While solving the below circuit:

enter image description here

The solution is as shown below: enter image description here

I noticed the following while analyzing it:

enter image description here

Why does the voltage across a pure resistance and the 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?

  • \$\begingroup\$ Imaginary or "Reactive" current lag is shared by the Inductor which causes it. \$\endgroup\$ Jan 14 at 20:43

Why does the voltage across a pure resistance and the current through it have imaginary parts?

The imaginary part tells you that the current flow through the resistor is lagging the voltage applied to the series network by a certain amount. As far as the resistor is concerned, both its voltage and its current are totally in phase but, relative to the applied voltage applied across the series network of R and L, they are both lagging.

What do these imaginary parts represent?

The imaginary part indicates by how much the current in both L and R (the same current) is lagging the applied voltage to that series network.

Your math is correct apart from forgetting to add 90 ° to the inductor voltage value

enter image description here

  • 1
    \$\begingroup\$ THANKS you made it crystal clear. \$\endgroup\$
    – OMAR
    Jan 14 at 11:11
  • 2
    \$\begingroup\$ here is what cleared it " As far as the resistor is concerned, both its voltage and its current are totally in phase" \$\endgroup\$
    – OMAR
    Jan 14 at 11:12

To make it perfecly clear, I want to make sure you do understand that no matter how many X, R or C elements you connect together in series, the current that flows through the mentioned in-series circuit is the same for every element.

That's specific to in-series connection as it has only one branch the current can flow, while parallel connected circuits might have different currents flowing through different branches.

In this case, the inductor X causes the current to lag in the whole circuit, that's the reason the imaginary part comes into play. Of course the same current also runs through the resistor R as it has no other choice.

Imaginary part of the current/voltage is used as a symbolic way to represent sine wave 'shift' compared to 'unshifed' wave. But that's oversimplification and please dont cite that during lectures.

gauss plane

In circuit with R only, the sine wave of current and voltage overlap each other as they are 'in sync'.

They de-sync once we introduce X or C into the circuit - see below.

current vs voltage sine waves shift

The shift happens due to the nature of energy stored/trapped inside coils X, and capacitors C, or other appliances/machines that do store energy, but can be reduced into symbolic representation of X or C (rotating engines do have high resistance R and reactance X)

Resistors R do not store any energy.

Reactive X elements store energy as electromagnetic field that's induced by the current flowing through them, see the picture to understand the electromagnetic field around coil that stores energy, literally "in air" or whatever there is around the coil. This field is loaded/deloaded 60 times a second due to 60Hz voltage pulses back and forth, we could say left and right (the voltage drives the current, which induces the field around coil). Imagine a baloon of electrons around our X growing/shrinking 60 times a second.

This as a whole shifts voltage sine wave ahead of current sine wave.

enter image description here

Reactive C elements store energy as electrostatic field (I might oversimplificate here again) that's trapping electrons in static field inside capacitor (or other capacitor-like appliance). Please see the picture to visualise energy storage on capacitor plates.

This as a whole shifts current sine wave ahead of voltage sine wave.

enter image description here

Energy that's stored in our circuit does not do any good for our us, as we want to use energy to do some work for us, like move an engine or heat the house.

All of that leads into the conclusion, that in real-life electrical systems, we try to keep power factor (cos fi), "the shift" (imaginary part of current/voltage compared to their non-imaginary counterparts) as minimal as possible. But how can we do that, having in mind that certain appliances/machines do need electromagnetic (rotors) or electrostatic fields (electronic devices) to operate?

We try to balance the ammount of X and C reactance in our system to be as close in value to each other as possible - that leads into two energy storage mechanisms exchanging energy inbetween themselves. Try to imagine current pulsating back and forth, while electrons trapped in a capacitor load the coil on the way back, and in couple milliseconds the voltage direction shifts (remember alternating current 60hz), and the coil loads the capacitor.

Voila, having the minimal lag between current/voltage in system that consists of several R, X, C elements, we're sure the pulsating "non-productive" reactive power is exchanged inside the system, not traveling past our energy meter making us pay for energy we don't use, the energy that's just "stored" in our system for 2 miliseconds each hertz :)

I'm far from scientific precision in this explaination, but I hope It'll help you in your electrical studies, good luck!


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