Engineering Finding the current in the circuit

[diredragon]

Homework Statement

In the first stationary condition of the circuit in the picture below the switch $p$ is closed and in the second it is open. The change of charge on the capacitors from 1st to 2nd condition is $ΔQ=30 mC$ Calculate the current $I_g$
Picture:

Homework Equations

3. The Attempt at a Solution [/B]
I will upload the picture of my work here an then explain what i did.

I labeled the steps so you can refer to them in your comments. I will also rewrite some of it here:
First i wanted to find the current through the branch with the switch when it is closed and i did that in the following way. I changed the circuit diagram into the Δ circuit as you see in the picture. Then using simple steps i found the current through the branch. But I'm clueless about what do i do now. Why did i need that current in the first place?
Could you give me a clue that is of the form which theorem to use and why. Thanks

 

[gneill]

Your ΔQ should have units μC rather than mC (it's a 1000x difference).

This is not a method of analysis that I've used before, but I can see how it works. It's essentially based on the use of the Superposition Theorem.

You have correctly determined that in order to produce the $ΔU$ of 15 V across the capacitor that the change in the middle branch current has to be 15 mA, and that the change is going to flow upwards through the branch (hence -15 mA for the $ΔI$ current source).

So it remains to determine what the branch current was to begin with (while the switch was closed). That you can determine in any number of ways, but you might consider nodal analysis to find the voltage of the top node and then determine the current through R3 using Ohm's Law.

Once you have that current (call it $I_o$) then you can determine the what the current $I_g$ has to be make the $ΔI$ your -15 mA.

 

[diredragon]

Your ΔQ should have units μC rather than mC (it's a 1000x difference).

This is not a method of analysis that I've used before, but I can see how it works. It's essentially based on the use of the Superposition Theorem.

You have correctly determined that in order to produce the $ΔU$ of 15 V across the capacitor that the change in the middle branch current has to be 15 mA, and that the change is going to flow upwards through the branch (hence -15 mA for the $ΔI$ current source).

So it remains to determine what the branch current was to begin with (while the switch was closed). That you can determine in any number of ways, but you might consider nodal analysis to find the voltage of the top node and then determine the current through R3 using Ohm's Law.

Once you have that current (call it $I_o$) then you can determine the what the current $I_g$ has to be make the $ΔI$ your -15 mA.

$ΔI$ that i have got is the current the flows through the branch with the switch when it is closed right?
I will now try using the nodal analysis taking the bottom nod as grounded one.
Basically then i have:
$V1(1/R1+1/R2+1/R3)=E/R1-I_g$
I have trouble with this. First how do i deal with two unknowns and what do i do know with the information that i have $ΔI$ Do i place it as a current generator in the branch with the switch when I am looking at the closed scenario?

 

[gneill]

$ΔI$ that i have got is the current the flows through the branch with the switch when it is closed right?

No, you've calculated the change in current (-15 mA) that occurs when the switch opens.

I will now try using the nodal analysis taking the bottom nod as grounded one.
Basically then i have:
$V1(1/R1+1/R2+1/R3)=E/R1-I_g$
I have trouble with this. First how do i deal with two unknowns and what do i do know with the information that i have $ΔI$ Do i place it as a current generator in the branch with the switch when I am looking at the closed scenario?

Apply nodal analysis when the switch is closed so that effectively Ig is suppressed and only the voltage source E will be active. That will yield your "initial" current in the branch. It should be some small current flowing down through the branch.

Then you need to find the value for Ig that will produce your -15 mA change.

 

[diredragon]

No, you've calculated the change in current (-15 mA) that occurs when the switch opens.

Apply nodal analysis when the switch is closed so that effectively Ig is suppressed and only the voltage source E will be active. That will yield your "initial" current in the branch. It should be some small current flowing down through the branch.

Then you need to find the value for Ig that will produce your -15 mA change.

What changes then from here?
$V1(1/R1+1/R2+1/R3)=E/R1-I_g$ I still have two nodes right?
I'm not sure how it changes things. I can try the superposition principle to find the current that goes through R3.
I3=I(E)+I(Ig)
When only E works the current i found to be 9/2 and when only Ig works i found the current to be 0 because of the short-circuit. (could you check this?)

 

[gneill]

When the switch is closed there is no Ig. It vanishes because it's shorted out and the short takes all its current.

You should use the more efficient way to write a node equation, as a sum of currents set equal to zero:

$$\frac{V1 - E}{R1} + \frac{V1}{R3} + \frac{V1}{R2} = 0$$

Solve for V1 and then the current in through R3.

 

[diredragon]

When the switch is closed there is no Ig. It vanishes because it's shorted out and the short takes all its current.

You should use the more efficient way to write a node equation, as a sum of currents set equal to zero:

$$\frac{V1 - E}{R1} + \frac{V1}{R3} + \frac{V1}{R2} = 0$$

Solve for V1 and then the current in through R3.

So basically the node equation is:
$V_1*(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3})=\frac{E}{R_1}$ and once the V1 is found i can solve for initial $I_0$ which is just $\frac{V_0}{R_3}$.
So i get:
$V_1=3V$ and from there
$I_0=3 A$ which is the intial current and it flows downwards. This is the flow through the branch with the switch. Now the switch opens. How do i then calculate I_g? First the switch is closed and the flow through the branch is 3A. Then it closes and we have that the change is -15. Which means that the amount of 15A flows in the other direction. 3-15=-12 so the $I_g=-12$ right?

 

[gneill]

Right. Except that your currents should be in mA, not A.

 

[diredragon]

Right. Except that you currents should be in mA, not A.

Got it, thanks gneill, you're the best :D

 

[gneill]

Cheers!