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Voltage Drop and Power Disipation in RL circuit

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    Voltage Drop and Power Disipation in RL circuit

    I need to refresh my memory on some basics.

    In an RL circuit consisting of a resistor and reactor......

    Is the voltage drop across the resistor a function of he total current (real and reactive) times the resistor value or just the real portion?

    Is the power dissipation across a resistor a function of the total current (real and reactive) times the resistor value (I^2R) or just the real portion?

    If the resistor and inductor both represent 50% of circuit impedance then if you measure RMS voltage drop across either of them at any point in time you will see a 50% voltage drop compared to the source voltage? However I believe if you look at these on a scope you will see that they are not in phase so the voltage drops are occurring at different times? Can someone please refresh my memory on this?

    Thanks

    #2
    Let's say you have a current flow that is 100% reactive driven. For example the circuit may be composed entirely of inductors and capacitors with no power out of any kind to heat or electro - mechanical energy conversion.

    The current flow itself in that circuit is very real. It may be graphed on the i or imaginary axis, but the moment you pass that current through a resistor, current is current and it does not matter that the flow is reactive power driven or comes from any other source (that also causes current flow). Reactive power only designates that the flow is into and out of an (perfect lossless) energy storage device.

    Inductance is the storage of energy in a magnetic field. Capacitance is the storage of energy in an electric field. The current flow associated with charging and discharging this *stored energy*' is very real as real as any other current flow from any other source. The actual quantity is by convention and usefulness graphed on the i axis and only unfortunately named as the imaginary axis.

    You may need your imagination to see this happening, but it is "*not* an imaginary quantity, only unfortunately named so.

    Pass that current through any resistor and E = IR and P = IV. Voltage drop across the inductor is V = L * di/dt and di/dt is a sine wave and frequency dependent. Voltage drop is proportional to the rate of change of current.

    I'm not sure what question was asked.
    Lasciate ogne speranza, voi ch'intrate

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      #3
      Originally posted by philly View Post
      I need to refresh my memory on some basics.

      In an RL circuit consisting of a resistor and reactor......
      [COLOR=#ff0000]I'm assuming the resistance is in series with the reactance. See the attached picture.
      [/COLOR]
      Is the voltage drop across the resistor a function of he total current (real and reactive) times the resistor value or just the real portion?
      [COLOR=#ff0000]It's the total current, it is a series circuit. The current is the same through all elements.[/COLOR]

      Is the power dissipation across a resistor a function of the total current (real and reactive) times the resistor value (I^2R) or just the real portion?
      [COLOR=#ff0000]The voltage drop across the resistor will be in phase with the current. Again, [COLOR=#ff0000]it is a series circuit. The current is the same through all elements.[/COLOR][/COLOR]

      If the resistor and inductor both represent 50% of circuit impedance then if you measure RMS voltage drop across either of them at any point in time you will see a 50% voltage drop compared to the source voltage? However I believe if you look at these on a scope you will see that they are not in phase so the voltage drops are occurring at different times? Can someone please refresh my memory on this?
      [COLOR=#ff0000]That is hard to do (each impedance being 50% of the total circuit impedance). One might say the reactive impedance does not point the same "direction" as the resistive impedance [/COLOR]

      Thanks
      Hopefully the pic does a better job explaining than the words
      Attached Files
      Without data you’re just another person with an opinion – Edwards Deming

      Comment


        #4
        190228-2052 EST

        philly:

        In a series circuit the current everywhere is exactly the same at any instant of time, the same no matter what the waveform, the same no matter how you want to measure the current, average, RMS, etc.

        Thus, the voltage drop across the resistance is e = R*i, and in phase with the current. Assuming the inductive component is pure inductance, then e = L*di/dt. If current is a sine wave, then for resistance e = R+I*sin wt, and for inductance e = L*w*I*cos wt. Note, cos x = sin (x + 90). Thus, voltage across an inductor in a series RL circuit leads the voltage across the resistor. Current thru the series circuit lags the source voltage. Thus, voltage across the resistance lags the source voltage.

        For RMS the vector sum of the R and L component voltages must equal the source voltage. These vector voltages form a right triangle. Thus, the sum of the two vector magnitudes is greater than the source voltage, unless R or L is zero. If R and Xl are equal, then the individual RMS voltages are 0.707 of the source voltage, and their magnitudes sum to 1.414 times the source voltage. The sum of the magnitudes is likely to be of little value in circuit analysis.

        Average power dissipation is Irms^2*R.

        It is hard to find a nearly ideal inductor. So take some fairly high Q inductor and a resistor and study these with a scope. You won't be able to directly measure the inductive voltage because the inductor has internal resistance. So it might be better to run experiments with a high quality capacitor. With a capacitor your phase shifts 180 deg compared to the inductor.

        .

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          #5
          Clarification without the fluff

          Originally posted by philly View Post
          I need to refresh my memory on some basics.

          In an RL circuit consisting of a resistor and reactor......

          Is the voltage drop across the resistor a function of he total current (real and reactive) times the resistor value or just the real portion?
          Yes: V= I X R
          Note: It doesn't help to think about current in terms of real and reactive components. As you have stated, its simply the total scalar current; but If you are interested in the voltage drop angle too, then you would use complex quantities: V= I X R

          Originally posted by philly View Post
          Is the power dissipation across a resistor a function of the total current (real and reactive) times the resistor value (I^2R) or just the real portion?
          Again its simply the total scalar current. Disregard thinking about things in terms of real and reactive components.

          P = I^2 X R; or
          P = V^2/R; Where V is the scalar voltage drop of the resistor calculated above.

          Originally posted by philly View Post
          If the resistor and inductor both represent 50% of circuit impedance then if you measure RMS voltage drop across either of them at any point in time you will see a 50% voltage drop compared to the source voltage? However I believe if you look at these on a scope you will see that they are not in phase so the voltage drops are occurring at different times? Can someone please refresh my memory on this?
          If both elements represented 50% of the ENTIRE circuit impedance then the full voltage drop would be seen across the terminals of the resistor/inductor series combination. This terminal voltage would be in phase with the source voltage (assumed to be ideal and at 0 degrees). The current would be lagging at 45 degrees, and each component would see 1/SQRT(2) of the source voltage (not 50%). Now each one of these 1/SQRT(2) voltage drops would be occurring at different times (having different angles). The resistor voltage drop would be 1/SQRT with an angle of 45 degrees lagging (in-phase with the current) and the inductor voltage would be 1/SQRT with an angle of 45 degrees leading (in-quadrature with the current).

          Hope that helps!

          Comment


            #6
            Deleted.
            Last edited by Sahib; 03-01-19, 10:42 AM.

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