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resonance

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The relay was designed to have its current limited by the inductance, not its resistance. When the relay is open, its inductance is small, so it takes more power, which is normal, but only lasts for a short time, so doesn't overheat the relay.


Could you please elaborate a bit?

Thanks a lot.

Have a look at the specification for this relay:-
https://docs-emea.rs-online.com/webdocs/1582/0900766b81582926.pdf

It says:-
Coil Power: 2VA
Coil Resistance: 8200 Ω
Coil Voltage: 230V ac

2 VA at 230 V is about 8.7 mA.
230 V across 8200 Ω is about 28 mA

The difference is caused by the inductance of the coil.

The coil's inductance will be much less when the coil is not energised, because the magnetic circuit is not fully closed.

On a contactor, the specification is clearer.

https://docs-emea.rs-online.com/webdocs/13df/0900766b813df0b7.pdf

It says:-
Inrush power in VA 70 VA at 20 °C (cos ϕ 0.75) 50 Hz
Hold-in power consumption in VA 7 VA at 20 °C (cos ϕ 0.3) 50 Hz

So the current consumption is 10 times as much at inrush. When the contactor is energised, the inductance is much more, which is why the current is much less and the power factors is so low.
 
Thank you for the clarification.

Assuming that your set up was quite similar to the one shown. The power coming out of 24 V transformer is AC and you had connected contact suppressor across R and W which, as you stated, was a mistake. It's the relay's coil which needed to dissipate its stored magnetic energy once the relay is turned off so connecting the contact suppressor was a better choice.

Anyway, it is shown in your linked datasheet for contact suppressor from post #57, http://www.farnell.com/datasheets/1719039.pdf , that the circuit is essentially a resistor and capacitor in series. A capacitor permits AC to pass through it. As the power from transformer was AC so using a contact suppressor across R and W doesn't make any sense at all. I'm just wondering why an experienced and wise person like you would do that! Or, it might be just that I'm too silly to see it differently! :)


Notes to self:
Difference between relay and contactor:

"A contactor is an electrically-controlled switch used for switching an electrical power circuit.[1] A contactor is typically controlled by a circuit which has a much lower power level than the switched circuit, such as a 24-volt coil electromagnet controlling a 230-volt motor switch.

Unlike general-purpose relays, contactors are designed to be directly connected to high-current load devices. Relays tend to be of lower capacity and are usually designed for both normally closed and normally open applications. Devices switching more than 15 amperes or in circuits rated more than a few kilowatts are usually called contactors. Apart from optional auxiliary low-current contacts, contactors are almost exclusively fitted with normally open ("form A") contacts. Unlike relays, contactors are designed with features to control and suppress the arc produced when interrupting heavy motor currents."
Source: http://en.wikipedia.org/wiki/Contactor

http://www.springercontrols.com/news/contactors-vs-relays/
**broken link removed**
Youtube Video

Thermostat wiring:
What does "R", "W" and "C" stand for?
R is power, W is heat, C is common.
http://www.wifithermostatreviews.com/the-thermostat-wiring-color-code-guide/
http://www.thermostastic.com/c-wire-issues-hacking-your-way-to-become-a-thermostat-wiring-pro/
http://sensi.emerson.com/en-us/support/adding-a-24-vac-external-transformer

Contact or arc suppressing:
**broken link removed**
 
Thank you for the clarification.

Assuming that your set up was quite similar to the one shown. The power coming out of 24 V transformer is AC and you had connected contact suppressor across R and W which, as you stated, was a mistake. It's the relay's coil which needed to dissipate its stored magnetic energy once the relay is turned off so connecting the contact suppressor was a better choice.

Anyway, it is shown in your linked datasheet for contact suppressor from post #57, https://www.farnell.com/datasheets/1719039.pdf , that the circuit is essentially a resistor and capacitor in series. A capacitor permits AC to pass through it. As the power from transformer was AC so using a contact suppressor across R and W doesn't make any sense at all. I'm just wondering why an experienced and wise person like you would do that! Or, it might be just that I'm too silly to see it differently! :)

Well a contact suppressor is normally connected across the contacts that have to be suppressed. On a contact breaker vehicle ignition circuit, the condenser (as the capacitor for that is called) is directly across the contacts, so it's not that stupid a place.

Having a contact suppressor directly in parallel with the contacts means that the inductance of the wires is also suppressed. In the normal opening and closing of the contact, an inductive spike can be very large, and whether or not the supply voltage is in series doesn't make any real difference.

Having a capacitor across the contacts will obviously result in ac leakage, but it has to be quite special circumstances where that leakage will only cause a problem when the relay is in one state and not when it is in the other.

Admittedly, I have heard of LED lights glowing when off due to capacitive leakage on the switch wires. Also, before I made this mistake with the suppressor, I had seen a relay with a 110 V coil stay operated due capacitive leakage that left 5 V across the coil, but I didn't associate that at the time.
 
Thank you!

I think having a capacitor with large reactance would help. The reactance is given as:
reactance111-jpg.118158


In the datasheet for a contact suppressor, https://www.farnell.com/datasheets/1719039.pdf , it can be seen that capacitance is rather low which gives large reactance or informally speaking capacitive resistance to AC. Having a large reactance would mean small AC leakage.

contact_suppressor_1a-jpg.118159
 

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