V_{23}}{R_3}[/tex] = 0.097 A or you find it by I3 = I1 - I2

if you have solved this problem using Kirchoof rule, please check if the result is same as above.
I think it should.

Note: in paralel, there is same voltage diff current.
I see it now, and I believe I did it like you did in the post above you, correct?


I agree that this feels like a totally different way of looking at it, and it is good to confirm it this way.
But just for fun, could you compare the numbers you have here with the numbers you had when using the voltage law?
You should find that you made exactly the same calculations!

I did, and indeed!
I even did my own little made up exercise to confirm the notion of "voltage drop" - which wasn't intuitive to me!

http://img14.imageshack.us/img14/5453/confirming.jpg [Broken]

By the way, the following questions (this isn't over yet, I told you I'm easing you in! :D ) is to calculate v2 that falls on the resistor R2, but we already know that from earlier calculations!

10-5.16 = 4.84V

The final question is to calculate the value of P3 that's on the resistor R3

Easy peasy!

P = V2/R
P = 4.842/50
P = 401 [W]

:) Done!

*blows kisses*


PS That was totally orgasmic!
 
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  • #22
Femme_physics
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BTW, I find that, using ohm's law and the voltage drop concept is a much simpler and easy way of getting to the answer than start doing all these algebraic sum = 0 equations.
 
  • #23
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I even did my own little made up exercise to confirm the notion of "voltage drop" - which wasn't intuitive to me!
Right!
I hereby proclaim you to be officially initiated in the circle of electronics gurus! :cool:

By the way, the following questions (this isn't over yet, I told you I'm easing you in! :D ) is to calculate v2 that falls on the resistor R2, but we already know that from earlier calculations!

10-5.16 = 4.84V
Should I still say it?
Well, it still feels like fun...

Yep! :smile:

The final question is to calculate the value of P3 that's on the resistor R3

Easy peasy!

P = V2/R
P = 4.842/50
P = 401 [W]

:) Done!
Oops!
Wrong result.
Well, I can't undo your initiation...
But could you please correct it?

*blows kisses*
Please keep doing stuff like that!
I makes me feel wonderful! :blushing:

PS That was totally orgasmic!
Dooly noted! :wink:
 
  • #24
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BTW, I find that, using ohm's law and the voltage drop concept is a much simpler and easy way of getting to the answer than start doing all these algebraic sum = 0 equations.
Yeah, I know it seems that way with circuits like this.
[edit] And I would have suggested it, but then, you were already off on Kirchhoff's laws which is also good! [/edit]

Please take a look at your previous circuit.
Could you have solved that one with voltage drops?

Using Kirchhoff's laws, and indeed using sets of equations and solving them, may at first seem daunting, but in the end you can solve way more problems.
And the chance of doing something wrong because of a mistake in reasoning is less.

I do agree that you should always try to do stuff with intuitive concepts too.
We need that because in the end the solution should always be intuitively acceptable.
It's similar to always trusting your visual queues!

And if there is a discrepancy, there's a job for *Physics girl* to ferret out the discrepancy! :smile:
 
  • #25
Femme_physics
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I hereby proclaim you to be officially initiated in the circle of electronics gurus!
Woah, feels..premature! Shouldn't I know about diodes, first? We haven't even started this!

But thanks..am honored :D

Oops!
Wrong result.
Well, I can't undo your initiation...
But could you please correct it?
Ah yes, with the decimal :) I got it I got it!
0.401 [W]

Damn it! I should really stop doing stupid stuff. I was just so happy seeing the finish line that I slipped!

I do agree that you should always try to do stuff with intuitive concepts too.
We need that because in the end the solution should always be intuitively acceptable.
It's similar to always trusting your visual queues!
Your mentoring is awesome! It permeates every fiber of my being and every thought in my mind as I'm solving problems. You're truly an inspiration.

And if there is a discrepancy, there's a job for *Physics girl* to ferret out the discrepancy!
Lol. I'm a superheroine too now? W00t! :D

Actually I'll be posting in an earlier problem we solved in a moment, there's something there still unclear to me.
 
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