Find the current in a circuit

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1. Feb 2, 2017

diredragon

1. The problem statement, all variables and given/known data
Problem details: (Look at the problem diagram)
First condition states that the switch is open. The second condition states that the switch is closed and during the transition time the charge $q$ flows through the branch on which it is labeled. Find the current generator Ig.
Picture:

2. Relevant equations
3. The attempt at a solution

I will show the details of my work and if anything is unclear or cannot be seen please tell me and i will upload a better photo. I will also describe and write some of it here:
I first calculated the voltage across the capacitor and labeled it like in the picture and transformed the circuit into a Δ-circuit where ΔI labeles a change in the current when the switch is closed through the branch with the switch.
Picture of work:

I have a problem now. I do not know how to continue. I cant transform the R7 and R6 into one as i will lose information about the voltage. So what can i do? Should i try node analysis or do you have a different idea?

2. Feb 2, 2017

Staff: Mentor

I'd be tempted to use mesh analysis. The resistor R8 is in series with the $ΔI$ current source so it is redundant and can be removed (it doesn't matter what resistance is in series with a current source, the branch current is fixed). That places $ΔI$ in parallel with R1, making it ripe for a source conversion (Norton to Thevenin) so you end up with two loops.

3. Feb 3, 2017

diredragon

I removed the R8 and made a source conversion like you said:
Picture:

Why could we remove R8? Don't we remove the information about the voltage across it then? How oculd we just remove a part of a circuit?
I then used a mesh current analysis and i came up with this:
Picture:

I dont know the mesh currents or the ET so what can i do?

4. Feb 3, 2017

Staff: Mentor

Placing a resistor in series with a current source does not change the current in the branch. If you think in terms of node equations, the nodes that the branch is connected to cannot be affected since the current contributed by that branch to those nodes cannot change. So that's one reason.

A current source will produce any potential difference across itself required to maintain its specified current. A resistor in series with a current source just means that the source will change the potential across itself to compensate for the potential drop that occurs across the resistor. The net potential difference across the entire branch remains the same (else the node potentials would have to change, which they can't because they are dictated by the current).

The only exception to this is if the resistor in question is used as a sensor to control a controlled source elsewhere in the circuit. Then the particular potential drop across the resistor is important. Even so, if the current source and resistor in the branch are fixed values, that potential difference will be fixed, too, and the controlled source can be replaced by a fixed source.

Note that your $E_T = ΔI R_1$. You can thus solve for the mesh currents in terms of $ΔI$. That gives you access to the potential drops on any resistors in terms of $ΔI$.

5. Feb 3, 2017

diredragon

So whenever i get a circuit the has a current source in series with a resistor i can remove it? When i want to transform into thevenin's equivalent or in any case?

Oh, i think i get it but just to check if this is what you mean:
Starting from the mesh equations i got in the last picture:
2)$I_{k2}=I_{k1}\frac{R2}{R2+R4+R5}$
So putting this into the first equation i get:
1)$I_{k1}=\frac{ΔIR1}{R1+R2+R6+R7-\frac{(R2)^2}{R2+R4+R5}}$
2)$I_{k2}=\frac{ΔIR1}{R1+R2+R6+R7-\frac{(R2)^2}{R2+R4+R5}}*\frac{R2}{R2+R4+R5}$
and since i have $ΔUc$ i can get $ΔI$ by
$ΔUc=-I_{k2}R5-I_{k1}R6$

6. Feb 3, 2017

Staff: Mentor

Yes. And you can do the same with resistors in parallel with a voltage source. Such a resistor will never change the voltage.
I haven't checked your derivations in detail, but yes, that's the idea. Plug in your component values and you should be left with an expression of the form:

$ΔUc = k ΔI$

7. Feb 3, 2017

diredragon

So i have $ΔI$ then. The change in the current with the switch. Now i need to get back to the original circuit and observe the two conditions, the switch closed and switch open. What do you suggest i do from there?
Since i need $I_g$ i need to find the currents that flow through the branhces that connect to $I_g$? Should i use mesh again or try superposition?

8. Feb 3, 2017

Staff: Mentor

I can't immediately see a preferred method to work with the original circuit, given what you're trying to determine. My first instinct would be to use nodal analysis since you're mostly interested in currents. I'll have to give it some thought.

9. Feb 3, 2017

Staff: Mentor

Okay, back from daydreaming about nodes and meshes (it's not as naughty as it sounds ).

You should have a numeric value for $ΔI$ at this point. Since $ΔI$ only flows when the switch is closed you only need to look at one version of the circuit (switch closed). And since it flows through R8 that sets the potential difference between the nodes that R8 connects to.

You also no longer care about preserving R5 and R6 for access, since they've done their job in determining $ΔI$. So you can now combine R4 and R5 ito a single resistance, and combine R6 and R7 and E6 into a current source and single resistor in parallel with Ig.

Draw your simplified circuit and see what you can make of it, labeling known potentials and currents. Can you choose a suitable reference node so that only one node voltage remains unknown?

10. Feb 4, 2017

diredragon

I do get what you mean but am unsure about the to current source transformation. I hope i did it right. Here is the photo of the initial transformations and labels and the labels and the ,,tryied to convert'' photo:
Initial transformations:

Tried to but not sure about the current part:

The rest pf the circuit up is the same.

11. Feb 4, 2017

Staff: Mentor

The new current source should end up in parallel with Ig.

R6 and R7 are both in series with E6. So combine them into a single resistance before transforming to a current source.

12. Feb 6, 2017

diredragon

Sorry for the delay, i solved it :)

13. Feb 6, 2017