AC RL circuit simulation with TINA & Multisim

Thread Starter

Khaled.Ibrahim

Joined Oct 11, 2016
4
For a RL Circuit having AC voltage source V=Vp Sin(ωt) and V=100 , R=1 Ω and L=0.1 H
The assignment is about plotting the current response for this circuit of the two instants of voltage zero-crossings and of the two instants of peak voltage with a given one cycle of sinusoidal wave.
I was able to get only the current response for the circuit while assuming frequency (f)=50 Hz but i don't know what plotting current response at zero and peak voltages mean or how to use Tina or Multisim to do.
I can choose the AC voltage source signal to be sine waveform with phase=0 but for Multisim i can only choose it to be sine or cosine so the result is different from Tina so is Tina the correct one or i'm wrong?
Anyone can help ?



TINA 9


Multisim 14
 

Danko

Joined Nov 22, 2017
767
I can choose the AC voltage source signal to be sine waveform with phase=0 but for Multisim i can only choose it to be sine or cosine so the result is different from Tina so is Tina the correct one or i'm wrong?
Anyone can help ?
Multisim 14
You can set any initial phase of V1 sine signal. In properties of V1, in "Value" tab, choose "AC analysis phase:" and type your ω value that you need, in degrees (0, 90, 180, 270...).
In Transient, "Analysis parameters" tab, set "Initial conditions" in "Set to zero".
 

Thread Starter

Khaled.Ibrahim

Joined Oct 11, 2016
4
You can set any initial phase of V1 sine signal. In properties of V1, in "Value" tab, choose "AC analysis phase:" and type your ω value that you need, in degrees (0, 90, 180, 270...).
In Transient, "Analysis parameters" tab, set "Initial conditions" in "Set to zero".
thanks!
When i change the phase (0,90,180,270) Current response doesn't change and is the same thing for Voltage , is that possible?
for 0 degrees

for 90 degrees

 
Last edited:

Thread Starter

Khaled.Ibrahim

Joined Oct 11, 2016
4
Sorry for my mistake. You need type your degrees not in "AC analysis phase:" but in "Phase:".
Ok but that changes the phase for the voltages source and for phase (0,180), (90,270) the current response would be the same is that right?
 

Danko

Joined Nov 22, 2017
767
Ok but that changes the phase for the voltages source and for phase (0,180), (90,270) the current response would be the same is that right?
Yes. Curves of current response at (0, 180) should be mirror reflection of each other. Same with (90, 270) pair.
 
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