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Why does a pure resistance has a an imaginary current and voltage?

Discussion in 'Electronics Homework Help' started by skyline1397, Jan 14, 2021 at 9:23 AM.

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  1. skyline1397

    skyline1397

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    Sep 13, 2017
    While solving the below circuit:

    [​IMG]

    The solution is as shown below: [​IMG]

    I noticed the following while analyzing it:

    [​IMG]

    how a pure resistance both the voltage across it and current through it have imaginary parts?
    What do these imaginary parts represent?

    Similarly for a pure reactive load (inductive) what does the real part of the voltage Vx mean?
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    10,771
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    Nov 17, 2011
    Because the circuit is not purely resistive. The imaginary part is added by the inductor in this case.
    The same current must go through resistor and inductor. See how this in influences your calculations and the real and imaginary parts.

    I'll also move this thread to where it belongs: homework.
     
    Arouse1973 likes this.
  3. skyline1397

    skyline1397

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    0
    Sep 13, 2017
    The current is the same. But why the resistor voltage has an imaginary part??
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    10,771
    2,427
    Nov 17, 2011
    What is the equation for a voltage across a resistor?

    When you look at the voltage across and the current through the resistor only, you will see that both are perfectly in phase. Since the current has an imaginary part, so does the voltage.
    Real and imaginary make sense only in combination with a reference frame that establishes a 0 ° phasor (or a purely real signal). Then you can determine the phase of other signals (voltages, currents) with reference to that 0 ° signal. When you look at a single signal, isolated from the others in the system/circuit, phase makes not much sense (if any at all).
    In your example the reference signal with phase 0 ° would be the driving voltage source. Only when looking at the voltage with respect to this 0 ° reference will you "see" the imaginary part of the voltage across the resistor. You will not "see" the imaginary part when you look at the voltage and current across and through the resistor only.
     
    Last edited: Jan 14, 2021 at 11:28 AM
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