# confused between impedance and reactance

Discussion in 'Mathematics and Physics' started by PG1995, Dec 6, 2011.

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Hi

Regards
PG

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2. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

I was waiting for you to discover this one with the 'w'

When we differentiate the sin function with respect to time like d(sin(w*t) )/dt we get w*cos(w*t). The leading w comes from differentiation that's all.
We could then convert the cos function into a sin function with a phase angle.

The impedance Z is not the same as the reactance usually given by Xc or similar.

Consider a number such as:
5j

which is simply 5 multiplied by the imaginary operator j.

Is this true:
5j=5 ???

Clearly it is not true. 5 is the reactance, while 5j is the impedance.
The impedance is a complex quantity in general, and so provides us with more information than just the reactance alone.
The general impedance is complex:
a+b*j

while the reactance is just a scaler (and is the amplitude of the impedance):
sqrt(a^2+b^2)

Note we can extract the phase angle from a+b*j, as well as the amplitude sqrt(a^2+b^2), but knowing only the reactance all we know is the amplitude and can not calculate the phase angle.

Typical reactances would be 5, 8, 23.4, 67, 1004.01, etc., all requiring only one number to completely specify.
Typical impedances would be 5+2j, 10+4j, 237.3+823j, etc., all requiring two numbers to specify. And note that if one part of the impedance is zero then it might simplify, but we still need to know that in order to properly specify the impedance: 4+0j, 0+5j, etc.

For the inductor alone, Z=j*XL.
For the capacitor alone, Z=-j*XC.

Sometimes the reactance is used in place of the impedance to simplify a calculation. This is an approximation when there are other circuit elements such as a resistor, but can sometimes lead to a very serious error so it's best to avoid at least until you've discovered a range of validity of using such an approximation. An example for you to investigate would be a simple line powered LED circuit using a capacitor and resistor in series to power an LED (with a reverse diode to protect the LED). You could try to determine the current through the LED using the reactance alone and using the impedance and compare results. You could look at higher powered LEDs this way too and see if anything changes as you go up in LED power. Experimentation like this in a simulator would give you great insight as to how things work. Ask questions, then try to answer them using the simulator. Match your equations to the simulator results.

I should have also mentioned that sometimes the magnitude of the impedance is also called the impedance:
Impedance=|Z|
and that is accepted in many areas of interest, but the complex impedance is the best when possible (and appropriate) because that tells us the most.

Last edited: Dec 9, 2011
3. ### PG1995Active Member

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Hi

I hope my understanding is not too much off the line and can suffice for the time being. Thank you.

Regards
PG

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5. ### PG1995Active Member

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Hi

Re: Power factor correction

Q1:
Please have a look on these circuits. Suppose that circuit #3 is being used. If an ammeter is connected to the circuit, it will read 1.410A and we can erroneously conclude that power being delivered to the load is 169.256W but that is not the case. On the other hand, a wattmeter would give us true value of 119.365W because it only reads true or real power and not the reactive power which keeps bouncing back and forth between the source and load without doing any real work.

Power companies always encourage power factor of unity because it create issues on their side such as it results in more I²R losses. Power factor correction method could be employed to make the power factor unity. The power factor correction method makes a load behave like a resistive load instead of a reactive load. If power correction method is applied to circuit #3 then it will ideally behave like 60Ω resistive load to the 120V source and still delivering same 'real' power of 119.365W.

Do I have it correct so far?

Q2:
Let's see how power factor correction method is applied to circuit #3. Please have a look here. Do I have it correct?

Q3:
In the example we saw that after power factor correction the current supplied by source reduces from 1.410A to just 1A. As the supply now just have to supply only 1A therefore firstly I²R losses are reduced and secondly now the unused 0.410A could be used by the source for some other load. Do you agree with this?

Q4:
I believe that the current between nodes "A" and "B" is 1A but the current between nodes "B" and "C" is still 1.410A. I mean that if an ammeter is connected between the nodes "B" and "C", it will read 1.410A. Do you agree with me?

Q5:
In real world most reactive loads are inductive in nature such as motors and transformers. Let's see how a power factor correction method works from a layman's viewpoint. Please have a look here.

The power factor correction method is based on concept of LC tank circuit. For more information about LC tank circuit, please refer to the links. The currents of capacitor and inductor in LC tank circuit are 180° out of phase. But what does this mean when we say that they are 180 degrees out of phase with each other? There might be different ways to look at it. Let's try this one. We will focus on the waveforms at the instant when they are crossing y-axis in Fig 1; notice the dots. Also have a look on Fig 2. The inductor pushes the current in counterclockwise direction and at the same time capacitor also 'pulls' or 'sucks' the current counterclockwise toward itself (thinking in view of Fig 2). These 'push' and 'pull' actions complement each other. Under ideal conditions this 'push' and 'pull' can go on forever. If there were no capacitor then inductor current would always lag the source voltage by 90 degrees assuming no load. In such a case, the source would need to reverse the direction of inductor current on its own and this requires energy which is stored in magnetic field of inductor. It's like stopping a moving car and it will take some energy to stop it due to its momentum before it can go in the opposite direction. In presence of a capacitor this job of reversing the direction of inductor current is performed by the capacitor. In other words, the capacitor acts like a local supply which is only there to supply/absorb reactive power to/from the inductor, and the main supply can only focus on supply current to the resistive load. Or, we can think that complementary action of capacitor and inductor takes away the affect of reactive power from the circuit and for the supply the load overall behaves like a resistive one. Do not forget that we are assuming that the system has been running for some time and it is in steady state. Also note that in actual circuit there is only a single current and no independent currents.

Do you think that I have it right to some degree?

Thank you.

Regards
PG

1: Apparent power, true power, and reactive power:
i: http://www.aspowertechnologies.com/resources/pdf/True vs. Apparent Power.pdf (good introduction)
ii: http://www.bayt.com/en/specialties/...we-change-the-reactive-power-to-active-power/ (good analogy)
iii: http://www.electro-tech-online.com/...wer-and-reactive-power-is-transferred.132176/

2: The following webpages about power factor correction are very useful and were used to write this post:
iv: http://www.blc.lsbu.ac.uk/eservices/Module1/Module 1f.html (good analogy on how inductor current lags voltage)

3: These webpages about power factor correction are somewhat good:
i: http://www.alcorn-energy.com/tech3.htm
ii: http://focuscos.com/power-factor-correction/
iii: http://www.4qdtec.com/water.html (analogy on how current leads voltage in capacitor)
iv: http://en.wikiversity.org/wiki/Power_factor
v: http://www.angelfire.com/jazz/pakspinning/powerf7.jpg (purely capacitive circuit and RC circuit)
vi: http://www.electronics-tutorials.ws/accircuits/ac-capacitance.html (RC circuit)

4: LC tank circuit:
i: http://www4.uwsp.edu/physastr/kmenning/Phys250/Lect26.html (good analogy)
iii: http://www.zap-tek.com/webpage/Elect/lsn_4/011_slow_tank.gif (GIF image)
v: http://www.animations.physics.unsw.edu.au/jw/LCresonance.html
vi: http://www.electronics-tutorials.ws/oscillator/oscillators.html

5: http://www.electro-tech-online.com/...3/?temp_hash=d0d7dae3e29a21fcee2032cfbae5eba1 (this explains how current, voltage and charge varies during charging and discharging of a capacitor)

6: http://www.electro-tech-online.com/...4/?temp_hash=d0d7dae3e29a21fcee2032cfbae5eba1 (this shows the negative and positive power in an RL circuit)

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6. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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PG:

I want you to look at this: http://doctord.dyndns.org/courses/Topics/Electronics/Mark_Sokos/Inductors_and_Capacitors_B.htm

For pf=1 you want the sum of the imaginary parts to be equal to 0, or Xc = Xl

From the article

And apparent power uses the resultant vector.

And from our buddies here:

Under the review:

Now there is an app note that I should try to dig up which shows how to measure the various powers on a fancy oscilloscope.

This discussion, says voltages and currents are sinusoidal. What if they are not? There is no phase angle per see, but there is a pf, real, reactive and apparent power.

Your "tank" argument dies, I think.

So, change your thinking a bit. How would you make a power meter using a microprocessor for non-sinusoidal currents. Let the voltage be periodic.

Last edited: Feb 21, 2015
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7. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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I've purposely hidden using a scope to calculate power of an arbitrary waveform.

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8. ### RatchitWell-Known Member

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To answer your first question. Lower case omega (w) is the angular frequency in radians/sec. One radian = 2 pi hertz. Therefore, the angular frequency of a sinusoid whose frequency is f is w = 2 pi f .

There is no difference between those formulas except you neglected to put the "j" operator in one of them.

No, that formula is not in phasor form.

Reactance is not equal to impedance except in limited circumstances. It is a constituent of impedance.

I would have thought that as long as you have been plugging away at electronics, you would be up to speed on the above questions by now. Did you slack off on your studies?

Ratch

PS Why do you display such an ugly avatar?

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9. ### DerStrom8Super ModeratorMost Helpful Member

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Impedance has two parts: a real part (just a number) and an imaginary part (a number preceded or followed by an i or j). The real part is your regular DC resistance and the imaginary part is your reactance, or AC resistance. Impedance is both parts together, ie. 30+j10 ohms. The 30 is DC resistance, and the 10 is the reactance. + suggests inductive reactance (AC resistance due to an inductor) and - suggests capacitive reactance.

Hope this helps!
Regards,
Matt

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10. ### Tony StewartWell-Known MemberMost Helpful Member

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If you want to consider an easy lookup for tank circuits, with variables like R,L,C,f, Q
For series Q=X/R
For Parallel Q=R/X
http://www.testecvw.com/carl/images/impedancenomograph.pdf

For quick and dirty look up values find the intersection of impedance to find the missing value on any of 3 axes.

11. ### RatchitWell-Known Member

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Yes, separate current and voltage measurements don't mean anything with respect to real power unless the phase between the two is also known.

No, that does not make sense. A PF of one results in the lowest resistive losses.

No, you calculated in the next question that a PF of one would make the circuit look like a 120 ohm resistor.

It looks good to me.

Yes, I do.

Yes, I do.

The inductor currents in any parallel circuit are always 90° out of phase with any resistance in parallel, not just tank circuits. Same for capacitive currents. Current is defined to lag by 90° with respect to resistive current in an inductive branch, and lead by 90° in a capacitive branch. That's what 180° out of phase with each other means. The energy of both L and C components are stored in their magnetic and electric fields respectively. If the stored energy is the same for both L and C, then they are in resonance. This means that when one component discharges all its energy, the other component is ready to accept it all. In a perfect LC circuit with no resistance, that can go on forever, even if there is no voltage across the parallel components. But, of course, that cannot happen in the real world.

Ratch

1: Apparent power, true power, and reactive power:
i: http://www.aspowertechnologies.com/resources/pdf/True vs. Apparent Power.pdf (good introduction)
ii: http://www.bayt.com/en/specialties/...we-change-the-reactive-power-to-active-power/ (good analogy)
iii: http://www.electro-tech-online.com/...wer-and-reactive-power-is-transferred.132176/

2: The following webpages about power factor correction are very useful and were used to write this post:
iv: http://www.blc.lsbu.ac.uk/eservices/Module1/Module 1f.html (good analogy on how inductor current lags voltage)

3: These webpages about power factor correction are somewhat good:
i: http://www.alcorn-energy.com/tech3.htm
ii: http://focuscos.com/power-factor-correction/
iii: http://www.4qdtec.com/water.html (analogy on how current leads voltage in capacitor)
iv: http://en.wikiversity.org/wiki/Power_factor
v: http://www.angelfire.com/jazz/pakspinning/powerf7.jpg (purely capacitive circuit and RC circuit)
vi: http://www.electronics-tutorials.ws/accircuits/ac-capacitance.html (RC circuit)

4: LC tank circuit:
i: http://www4.uwsp.edu/physastr/kmenning/Phys250/Lect26.html (good analogy)
iii: http://www.zap-tek.com/webpage/Elect/lsn_4/011_slow_tank.gif (GIF image)
v: http://www.animations.physics.unsw.edu.au/jw/LCresonance.html
vi: http://www.electronics-tutorials.ws/oscillator/oscillators.html

5: http://www.electro-tech-online.com/...3/?temp_hash=d0d7dae3e29a21fcee2032cfbae5eba1 (this explains how current, voltage and charge varies during charging and discharging of a capacitor)

6: http://www.electro-tech-online.com/...4/?temp_hash=d0d7dae3e29a21fcee2032cfbae5eba1 (this shows the negative and positive power in an RL circuit)[/QUOTE]

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12. ### PG1995Active Member

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Thanks a lot for helping me with the queries, Ratch.

Now I look at it again carefully, I feel that I don't really understand it.

This is the calculation which was made. We have a 60.319Ω inductor and 60Ω resistor in series and it gives total series resistance of 120.319Ω. As was calculated the current supplied by AC source is almost 1A and 0.410A is supplied by the capacitor. This is how I was thinking previously. I was under the impression that inductor is totally 'powered up' by the capacitor and the AC source is only concerned about supplying power to the resistive load. In other words, I was treating the interaction of capacitor and inductor in a way as if they are really absent from the supply's perspective and they both constitute an independent system of their own. But Ohm's law fails, i.e. 1^2*60 (power dropped across load)120*1 (power generated by supply), where "60" is resistance of load, "1" is current supplied by the source and "120" is RMS voltage of the supply. Total power generated by the source is 120W but only 60W is dropped across the resistor which means the remaining 60W is dropped across the inductor. What are those 60W doing in the inductor?

Please also note that I do understand that overall resistance of the system is 120Ω as is calculated here in detail.

Where am I going wrong? Could you please guide me? Thanks

Regards
PG

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Regards
PG

15. ### JimBSuper ModeratorMost Helpful Member

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My poor little brain cannot follow the twists and turns of this thread, I must go back to basics.

A perfect resistor has resistance.

A perfect capacitor has reactance.

A perfect inductor has reactance.

When a component/circuit has resistance and either inductance or capacitance, or both, then that component/circuit has impedance.

Example:
If we have a series circuit, with a 30 Ohm resistor and a capacitor with a reactance of 40 Ohm at the frequency we are interested in, then the impedance will be:

Z = Sqrt(30^2 + 40^2) = 50 Ohms.

If we have 100 volts across this series circuit, the current will be 2 Amps.
The voltage across the resistor will be 60 volts, and across the capcitor 80 volts.

Does this start to clear your confusion?

JimB

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16. ### JimBSuper ModeratorMost Helpful Member

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Continuing the example.

If we re-connect the R and C, so that they are in parallel, and we put this same 100v supply across them,
The current in the resistor will be 100/30 = 3.333 Amps. That current is in phase with the supply voltage.
The current in the capacitor will be 100/40 = 2.5 Amps. That current is leading the supply voltage by 90 degrees.
The total current from the supply will be Sqrt(3.333^2 + 2.5^2) = 4.166 Amps

If we add an inductor in parallel to our circuit, and that iductor has a reactance of 40 Ohms, the current in the inductor will be 100/40 = 2.5 Amps. That current is lagging the supply voltage by 90 degrees.

The total current taken from the supply by the R C and L will be....
This is where phasor diagrams come in useful. But I am not drawing one at this time (03:07 in the morning).
The inductive current and the capacitive current are equal and opposite and they cancel each other out. The supply current is just the resistive current of 3.333 Amps.
There will be a circulating current of 2.5 Amps around the inductor and capacitor, but these reactive currents cancel out and do not affect the supply. Our circuit has a power factor of 1.
The circuit is resonant.

And so to bed.
JimB

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17. ### RatchitWell-Known Member

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You had 1.41 amps existing in both the inductor and resistor which was supplied by the power supply. Adding a capacitor did not change any components in the L-R branch, and the voltage was not changed across this branch. Therefore, the current existing in the L-R branch after the capacitor was added remained at 1.41 amps. Afterwards, however, the energy in the L and C swap with each other twice every cycle, and the power supply only needs to supply 1 amp. In other words, the PS only has to supply the losses from the resistor.

Ratch

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18. ### PG1995Active Member

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Thank you, JimB, Ratch.

Mathematically, I understand it but conceptually I'm somewhat confused.

The 120V supply is supplying 1A which means it's outputting 120W. The power dissipated by resistive load is also 120W because the current passing thru it is 1.41A.

The current supplied by power supply is 1A but the current passing thru load is 1.41A. Where does extra 0.41A come from? I believe that it comes as a result of power/energy swap between L and C because this power/energy has to travel thru the load and power is product of voltage and current. The voltage being dropped across load is 84.6V. I'm having difficulty visualizing how all this is taking place. Could you please guide me? I have also attached Multisim13 circuit. Thanks.

Regards
PG

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19. ### nsaspookWell-Known Member

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You have to see the entire circuit as a system. Looking at one component or value works in a static world but as soon as you start to look at dynamic or changing energy flow it doesn't as all parts depend on each other. The problem IMO is how electricity is taught in the beginning that tends to build a narrow mental image of circuits. You need to look at energy flows in circuits not just as voltages and currents but as fields in motion around the components in your mind if you want to visualize circuits.

This won't answer your problem of visualizing "how all this is taking place" quickly but it might in the long run.
http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px263/lectures/sefton.pdf

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20. ### RatchitWell-Known Member

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One amp of current comes from the power supply, and 0.41 amps comes from one of the plates of the energized capacitor. Together they supply 1.4 amps to the L-R branch. The opposite plate of the capacitor receives 0.41 amps with the remaining 1.0 amp returning to the power supply.

Ratch

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