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Wled mit Alexa verbinden?

BySonoffemre

Hallo, habe kürzlich mit Wled geholt und kann es nun nicht mit Alexa verbinden hat da jemand Erfahrung ? Mein Controller: Gledopto Wled esp32 with mic Version : v.0.13.4 Toki+Sr (WLEDSR_0.13.4 soundReactive_esp32dev.bin) build 2309021 Mein Alexa: läuft über Sonos soundbar in den Einstellungen vom Wled Controller steht aktuell Alexa Voice Assistant This firmware build does…

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guenterhalt
guenterhalt
2 years ago

sometimes it is very good to redraw such a circuit.

  .--- R1-- R2-- R3---.
Quelle                |
  '-------------------'

The current must therefore be through all 3 resistors, abbreviations are not available.

From the point of view of the voltage source there is only one resistance and it must have the value of the sum of all three. o.k?

You know the formula for the relationship between voltage, current and resistance.

U = I * R . You can change it (as in maths).

I = U / R

Set the values and reckon. Are you coming to 0.3A?

What a potential is, you know?
This is a voltage related to a striking point (measured or calculated from this point).

In the circuit diagram this is the point with the sign for earth (in the drawing point D)

Now calculate, the current you have calculated how large the voltage from D is at point A. That’s just 9V?

When 0.3 A flows, the voltage across R1 will be 0.3*10 V. Also via R2 and R3 are the 3V .

You can start from above or below.
Between D and B, the voltage 3V must be less than the voltage at the battery.

There’s potential as a 6V. The rest you can now.

2
RareDevil
RareDevil
2 years ago

Note the basics of the series connection (Calculate voltage and current with respect to resistors, equivalent resistance)

Applying Ohms Law (U=R*I), reconfiguring it…

The rest results from this. Shields, if necessary, what exactly is unclear to you. “I don’t know” doesn’t count, there are specialist books and the Internet that teach these basics. 😉

1
ULUFL
Author
ULUFL
2 years ago
Reply to  RareDevil

Thank you, then if I want to calculate the voltage at the individual intervals for R just use 10 ohm? How do I calculate the potentials that are present in the U=R*I formula

0
RareDevil
RareDevil
2 years ago
Reply to  ULUFL

For the voltage at a single resistor, you must count on the individual value of the resistor and the current through the circuit. For example, the current, the total voltage and the substitute resistor, or with the partial voltage at a resistor and the nominal value of the resistor.

The voltages of the resistors can also be put into proportion, as these are proportional to the values. for example:

R1=1 Ohm; R2=5 Ohm; R3=10 Ohm; at 32V.

In this case, for example, the voltage at R1 1/16*32V=2V would then be, since all resistors together yield 16 ohms, and R1 has only 1 ohm. For R2 it would be 5/16*32V=10V and 10/16*32V=20V

If the potentials are now considered in the same circuit as in the task position, point A 32V (Since the source voltage was 32V). 2V drop across R1, so at point B 2V is less, so 30V… About R2 fall off again 10V, with C remaining only 30V-10V=20V. Over R3 20V decreases, so D 0V is also the reference potential… It’s all right.

You only have to apply this to your task.

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ULUFL
Author
ULUFL
2 years ago

Thank you.

0
RareDevil
RareDevil
2 years ago

That’s right, with my example. If you already have the current, you can also calculate the partial voltage dropping thereon via the individual resistors. There are often several ways to get to the destination. Which one chooses is possibly a question of preference, or of the given values, which is just simpler to go on…

1
ULUFL
Author
ULUFL
2 years ago

If you can also use the U=R*I formula for U at the individual residues, since I is constant and then for R simply the value of the residue you are using is I=U/R=32/16=2 and then for U at R2 count U2=I*R2 =2*5ohm =10V

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ULUFL
Author
ULUFL
2 years ago

Thank you

0
heilaw
heilaw
2 years ago

You can’t calculate anything for resurrections.

0