The RLC Circuit:

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Re: The RLC Circuit:

Originally posted by crossman:
phase angle is 45 degrees which can be found by using trig with the triangle
Click to expand...
I concur. But I'm getting -45 degress for the phase angle. Using RPN on my trusty HP 48G Calculator (Polar Form):

Total Impedance (Zt): [20(Angle 0) x 20(Angle 90)]/[20(Angle 0) + 20(Angle 90) = 14.1421(Angle 45)

Total Current (It): 120/14.1421(Angle 45) = 8.485(Angle -45)

[ February 07, 2005, 01:43 PM: Message edited by: rr ]
 
Re: The RLC Circuit:

I think you could go either way on the angle... call it 45 lagging or -45 in my book, either one means the same thing to me.

Does that sound right?
 
Re: The RLC Circuit:

Originally posted by crossman:
I think you could go either way on the angle... call it 45 lagging or -45 in my book, either one means the same thing to me.

Does that sound right?
Click to expand...
Whoops....didn't see the "lagging" comment on your original post. As Ed McMahon might say: "You are correct sir!" :D
 
Re: The RLC Circuit:

Crossman, - anything is not equal to + anything except zero which is not anything; it is nothing.

If Z has an angle of +45, the current carries an angle of -45; that is, the current lags.

Now what happens if we have only the 20 Ohm R, and the capacitor?

I did a tiny bit of work on the first hp scientific calculator, so anything that comes from hp has to be right.
 
Re: The RLC Circuit:

Originally posted by rattus:
Crossman, - anything is not equal to + anything except zero which is not anything; it is nothing.
Click to expand...
:confused: DO WHAT???? :confused:

Originally posted by rattus:
If Z has an angle of +45, the current carries an angle of -45; that is, the current lags.
Click to expand...
DO WHAT again?? I am seeing that impedance, reactance, and resistance are all in phase with the source voltage in a parallel circuit.

What am I missing?
 
Re: The RLC Circuit:

Crossman, the rules are the same. Just compute the real and reactive currents separately and add them vectorially. Then,

I = V/R + V/jwL

= V(1/R -j/wL)

= 120(0.05 -jx0.05)

= 6 - j6

= 8.48 @ -45

[ February 07, 2005, 03:40 PM: Message edited by: rattus ]
 
Re: The RLC Circuit:

okay, but those are currents

what about the actual vectors for the ohms of each component? Where do they sit on the coordinate system? What directions?
 
Re: The RLC Circuit:

Crossman,

Resistance carries a phase angle of zero, but we write it a simple, "R".

Inductive reactance is written as

jwL equvialent to wL @ 90.

Capacitive reactance is written as

1/jwC = -j/wC equivalent to 1/wC @ -90.

In this example you must take the parallel combination of R and jwL to obtain the impedance.

It is easier just to write the "admittance" which is the inverse of impedance as,

Y = 1/R - j/wL Siemens, used to be Mhos

= .05 - jx0.05 Siemens

= .0707 @ -45 Siemens

Z = 1/Y = 14.14 @ -45 Ohms

I = V/Z = VxY
 
Re: The RLC Circuit:

Is the "j" you are using the "unit vector" of the y-axis as used for vectors in Physics?

I guess my desire is to see an actual drawing of the vectors for inductive reactance, capacitive reactance, resistance, and impedance for a parallel LCR circuit
 
Re: The RLC Circuit:

Crossman,

A recent post by Ed MacLaren explained the j operator, but this forum is not a good way to convey such information. You need to find a good text and study it all night.

You can find such stuff on the Internet too.
 
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