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I mean how are they physically different? What does it mean to have imaginary numbers in the reactance?

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6 Answers 6

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Reactance only refers to the imaginary component, whether it is on its own or paired with a real resistance.

Impedance just implies frequency dependency. So it refers to both the imaginary reactive component and the real resistance, together, if the real resistance is present.

You know how a resistance removes energy from the circuit and then dissipate it as heat thereby losing it permanently? That's real because it's real energy truly removed. The reactance is imaginary because it removes energy from the circuit at that moment but doesn't actually dissipate it. It just takes it out of circulation and stores it, releasing it back into the circuit at a future time. So the energy is removed from the circuit, but only in an "imaginary" sense if you want to somehow put it into words.

The sign of the imaginary component is the lag-lead in the voltage and current which is opposite between capacitances and inductances. You see this in time-domain graphs and this translates to the complex numbers being the way they are in phasor representation when rotating around the circle. In a sense I guess you could say it describes whether energy is being removed from the circuit and stored by accumulating current or by accumulating voltage.

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  • \$\begingroup\$ what is the relevance of imaginary numbers in reactance? \$\endgroup\$
    – FISqrt
    Jan 27 at 16:59
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    \$\begingroup\$ @FISqrt You know how a resistor removes energy from the circuit and then dissipate it as heat thereby losing it permanently? That's real because it's real energy truly removed. The reactance is imaginary because it removes energy from the circuit at that moment but doesn't actually dissipate it. It just takes it out of circulation and stores it, releasing it back into the circuit at a future time. So the energy is removed from the circuit, but only in an "imaginary" sense if you want to somehow put it into words. \$\endgroup\$
    – DKNguyen
    Jan 27 at 17:00
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    \$\begingroup\$ Imaginary numbers are a mathematical convenience which allow otherwise impossible calculations using real numbers. They don't have any real world interpretation outside the calculation as such. \$\endgroup\$
    – Ian Bland
    Jan 27 at 17:01
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    \$\begingroup\$ @FISqrt That's current and voltage lag-lead which is opposite between capacitances and inductances. You see this in time-domain graphs and this translates to the complex numbers being the way they are in phasor representation when rotating around the circle. \$\endgroup\$
    – DKNguyen
    Jan 27 at 17:04
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    \$\begingroup\$ @FISqrt Negative reactance means the current is 90 degrees behind the voltage - in other words, changing the voltage produces a gradual change in current. Positive reactance means the current is 90 degrees ahead of the voltage, or actually (since we don't have time travel!) that the voltage is 90 degrees behind the current, which means that changing the current produces a gradual change in voltage. You should already recognize that behaviour from what you know about capacitors and inductors - this is just a way to put equations on that behaviour for AC. \$\endgroup\$
    – Graham
    Jan 30 at 9:54
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In a comment on another answer you clarified your question,

why reactance is multiplied by an imaginary number?

It means that when you apply an AC voltage to a reactive element, the current that is produced is out of phase with the voltage.

Or if you force an AC current through the reactive element, the voltage it produces is out of phase with the current.

The exact phase relationship depends on the angle of the impedance. That is, it depends on both the reactance and the resistance as components of the impedance.

As others have pointed out in comments, the consequence of the voltage and current being out of phase is that (depending on the phase) some or all of the power delivered to the element during part of the AC cycle will be returned to the circuit on other parts of the cycle.

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Impedance is a complex number. Its real part is called resistance and its imaginary part is called reactance

$$Z=R+jX $$

what does it mean to have imaginary numbers in the reactance ?

There is not. Reactance is always a real number

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  • \$\begingroup\$ I didn't explain it well in my question , but I meant why reactance is multiplied by an imaginary number? \$\endgroup\$
    – FISqrt
    Jan 27 at 17:10
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    \$\begingroup\$ @FISqrt That's more of a math book keeping thing. You don't even have to think of it as an imginary number. You can think of it as a variable name that just tags the term to identify it and stop it from improperly mixing with other terms. You see the same thing in mechanics with i,j,k when talking about cross products and vectors. but that's in 3D, not 2D. \$\endgroup\$
    – DKNguyen
    Jan 27 at 17:11
  • \$\begingroup\$ @Carl, he didn't ask for this answer. \$\endgroup\$
    – Sadat Rafi
    Jan 27 at 17:12
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    \$\begingroup\$ @FISqrt The trick of making reactance imaginary is that it makes everything you know about resistive circuits work for impedances. Series impedances add, just like resistances. Parallel admittances add, just like conductances. So, if you know how to analyze a resistive network using real numbers, you can carry that knowledge over to networks that have capacitors and inductors by doing the same kind of calculations with complex numbers. \$\endgroup\$
    – John Doty
    Jan 28 at 23:51
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Multiplying or dividing by j is the way to denote a + or - 90-degree phase shift, respectively. It is the rectangular coordinate version of saying that some component of the current is out of phase with voltage. Phasor notation is the polar coordinate version of the same thing.

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It's most intuitive to understand impedance when you look at it in angle notation $$Z=|Z|\times[\cos(\phi)+j\sin(\phi)]=|Z|\angle\phi$$ The magnitude of the impedance |Z| describes how current flow is impeded (same way as with regular resistance). The argument ϕ describes how the current is displaced in time - phase shifted.

For example, say you have 5 volt peak-to-peak input that you pass through a 5∠60° ohm impedance, you will get 1 amp peak-to-peak current that is phase shifted by 60°.

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In circuits with primarily capacitive loads, current leads the voltage. This is true because current must first flow to the two plates of the capacitor, where charge is stored. Only after charge accumulates at the plates of a capacitor is a voltage difference established.

When you apply power to an Inductive or capacitive load, the term phase difference appears.For example, you are providing power to a capacitive load. First, it will get charged and draw current later; the voltage will rise. You can't explain this angular difference in normal real number mathematics.

If you take voltage waveform as a reference and measure how much angle current is behind, then you will get vector relationship like \$i∠\theta\$ (current at an angle theta). According to Euler's formula this can be written as \$i(cos(\theta) + j*sin(\theta))\$. The complex number notation 'j' is simply used to represent the 'Y-axis' in the vector diagram.

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