[SOLVED] Ac Sweep for a single frequency

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Hello
here is my program and settings
**broken link removed**
**broken link removed**
As you see the simulation settings I want to see voltage output of each nodes @60Hz.but here is my output file information:


but I got voltage of each node zero.
what is my mistake?
 

hi,
An AC sweep requires a Start/Finish frequency which are not the same value, also more than one point.

For a fixed frequency use a Time sweep, with V1 set to 60Hz.

E
 

Hi,

... maybe the voltage readings are DC readings.
Try to use RMS redaings.
Or use a scope.


Klaus
 

Thanks for you replay
could you tell me how can see Ac voltage such as Dc voltage on schematic simulation?or how can see the Stady state ac voltage?
Thanks
 

Pspice RLC AC analysis

Hello
here is my pspice code and circuit that I want to simulate.


Code: 
Hello the world
V 1 0 AC 1V
R 1 2 50
L 2 3 .02H
C 3 0 150nf
.ac LIN 4901 100Hz 5kHz
.Probe
.END

as you see the probe windows we have this output(Resistor current and Capacitor voltage)



So we do some calculation
|I|=.02 in resonance frequency (2.9Khz)
|Xc|=1/(2900*150*10^(-9))=2298.8
|Vc|=|I|*|Xc|=.02*2298.8=45.97

but according the V(3)(Capacitor voltage) we have 7.3 at resonance frequency.
other calculation:
we know the Capacitor voltage at resonance frequency is about |VC|=Q*|Vin |

|Vc|=|I|*|Xc|=(|Vin|/R)(1/w0C)=|Vin|*Qc

w0=2900hz
QL=(w0*L)/(R)=1.16
|VC|=Qin*|Vin=1.16*1=1.16

Or

w0=2900hz
QC=(1)/(w0CR)=45.97
|VC|=Qin*|Vin=45.97*1=45.94


If there is no problem could you tell me what are my mistakes?

- - - Updated - - -

Sorry I found my problem
I didn't substitute the resonance frequency into w=2*Pi*F
 

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