Finding the currents and voltage using mesh analysis

[husny540]

I'm sorry for the delay because my internet was out of service for a few hours By the way i too drew the same circuit as of yours but thanks for uploading it for me.These are my equations, can you please say whether are these correct or not? I'll first of all use only the symbols as you said.for I_1 loop : V3=I_1*R12 + (I_1+I_3)*R8 For I_3 loop : V2-V4=(I_2+I_3)*R11+I_3*R9+(I_1+I_3)*R8, For I_2 loop : V1-V4=I_2*R13+(I_2+I_3)*R11. Am i correct?

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Are my equans wrong or am i supposed to submit you the answers with the current values as numbers itself?

 

[KerimF]

Since you wrote the equations we can verify them together as follows.

If you like writing each equation as sum of V's on the left side and sum of I*R's on the right side, we can follow the following rules:

(1) The voltage sign of a source is positive if the loop current flows from the minus terminal to the positive terminal of its source. It is negative otherwise.

(2) If all loop currents are made of the same direction (clockwise or counterclockwise) and a resistor R is common with an adjacent loop , the voltage on R would be [ + ( I - I_n ) * R ], where I is the current of the equation's loop. Otherwise it is (+ I * R).

By following the above rules, we get:

For I_1 loop : V3 = I_1 * R12 + ( I_1 - I_3) * R8

For I_2 loop : V1 + V4 = I_2 * R13 + (I_2 - I_3) * R11

For I_3 loop : - V2 - V4 = (I_3 - I_2) * R11 + I_3 * R9 + ( I_3 - I_1 ) * R8

As you noticed, you did a few errors related to sign.
Now we can re-write these 3 equations after substituting the given values.
And we get 3 equations with 3 unknowns. I think your learnt how to solve them.

Note:
My internet, also electricity and water, may anytime be out of service because my city lives right now a real war; surrounded by about 150,000 foreign mercenaries, could you guess its name

 

[husny540]

I'll solve the equations and submit you the answers to check whether are those correct or not. yes i have done a mistake in signs, thanks for correcting me. You see i have posted a question to this site on 'Finding the impedance and the quality factor in RLC circuit'. can you help me on that too, i have done certain calculations on that and i have included that in the post. I hope you can help me because you've helped on the thevenin circuit, and now you are heping me on this question, hope you can help me on the impedance question as well because i'm having some related questions to that.Note:You are from syria and the mercenaries are from the west,am i right?

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I found the values for I1,I2 and I3. I_1 = 0.225mA, I_2 = 9.333mA, I_3 = -4.527mA are these values correct. i calculated these by doing a series of calculations to attain these values.have i given the correct current values? If so is the voltage across R6 = 9.333kV. I'm doubtful of this because is this value possible?

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Also i need your help in answering this question. At the beginning the capacitors were replaced with their equivalent sources.how can we justify this replacement using the knowledge about capacitors and AC signals(including the mathematical calculations)?

 

[KerimF]

the voltage across R6 = 9.333kV

Please remember that if current is in (mA) and resistance in (Kohm), voltage is in (V).

For I_1, I_2 and I_3, I got them using the simulator (LTspice) then I verified the equations using excel. The result is on the attached "DCproblem_01_step4.png".


About the capacitor reactance, we know it is 1/jwC in case of sinusoidal current/voltage, where w = 2 * pi * F.
When the current and voltage are DC, it is equivalent to F = 0 Hz or w = 0. In this case, the value of 1/wC = 1/0 = infinity.
Is this what you were looking for?

 

[husny540]

Seems to be that my answers are incorrect.i did the calculations using pen and paper, whatever it is i should have got them correct. Thanks for helping me.Now about the capacitance, your justification seems to be correct and the question says to justify the replacement of capacitors in respect to the AC signals and the knowledge of capacitance. so is this explanation enough or is there more to it?I have to simulate this using ORCAD 9.2 and i installed it and the final step is to go to ***** and run the folder PDXOrcad right?but i cant find it under *****, if so how am i to proceed because the software is not opening without this step.can you direct me because this is related to this question?

 

[KerimF]

Sorry, I don't have ORCAD and I couldn't get what you wrote

i cant find it under *****,

About the capacitor, when its electrodes are fully charged (DC steady state) no more current can flow thru it (as if it is an open circuit).
In case of AC, its electrodes are charged and discharged continuously so there is always a current (positive and negative, alternatively) flowing in it. And if the current is sinusoidal:
Vc / Ic (peak value or RMS) = 1/wC
Note: The 1/j in the expression 1/jwC (which is equivalent to -j/wC) shows that Vc(t) lags Ic(t) by T/4 or 1/4F.