I need some assistance with a problem in which I need to solve a circuit using the node voltage method. I understand the basics of this approach, but I only know how to do it for resistors, not to mention some of the currents and sources need to be expressed in their complex form.
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normally you'd short the voltage sources separately and superimpose. so for each voltage source separately you'd combine the complex impedances and then you add it all up. here the sources are complex too, adding to the fun.
thing is, if you only know how to do this for resistors and are new to phasors/complex stuff, then this is a real baptism of fire.
That's one way. But i believe the circuit you have is a digital to analog converter. So when you have a ladder network like this, the easiest for me to do this is, source transformation. So first, you'd have all the resistors, capacitors, and inductors in impedances. Then solve it exactly the way you'd solve it with just resistors. Just make sure you keep track of j. And you'd only need to figure out how to write two different sources in phasor form, and then after that, treat it like a DC source. SOrry if that sounds very confusing. Let me know if it does.
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I highly recommend that you learn how to handle phasors, and complex number converesions on your calculator. Reduces calculation error by about 75%. haha
The math can be labor intensive, suggest Wolfram or hopefully your school has mathlab or equivalent. I am looking at this problem and come up with a single node, but quite a combination of complex terms.
|dw:1434113499159:dw| Wanted this drawing so I could see without leaving this window..
We know that by Kirchoffs Law, that I1 + I2 + I3 + I4 + I5 = 0. I choose to express it this way: I2 + I5 = I1 + I3 + I4 They are both stating the same thing. I intend to sum up I2 and I5 then sum up I1 +I3 + I4. I intend to use Vn as the node voltage. There will be 5 unknowns (I1 thru I5) and Vn. I may go down in flames attempting this, but I intend to try, but not today lol.
I2 = (Vn - 10J) (Volts)/J1 (Ohms)=(-VnJ-10)A or -(10 + JVn) A
I5 = (Vn - 20J)(Volts)/(4 + J3) (Ohms)
I1 =Vn (volts)/J3 (Ohms)=-jv/3 A
I3 = Vn (volts)/3 (Ohms)=(1/3)Vn A
I4 = (Vn - J20)(Volts)/(1-J2) (Ohms)
Results in Amperes
In retrospect, when these currents are summed up to zero, I will actually have one unknown, and that is Vn. If it wasn't for the complex values, this would be a simple problem. I may be looking at this the wrong way, we will see.
For I5 I have come up (Using Wolfram) (.16 - .12j)(Vn-20j)
For I4 I have come up with (.2 + .4j)(Vn-20j)
I don't know for sure how to handle such values.
@Michele_Laino I have become entangled with handling complex values. Especially where using a variable (Vn). I don't know how to convert E=10j[V] to a phasor, I think it would be
7.07@90 degrees ?? also how do i convert an expression like Vn - j20 to a phasor?
this problem is above my paygrade lol.