Find impedance of a 2nd ODE circuit (PSPICE)

[Evales]

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 S²+[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 realized 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: -15.35-2.44j

Anyone got any ideas?

 

[gneill]

The impedances of inductors and capacitors are, respectively, ##j\omega L##, ##\frac{1}{j\omega C}##. Add up your impedances for a series circuit.

 

[Evales]

Thanks but you didn't respond regarding the discrepancy that I found in my lab instructors results. Please just tell me, I handed it in yesterday anyway.

 

[gneill]

Thanks but you didn't respond regarding the discrepancy that I found in my lab instructors results. Please just tell me, I handed it in yesterday anyway.

Make sure that the exponents of e in your calculations are expressed in radians, not degrees.

 

[DeeNos]

I would suggest that the first step in finding the impedance of a 2nd ODE circuit is to understand the basic principles and equations involved. In this case, the impedance of a circuit is defined as the ratio of the voltage to the current, Z = V/I. In order to calculate the impedance, we need to know the values of both the voltage and the current at a given point in time.

In the circuit diagram provided, we can see that there is a sinusoidal AC source with an amplitude of 240, frequency of 50, and a phase of 90. This means that the voltage source can be represented by the equation V(t) = 240sin(100πt + π/2). Using this equation, we can calculate the voltage at any point in time.

Next, we need to find the current in the circuit. This can be done by using Ohm's law, which states that the current is equal to the voltage divided by the resistance, I = V/R. In this case, we have a circuit with a resistor (R1) and an inductor (L1), so we need to use the complex impedance formula Z = R + jωL, where R is the resistance, ω is the angular frequency (2πf), and L is the inductance.

To find the impedance of the circuit, we need to first find the values of R and L. The resistance can be found by using the given value of R1, which is 15.44Ω. The inductance can be calculated using the formula L = ωL1, where L1 is the inductance value of the inductor (30μH) and ω is the angular frequency (2πf).

Now, we can calculate the impedance using the formula Z = R + jωL. Plugging in the values, we get Z = 15.44 + j9.42Ω. This is the impedance of the circuit at any given point in time.

To verify the results obtained from PSPICE, we can do some hand calculations by using the voltage and current values at a specific point in time. As mentioned in the lab instructions, we need to find the voltage and current at the second peak from the right of the probe. Using the voltage equation V(t) = 240sin(100πt + π/2), we can calculate the voltage at this