Electric Circuit problem

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Jkarir

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Yesterday i had exam at electronics and i failed at it, the reason was i didn't know how to solve correctly electric circuits. Our exam contains 2 theories and 2 circuits electric problem, one is a continuous circuit and the other a alternative circuit.

We usually solve circuits with Kirchoff Laws/The Branch-Current Method.
But at the first step i did something that i didn't quite understand.
Below is a photo how the circuit was looking, someone told that all the elements which are on the conductor line must be cut from the circuit and also the circuit must be disconnected from the source, after that i should apply the The Branch-Current Method/Kirchoff.

I started all wrong resolving the circuit for the fact i didn't know to disconnect and cut what?? i still don't know how the circuit will look after i do this in order to solve it:shock:
 

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Hello Dear Jkarir
What do you want to find from your circuit exactly ?
at the right of your circuit i can see some things like hair ! are they inductors ?
So where is the value of the elements ?
Best Wishes
Goldsmith
 
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    Jkarir

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Hi, sorry for the 'funny' circuit, i scanned here how the circuit looked, there aren't the exact values that i had on my test paper but they were like this, it was asking for the values of currents and voltages.
The first thing that i started to do is to solve it with the Kirchhoff Laws. My teacher said that it's wrong because the capacitor has a switch function and also that i forgot to disconnect the circuit from the power source(another person told me that all the elements that are on the line of the capacitor must be cut from the circuit) i quite didn't understand what he mean with this, i am able to solve simple circuits with Kirchhof Laws/Branch current method, but this circuit for me is a complex one, thanks for the reply.

**broken link removed**

If a skipped a value or anything sorry.
 
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Are you Ok my friend !? your attachment isn't available !:grin:
I don't know , why it deleted , but can you draw it with your hands and attach here , please?
 
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    Jkarir

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where is the value of capacitor and where the independent current source is depends of ?
Any where , it is very simple example : those inductors are in series together . so you can simplify , it to an inductor with the value of 35 henry . and at DC , the inductor after transient time will be short circuit , and the capacitor will be open circuit . so if you don't want calculate transient analysis , you can simplify the circuit to this :

So i think you can find the currents of this simple circuit .
Good luck
Goldsmith
 
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    Jkarir

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Thanks for your example, but if i had another resistor R5 between those two inductors, in which place it should be after short circuit? It will be moved together with the other resistors in your draw?
 

is your mean in series with the inductors , or connected between them to the ground ?
 

Ok you can consider that those inductors are like a short circuit , but you'll have that resistor , and you can solve this circuit again . simply!
the voltage across this resistor is known and it is equal to the voltage of dependent voltage source .
By the way , you can solve it with node and potential way ! it will be very simple . but you should tell me that , how this source is dependent , and it depends on where .
Respectfully
Goldsmith
 

Our teacher didn't provide more information about the circuit, only that the capacitor has a role of a switch, but your solution is correct i think, it was about disconnecting. Ok, if i understand correctly, does the circuit below has the right solution? First circuit is the circuit provided and below i made the solution, is it correct?
 

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Again Hi
Why you added that 50 volt DC voltage at right of the circuit ?

---------- Post added at 11:46 ---------- Previous post was at 11:42 ----------

suppose that L is short circuit . ok ? and the capacitor is opened after transient time . so 10 k ohms will be in parallel with the 50 volt .( my mean is that 10 kohms , that is in series with the inductor .
By the way , did you change the example ? last example solved ?
Regards
Goldsmith
 

Hello, yes, i changed the exemple to be sure i understand what you mean with your previous example, my example is below attached and i tried to make it open circuit in previous photo

I want to now how the circurt will look after DC and short circuit, thanks

 
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So , see below , please: ( i supposed that you have 2 voltage source ) . the middle voltage source is alike with capacitor ! when you want to draw , elements at your circuits you should pay more attention to them and with enough care .
( i created some change in your original example )

Wish you the best
Goldsmith
 

Well, this is the main problem I have, i'm not sure how the circuit will look after the shortages of inductors;

So basically I short all the inductors from the electric circut and take out the capacitor making a new open circuit without inductors or capacitors, and i apply Kirchhoff Laws/Branch current method to the new circuit(without inductors/capacitors)to solve the circuit? did i understand correctly?
 
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Yes Dear Jkarir. you correctly understood it .
But don't forget that this law , is just for DC and after transient time .
Good luck
Goldsmith
 

Ok, when you say DC it's Direct Current, but at alternating current circuits after i tranfer all the circuit elements in Complex numbers can i apply the Kirchhoff Laws/Loop Current Method to the circuit?, for e.g. i have this:

From what i know first thing that we do is putting the elements in complex using this formulas: ZResistor=R, ZInductor=jωL, ZCapacitor= 1/jωC, i do this for every element, where ω=l*Pi*freq.
In my circuit below if R1=2kOhm, in complex it's ZR=2kOhm, if the value of the inductor is L1=23H, in complex will be Zi=23jω? is it the same thing for the capacitors, if there one in the circuit does it get cut also from it? Thanks for you advice.
 
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where ω=l*Pi*freq.

I think your mean was 2*pi*F?
And about :
"Ohm, if the value of the inductor is L1=23H, in complex will be Zi=23jω"
No it is complex . where is the resistance effect ? Z=a+jb a is real and b is the coefficient of imaginary part .
consider about a series resistance with an inductor . R=10 ohms and L= 50H . ok ? So what will be the impedance ? ( suppose that omega = 1500)
So lets start to solve it : Z has a real part ( 10) with imaginary part (j1500*50) ok according to the standard format
(a+jb) it will be 10+j75000
and about the capacitor : XC=1/jwc if you do it : (1/jwc)*j/j so because j is -1 under radical , thus : XC=-j/wc .
I hope you understand it clearly.
Good luck
Goldsmith
 

Yes the math is right, according to my electrotechnics manual this are the steps of resolving with Kirchhoff Laws:
1)all the elements of the circuit are replaced with theire imaginary's in complex
2)A way of the current is selected on every brench
3)no. of nodes, brench and loops
4)p=N-1 Kirchhoff I and l=b-n+1 Kirchhoff II ecuations, which will result a system of ecuations where theire solutions will be the currents to the every brench.

This is the methodology to resolve alternating circuits after i switch them to complex, but does the same thing happen like in DC? The inductors are shorted and the capacitors makes the circuit open, ready to apply the Kirwchhoff laws?
 

Again Hi
Af transient time , the inductor is shorted and the capacitor is opened , and you can use the kirchhoff law simply .
Regards
Goldsmith
 

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