Calculating Apparent, Reactive, and Active Power for Receiver with I=5A, Pn=1000W, Un=230V

Question

Hello, As I do not have any electrician I know to help me, I am asking you, dear colleagues, to check the task. A current I = 5 A flows through the receiver with the rated power Pn = 1000 W and the rated voltage Un = 230 V. Calculate the apparent and reactive power dissipated in this receiver. I...

kozi966 · Accepted Answer

That's right. This is not correct. The formula for calculating the individual powers is a quadratic relationship. var - reactive volt-amp (lowercase).

telecaster1951 · Accepted Answer

There is only one way to solve this task. We consider the rated power as active power. The product of the effective values is the apparent power. Find the formula for reactive power yourself.

DiZMar · Answer

The rated power has nothing to do with apparent, reactive or active power under specific operating conditions. Read the rest on the web.

Hold_on1 · Answer

Edited the topic, maybe it's okay now, will someone have a look?

DiZMar - on the Internet I found information that the rated power is active power (i.e. the one that is converted by the device into a different type of energy, e.g. kinetic)

DiZMar · Answer

You have not read it carefully. Rated power is the power with which the device can operate under rated conditions (typical for a given device). The power at a given moment depends on the conditions in which it actually works, which are not necessarily rated. In the case of electrical devices, the main condition is the supply voltage, but also other conditions, such as ambient temperature, etc. Hence, the rated power does not matter for the calculations in the given conditions. Added after 9 [minutes]: In your case, as you can see, the real power is 1150W, which is 15% more than the rated power, which may damage the device. On the other hand, with a rated voltage of 230V, a current of 4.35A should flow, not 5A. The task is contradictory in itself.

jack63 · Answer

I do not know what the original content of the task looks like, so it is difficult to judge honestly, but I suspect it lacks information about the shape of the Un voltage. The fact that it is 230V (it is not known whether DC or AC) does not determine its shape at all.
This task only makes sense for periodic voltage and current. However, it is possible to calculate only assuming a sinusoidal (!) Voltage and current waveform.
Treat powers like vectors, though they're not, and add them geometrically. You have a power triangle. Read the rest.

telecaster1951 · Answer

Gentlemen, please ... What mushroom such texts for? Maybe you still tell the guy to calculate the integral for the effective value? Great, brilliant, we know. This is one common school assignment. Cool, cool. I understand that when I give students a problem where 10 amps flow through a 2 ohm resistor, I should be accused of trying to arson. A school assignment does not have to be correctEven it should be abstract. This is the entire content of the quest?

Hold_on1 · Answer

DiZMar - I know what the names of the individual mathematical operations are, I just "skimmed", I meant the difference VA - W. And instead of answering a short question and specifying the unit, you smeared me a few lines of criticism ...

This is the whole content of the task, I added a screen of the task to the first post ^^

PS This is not homework, but I learn electrics "on my own" from the Internet because I consider it an interesting and PRACTICAL direction (my friend has the engine controller in the car, he undressed, he soldered something and works - if he sent PLN 500 to the site, not his, and the driver used is at least PLN 350 and it is not sure how much it will work.

.

jack63 · Answer

After all, I wrote to the author of the topic about the power triangle. I don't think he gets what all the helpers are about. And for what mushroom do you write about integration? I didn't write a word about him ... The task is sloppy and must be guesswork. But what for? Is adding AC to 230V or climbs about the home network / LV a sin?

telecaster1951 · Answer

And is it the author's fault that such a task is? As someone writes about 230V in a school task, it is a network by default. Don't look for a hole all over.

Hold_on1 · Answer

As indicated by telecaster1951, I tried to solve the task again, this time using the above-mentioned power triangle. The result was really big, 567.89 var, okay?

telecaster1951 · Answer

Since when is reactive power in VA?

Hold_on1 · Answer

active power in W. apparent power in VA reactive power in var already corrected, besides solved well?

jack63 · Answer

Yes. Now imagine what is the substitute diagram of the receiver. Such an exercise for understanding.

Hold_on1 · Answer

Under the term substitute diagram is an ordinary drawing? Is it something like that?

jack63 · Answer

Not really. You have a contradiction in the descriptions of the source. On resistance gives off only active power, so .... where is the calculated reactive ???

Hold_on1 · Answer

So I understand that the calculations are good and the diagram is wrong.
What contradiction do you mean?
"Only active power is emitted on the resistance, so ... where is the calculated reactive power ???" - how to draw this reactive power, I don't understand anything, someone will explain to me?

jack63 · Answer

Yes. You have drawn a DC voltage source (battery symbol) with +, - and next to it you have written: ~ 230, suggesting an alternating voltage. One contradicts the other. The power cannot be drawn on the schematic diagram! Items are placed on it. who can give or take this power. An ideal resistor, as it is in the diagram, takes active power, that is, the product of the effective values of current and voltage, when their waveforms are in phase! Read what causes reactive power and what elements generate it.

Hold_on1 · Answer

1. What is an ideal source of energy, that is, there is no own resistance (e.g. in a battery), but what is an ideal resistor? I read somewhere that the resistors are made with a certain accuracy of 1-5% of their value, i.e. a 1k ohm resistor can actually be ABOUT 1000 ohms +/- 1-5%

kozi966 · Answer

I see an indefinite element in the diagram. It may well be Z (impedance). The author made the mistake of drawing a source of direct voltage, not alternating voltage. Hence the interpretation that it is resistance. It was probably about instantaneous values for which the sign is the same - specifically the product of these values. Here is the explanatory topic (picture): One that has no inductance and no capacitance. An actual resistor is typically a wound element (like a coil) that has inductance and (interturn) capacitance. Not ohms, just talk. Same for volts, amps, and watts. Coil / capacitor. If you mark the current element with equivalent impedance, it will not be missing.

Hold_on1 · Answer

Thanks for the answer, now is a good drawing?

jack63 · Answer

Unfortunately not. First of all, we mark the inductance, that is the coil around the world with the letter "L", and the capacitor "C".
Secondly, an ideal inductive element only consumes reactive power, i.e. it basically shifts the current relative to the voltage in phase by a quarter of a period.
There must be two basic elements connected to the voltage source.

kozi966 · Answer

Better, but it's not C1 a L1.
For this you need to know data such as the frequency of the supply voltage expressed in hertz, the inductance of the element expressed in henry and possibly the resistance of the element (if it is not ideal) to calculate the impedance and then the powers.

Hold_on1 · Answer

I also tried to calculate the impedance (because I understand that resistance = direct current, impedance = alternating), assuming that we have an ideal voltage source and the cables connecting the source to the receivers do not constitute any resistance.

Unless the coil impedance is counted from the power triangle in the same way as the reactive power counted, then the coil resists 26.32 ?

kozi966 · Answer

Not completely. Impedance is a substitute form, in a specific case the impedance may equal the resistance. When the receiver is purely resistive then Z = R. Unfortunately, the task on the page is wrong. Due to the fact that the reactance cannot be subtracted from the resistance, and additionally, when dividing the effective values (without knowing the angle), i.e. for the apparent power data, we can only calculate the impedance modulus. Knowing additionally the active power, we are able to know the phase shift angle. Therefore, we are able to count all the necessary data. According to my calculations, this coil has a resistance of XL = ~ 22.72 ohms. The resistance of this circuit is R = 40 ohms. The phase shift angle is approximately 30 degrees.

Hold_on1 · Answer

Listen, I have an urgent question, deviating a bit from the topic, I connected the red LED to the 9V batteries through a 1k ohm resistor and the second one burned out, what is going on? from the right of the ohm there is a current of 9mA and they are on fire, why? the connection diagram below, why are they both lit? PS I know that the longer leg of the diode is a plus

kozi966 · Answer

First of all, show us this resistor.
Second, give us the parameters of the diode you bought.

Hold_on1 wrote:
why are they both?

Because you exceeded their maximum operating current, is it logical?

By the way, it doesn't matter where you stick the resistor (behind or in front of the diode). The current in this circuit is exactly the same at any point.

Hold_on1 · Answer

I bought the whole set for the course Forbot - basics of electronics part 1, photo of the set used below. Burned LEDs on the left

Calculating Apparent, Reactive, and Active Power for Receiver with I=5A, Pn=1000W, Un=230V

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