Hi, I'm trying to graph the series in LT SPICEresonance, making a AC sweep analysis.
SIMPLE RCL circuit in series
*
VS 1 0 SIN(0 10 166.81) AC 100
*
RESISTOR 1 2 4
CAPACITOR 2 3 10uF
Linduc 3 0 100mh
*
* ANALYSIS
.AC DEC 9 10 1MEG
*.TRAN 10s 10.02s 10s
* VIEW RESULTS
*.PRINT VP(2) vm(1,2)
*.PLOT AC VM(2) VP(2)
*.PRINT TRAN V(1) V(2)
*.PLOT TRAN V(1) V(2)
#autoplot 1 v(1,2) v(2,3) v(3)
#autoplot 2 i(RESISTOR) i(VS)*-1
#autoplot 3 vp(1,2) vp(2,3) vp(3)
.probe
.END
My question is : at resonance the current must be of I=100v/4Ω≈25A. So the voltage in resistance= 25A*4Ω=100volts, capacitor or inductor should be ≈3900 Volts.
The resistor is not at 0° so there is no 100Volts.
The link example is below, and I would aprecciate your help. Guess LT SPICE have a bug?. Don`t think so.
Below is the link with example the example. I make the math again & is correct!. SO why lt spice is not showing the correct graph?.
THANK YOU VERY MUCH
http://www.electronics-tutorials.ws/accircuits/series-resonance.html
SIMPLE RCL circuit in series
*
VS 1 0 SIN(0 10 166.81) AC 100
*
RESISTOR 1 2 4
CAPACITOR 2 3 10uF
Linduc 3 0 100mh
*
* ANALYSIS
.AC DEC 9 10 1MEG
*.TRAN 10s 10.02s 10s
* VIEW RESULTS
*.PRINT VP(2) vm(1,2)
*.PLOT AC VM(2) VP(2)
*.PRINT TRAN V(1) V(2)
*.PLOT TRAN V(1) V(2)
#autoplot 1 v(1,2) v(2,3) v(3)
#autoplot 2 i(RESISTOR) i(VS)*-1
#autoplot 3 vp(1,2) vp(2,3) vp(3)
.probe
.END
My question is : at resonance the current must be of I=100v/4Ω≈25A. So the voltage in resistance= 25A*4Ω=100volts, capacitor or inductor should be ≈3900 Volts.
The resistor is not at 0° so there is no 100Volts.
The link example is below, and I would aprecciate your help. Guess LT SPICE have a bug?. Don`t think so.
Below is the link with example the example. I make the math again & is correct!. SO why lt spice is not showing the correct graph?.
THANK YOU VERY MUCH
http://www.electronics-tutorials.ws/accircuits/series-resonance.html