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Happy New Year! This month's challenge is based around electrical engineering, specifically an AC circuit.
An AC electrical circuit consists of the following components connected in series:
• A 120 Volt 60 Hertz power source
• A 50 Ohm resistor
• A 0.20 Henry inductor
• A 10 micro Farad capacitor
Calculate the following for this circuit:
• The reactances of the inductor and capacitor
• The impedance in the circuit
• The current in the circuit
• The power dissipated in the circuit
• The phase angle of the circuit
Using an XY Plot or a Chart Component, graph the following in the same plot or chart over a minimum of 2 cycles:
• The current in the circuit.
• The voltage in the circuit.
• The voltage drop across the resistor.
• The voltage drop across the inductor.
• The voltage drop across the capacitor.
You can use the values from the calculation challenge, or you can use generic values. The point is to communicate that the current, total voltage, and voltage drop across the resistor are in phase; that the voltage drop across the inductor leads the current by a quarter cycle; and that the voltage drop across the capacitor lags the current by a quarter cycle. You may want to use 2 y-axes, one for current and one for voltage.
Using the advanced input controls like sliders, buttons, check boxes, etc., allow people to change one or more values in the Calculation Challenge to see how that affects the calculated values.
Find the Mathcad Community Challenge Guidelines here!
Disclaimer: I am not an electrical engineer. I apologize in advance if any of my terminology offends your sensibilities.
I'm no electrical engineer, but here's how I tackled the challenge:
Just noticed that my calculation of the cumulative average, used in the calculation of dissipated power, should have a divisor of k+1, not just k. This is because Mathcad's origin is 0 not 1. However, the correction makes no significant difference!
Alan
1. Add phase shift of sine wave.
2. Add transient responces of 3 waveforms, with all 120VRMS.
3. Add rectifier radio bottuns.
Transient responce of sine wave with phi and rectifier can not calculate by Laplace Transform method. So, need more Mathcad skill to solve it.
Thanks exsellent answer. How about cosine or rectangular waves, that not start from 0 values.
On my sheet, last page, change the input signal v.r to you liking, and its Laplace transform v.s.
Then see what happens.
Success!
Luc
In your last page vs is fixed to sine wave. So nothing change for output signals, except an input voltage waveform.
I said to change the signal AND its Laplace transform...
Then a square wave function might be:
Its Laplace transform (from literature) is:
And the waveform results are:
Not too bad, as far as the low frequency signals go (circuit current, resistor and capacitor voltage). But the inductor voltage goes way off after the first half period. That's due to the invlaplace function. See how it transforms the square wave back:
Success!
Luc
I want to say the noize after the half cycle, such as typically vL, as you sayed.
One of the confusing arise from s. It is sometimes Laplace's and unit s. But Mathcad treat it in your sheet and show the plot.
So, this is not the reason of noize. The Laplace transform treatment of vL=L*di/dt sometimes includes delta(t), the impulse function. Mathcad can not plot delta(t) and plot result may faild. Using unit step function to make rectangular wave cycle may be 2 cycle is OK. More than 4 cycles Mathcad can not show the results because it is to long to show. Also using long time to calculate.
Therefore, only using Laplace transform is limited to use it. We need another method to solve it.
The resulst of cosine half rectifiered wave and rectangular waves are must be below.
Ver. 3 can chage each of R, L, C to short it. Applied voltage has angle Z. Therefore, RL circuit can start no transient responce.
You can see RC, RLC or LC circuit's transient responces.