RLC Complex Help

Discussion in 'Homework Help' started by hotrocks, Nov 28, 2013.

  1. hotrocks

    Thread Starter New Member

    Jun 29, 2010
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    Please find attached my working out for an RLC circuit using complex notation. Im pretty sure the answer in part A is correct but im told the answer to part B with the capacitor loaded should be around 31.8V.

    I cant figure out where im going wrong

    Thanks to all in advance.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Let's consider the extent of the impact of the load resistor we can reasonably expect.

    The other impedances in the circuit have magnitudes of 10Ω, 31Ω, and 318Ω. The load has an impedance of 10000Ω. Does it seem like a load, placed in parallel with the -j318Ω impedance, that is so much larger than the other impedances is going to have much of an effect at all result?

    More to the point, does it seem at all reasonable that the output is going to up go from 11V to 32V by loading it down with a resistive load?

    Your loaded magnitude goes from 11.08V to 11V, which is actually more of a change that I would expect, but it isn't unreasonable. However, does it make sense that your angle would change by 20 degrees but the amplitude would only change by 0.8%. Not saying it's not possible, but it seems unlikely.
     
  3. hotrocks

    Thread Starter New Member

    Jun 29, 2010
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    Thanks for the reply. I was wondering if there was any maths guru`s out there who could confirm the theory and working out.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The value of the argument for the capacitor voltage in the first instance is incorrect. Also keep in mind your results are in peak values. The usual practice is to use RMS values, though obviously this is not mandatory. So if you are giving answers in peak values you should explicitly state this to be the case.
    Re-check your answers regarding the angles.
    As a quick check keep in mind the majority of the source voltage is across the capacitive reactance. It's unlikely then that this voltage would lead the source by 178 degree.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    What, exactly, are you looking for? I've given you some pretty specific things to look at. Please look at them.

    Or, let me be a bit more explicit -- their answer to the second part is wrong. Your's is close, but not quite right.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Well, let's consider the simpler case of a source connected in series with an inductor and a capacitor. Since the load is purely reactive, the current in the source has to be either 90 degrees ahead or 90 degrees behind the source voltage, depending on which reactance has the greater magnitude. Since this is the same current that is flowing in both the inductor and the capacitor and since the voltage across the inductor will lead the current by 90 degrees while the voltage across the capacitor will lag the current by 90 degrees, the voltage across one of them will be in phase with the source voltage while the voltage across the other will be 180° out of phase with the source voltage.
     
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