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See the circuit diagram attached.
The voltage source is a sinusoidal AC source with amplitude = 240, Frequency = 50, Phase = 90.
Essentially I have a lab report and I was wondering what sort of equations are required to find the impedance of the circuit. We're not really told if we're actually able to calculate this by hand (We've only just started 2nd ODE circuits)
The whole report is trying to prove that PSPICE produces correct answers.
See the probe diagram attached.
What our pspice instructions are, is to go the to second peak from the right of the probe and get the peak of the voltage over A [V(A)] and and go the the peak of the current over R1 [I(R1)], then using the values from those points and the phase difference substitute the values into:
Z = [itex]\frac{V}{I}[/itex] to find the impedance.
So for my diagram I have:
Z = 239.931e^{j0}/15.455e^{35.33j}
I understand that these values are inaccurate (our lab sheet tells us this as well) however is there any way to do comparable hand calculations (not using the graph) that would give similar results?
I tried using S^{2}+[itex]\frac{10}{300mH}[/itex]S+[itex]\frac{1}{300mHx30uH}[/itex]=0
To find the voltage at time t however when I used that value to get a current for the circuit I got a non complex number.
Any recommendations are much appreciated.
ALSO
Just realised that the lab notes (which have different results to mine, also known and okay) state that:
240e^{0xj}/15.44e^{34.4xj}= 12.9  8.8j
I'm totally not getting that when I plug it into my calculator.
I get: 15.352.44j
Anyone got any ideas?
The voltage source is a sinusoidal AC source with amplitude = 240, Frequency = 50, Phase = 90.
Essentially I have a lab report and I was wondering what sort of equations are required to find the impedance of the circuit. We're not really told if we're actually able to calculate this by hand (We've only just started 2nd ODE circuits)
The whole report is trying to prove that PSPICE produces correct answers.
See the probe diagram attached.
What our pspice instructions are, is to go the to second peak from the right of the probe and get the peak of the voltage over A [V(A)] and and go the the peak of the current over R1 [I(R1)], then using the values from those points and the phase difference substitute the values into:
Z = [itex]\frac{V}{I}[/itex] to find the impedance.
So for my diagram I have:
Z = 239.931e^{j0}/15.455e^{35.33j}
I understand that these values are inaccurate (our lab sheet tells us this as well) however is there any way to do comparable hand calculations (not using the graph) that would give similar results?
I tried using S^{2}+[itex]\frac{10}{300mH}[/itex]S+[itex]\frac{1}{300mHx30uH}[/itex]=0
To find the voltage at time t however when I used that value to get a current for the circuit I got a non complex number.
Any recommendations are much appreciated.
ALSO
Just realised that the lab notes (which have different results to mine, also known and okay) state that:
240e^{0xj}/15.44e^{34.4xj}= 12.9  8.8j
I'm totally not getting that when I plug it into my calculator.
I get: 15.352.44j
Anyone got any ideas?
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