Find the current in the circuit

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1. Dec 31, 2016

diredragon

1. The problem statement, all variables and given/known data

($R_5=8\Omega$)
a) Calculate the current $I_{g2}$ so that the power of the generator $I_{g1}$ equals $P_{ig1}=6.25W$
b) What is $I$ then?
2. Relevant equations
3. The attempt at a solution

First step: I change the last two branches with an equivalent thevenin's system.
$R_{t4}=\frac{R_4R_3}{R_2 + R_3}=12\Omega$
$E_{t4}=\frac{-E_4R_3}{R_3+R_4}=-15V$
(I have set the thevenin generator to have + sign pointing up)
Second step: I do the same with the remaining two right branches ( The thevenin system and the $E_6$ )
$R_{t6}=(R_{t4}+R_5)||R_6=10\Omega$
$E_{t6}=\frac{E_{t4}R_6+E_6(R_5+R_{t4})}{R_{t4}+R_5+R_6}=-2.5V$
Third step: Change the left part into the thevenin system
$R_{t1}=R_1=5\Omega$
$E_{t1}=I_{g1}R_1=5V$
DO you think that this is the right way to start and can you hint at what i should do in this problem? I'm thinking since $P_{ig1}=6.25W=U_{ig1}I$ that that power must equal the power of the generator i created to replace this? Is that right? So i can get the current that goes through the generator $E_{t1}$ Its $I=\frac{6.25}{5}=1.25mA$? Is this good?

2. Dec 31, 2016

Staff: Mentor

I like your change of the right hand network to a Thevenin equivalent. Rth = 10 Ω and Vth = -2.5 V looks good.

I don't like so much your idea of losing the current source $I_{g1}$ to an equivalent circuit. Usually you want to keep intact any components that you'll need to know the current through or potential across.

You know the current that the source produces is 1 A, so all you need to do is fix the potential across it to establish the desired power. That's easily done from the given P = 6.25 W. That establishes node voltage V1 in the reduced circuit:

You should be able determine the currents and potential drops almost by inspection from that point.

3. Dec 31, 2016

The Electrician

You aren't keeping track of your units. 6.25 watts/5 ohms doesn't equal current, and even if it were current, where did you get milliamps, rather than amps?

Edit: I took the 5 to be resistance, so you do have power/voltage, but I still don't see how you get milliamps.

Last edited: Dec 31, 2016
4. Jan 1, 2017

diredragon

Yeah, i wrote in wrong, its suppose to be just amps.
Did you mean for the second generator in the picture to write $I_{g2}$ instead of $I$?
Voltage is then easy found by $U=P/I$ which gives $6.25V$. Then to find the voltage across the $I_{g2}$ i calculate the drop which is given by expression
$6.25V-I_215\Omega + 15V=U_{21}$
From the way the current splits i see that $I_1+I_3=1A$ and $1A=I_1+I_2$ Did i do something wrong here couse i get that $I_3=I_2$ COuld this be done more easily or do i have to apply loop current or node method? If so how, its unclear to me what is happening and how to calculate the current that splits.

5. Jan 1, 2017

Staff: Mentor

I just used $I$ to represent the unknown current. Yes it's identical to $I_{g2}$, but has fewer characters to lug around through equations .
$I_1+I_3=1A$ is not correct. How did you arrive at that equation?

If you know V1 then you should be able to determine $i_1$ directly thanks to Ohm's law. KCL at node V1 then gives you $i_2$. Continue from there.

6. Jan 1, 2017

diredragon

Oh, i did not see that xD! $I_1=6.25/5=1.25$
$Ig_1=I_1+I_2$
$I_2=-0.25A$
Then i have $6.25+15V-I_215\Omega=25V$
Using the second rule i get that $-10I_3+2.5+25=0$, $I_3=2.75A$ and $I_2+Ig_2=I_3$ so $Ig_2=3A$ right? Now for the part b) where they ask for te current $I$ from the original picture. The voltage across and the current $I_g2$ are still the same right?
So $E_6-I_{6}R_6=25$ and $I_6=-0.75V$ so $I_5=2A$ and that must be the same as current $I$ because it must come back the other way exactly as it went in. This is good according to the book results, is there anything you might suggest i do differently?
I have two remaining questions, first what did u use to draw the circuit xD and second why you think it might have been more difficult to not change the right part to thevenin? Because generally we shouldnt change the part that was given in order not to lose information?

7. Jan 1, 2017

Staff: Mentor

Yes, all right so far.
Everything looks good. Your methods are fine.
I use Visio to make my drawings. It's a really old version that I've used for many years, and I made my own template for electronics.
It's generally easier to simplify what you can ahead of time as it means fewer simultaneous equations to deal with. If you're looking for a particular value that depends upon a given component or configuration, don't "lose" that component in the simplification. Everything else is fair game. If there are several different things you need to find and they're in different parts of the circuit, it's okay to do separate analyses with different simplifications that preserve just the portion of interest in each case.

In this circuit you were given the power associated with one component and wanted to find the current of another. So those two components should be left alone to preserve their operating parameters. But the rest of the network could be simplified so you wouldn't have to deal with all the currents of the separate branches at the same time.