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KingLee Aug 25th 2017 09:41 AM

How to calculate the current across two points ? PLEASE HELP
 
How to calculate the current across two points in this :
https://lounge.kth.se/RequestHandler...17508f458bd73b

So what i did is i used Ohms law to find the total current across the whole circuit which i computed to be 3 A and i am given the EMF of the battery and asked to neglect the internal resistance. So i am stuck at calculating the current across point A-B. Please help

ChipB Aug 25th 2017 10:12 AM

Next step is to calculate the currents through R1 and R2. You know that the voltage drop across these two resistors is the same (do you see what that is?), so you can solve for I1 and I2 using:

$\displaystyle R_1 I_1 = R_2 I_2$

and:

$\displaystyle I_1 + I_2 = I_T$

Can you take it from here?

KingLee Aug 25th 2017 01:28 PM

Quote:

Originally Posted by ChipB (Post 36754)
Next step is to calculate the currents through R1 and R2. You know that the voltage drop across these two resistors is the same (do you see what that is?), so you can solve for I1 and I2 using:

$\displaystyle R_1 I_1 = R_2 I_2$

and:

$\displaystyle I_1 + I_2 = I_T$

Can you take it from here?

Thanks for your reply really appreciate it.
I am not after \I_t\ what i am after is \I_{ab}\ here is my work:

since \R_1\ is parallel to \R_2\ and \R_3\\R_4\, hence:
\R_t_1\=\frac{\frac{1}{R_1}\cdot \frac{1}{R_2} }{\frac{1}{R_1}+ \frac{1}{R_2} }\ = 2\Omega[math]

Applying the same for R_3 and R_4 we get that the resultant R is 2 as well hence:
I=\frac{V}{R}= \\frac{12}{2+2)}
which gives that I_T =3 A

but what i am after is the current floating between AB

kiwiheretic Aug 25th 2017 03:24 PM

1 Attachment(s)
I wouldn't try to solve this using resistor series / parallel formulas. That's making it difficult.

Basically what ChipB was saying except he was probably drawing the current arrows a little differently (see my diagram).

In the case shown where I redrew your diagram to make it a little clearer.

$\displaystyle R_1 I_1 = R_2 (I-1_1)$

$\displaystyle R_3 I_2 = R_4 (I - I_2) $

$\displaystyle R_2 (I-I_1) + R_4 (I - I_2) + E_{battery} = 0$
(Don't think you need the 3rd equation because that loop doesn't go through AB but included for completeness)



Basically its just relying on the voltages added up around a circle should always equal zero. This gives you a set of linear equations which you can solve for I1 and I2 in terms of I.

KingLee Aug 25th 2017 04:21 PM

Quote:

Originally Posted by kiwiheretic (Post 36761)
I wouldn't try to solve this using resistor series / parallel formulas. That's making it difficult.

Basically what ChipB was saying except he was probably drawing the current arrows a little differently (see my diagram).

In the case shown where I redrew your diagram to make it a little clearer.

$\displaystyle R_1 I_1 = R_2 (I-1_1)$

$\displaystyle R_3 I_2 = R_4 (I - I_2) $


Basically its just relying on the voltages added up around a circle should always equal zero. This gives you a set of linear equations which you can solve for I1 and I2 in terms of I.

Thanks a lot for your answer, but how did you derive this formula? what law are you using?

kiwiheretic Aug 25th 2017 04:30 PM

Quote:

Originally Posted by KingLee (Post 36763)
Thanks a lot for your answer, but how did you derive this formula? what law are you using?

I would say conservation of energy. The voltages (potential energy per unit charge) around a loop must sum to zero. Ie V1 + V2 +V3 + etc = 0. In my examples I knew that R1 I1 + R2 (I - I1) = 0 but I rearrange by putting one term on the other side as R1 I1 = R2 (I1 - I). That's why I draw currents as circular arrows to help me keep the signs of the current straight in my head. If you calculating the voltage in the direction of the current I treat it as positive but if you're going against the flow I treat the voltage drop as negative and that's how they all sum to zero (conservation of energy) and (work done by a conservative force around a loop must be zero) are the laws I think about.

Pmb Aug 25th 2017 04:47 PM

Quote:

Originally Posted by kiwiheretic (Post 36764)
I would say conservation of energy.

which is formally known as Kirchhoff's voltage law

See: https://en.wikipedia.org/wiki/Kirchh..._law_.28KVL.29

KingLee Aug 25th 2017 05:49 PM

Quote:

Originally Posted by kiwiheretic (Post 36764)
I would say conservation of energy. The voltages (potential energy per unit charge) around a loop must sum to zero. Ie V1 + V2 +V3 + etc = 0. In my examples I knew that R1 I1 + R2 (I - I1) = 0 but I rearrange by putting one term on the other side as R1 I1 = R2 (I1 - I). That's why I draw currents as circular arrows to help me keep the signs of the current straight in my head. If you calculating the voltage in the direction of the current I treat it as positive but if you're going against the flow I treat the voltage drop as negative and that's how they all sum to zero (conservation of energy) and (work done by a conservative force around a loop must be zero) are the laws I think about.

Would please show how you work it all out, with some details about everything you use, please, i have been sitting with this question for a week now, i read the book over 10 times, it just wont enter my head

kiwiheretic Aug 25th 2017 06:24 PM

Quote:

Originally Posted by KingLee (Post 36769)
Would please show how you work it all out, with some details about everything you use, please, i have been sitting with this question for a week now, i read the book over 10 times, it just wont enter my head

It's basically just linear algebra. I'll write all the equations out for you that's a little easier to look at:

$\displaystyle R_1 I_1 - R_2 I + R_2 1_1 = -R_2 I + (R_1 + R_2 )I_1= 0$

$\displaystyle R_3 I_2 - R_4 I + R_4 I_2 = -R_4 I + (R_3 +R_4) I_2 = 0 $

$\displaystyle R_2 (I-I_1) + R_4 (I - I_2) + V_{battery} = (R_2 + R_4) I - R_2 I_1 - R_4 I_2 = -V_{battery}$

In terms of linear algebra and matrices this is just:

$\displaystyle \begin{bmatrix}
-R_2 & (R_1+R_2) & 0 \\
-R_4 & 0 & (R_3 + R_4) \\
(R_2 + R_4) & -R_2 & -R_4
\end{bmatrix}
\begin{bmatrix}
I \\
I_1 \\
I_2
\end{bmatrix} =
\begin{bmatrix}
0 \\
0 \\
-V_{battery}
\end{bmatrix}
$


Once you solve this matrix equation you will have I, I1 and I2 in terms of R1, R2, R3 and the battery voltage. Then current through AB is just I1 - I2 (or I2 - I1 depending on sign).

Someone should probably check my algebra in case I messed up signs somewhere and you should probably get used to doing that yourself anyway.

In this case it's probably just as easy to solve via substitution without all the matrix hullabaloo but you get the idea.


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