Choosing correct relay for LED power supply rating

Question

I have a 500W 24V power supply for running 24V LED strips. The power supply says it's rated for ~21 amps (500W / 24V). Assume input is US 120V household power.

I want to control the 120V supply with a relay driven by 5V Arduino logic signal.

I'll be driving 4x 24V 96W LED strips.

From A = W / V:

A = (4 * 96W) / 24V = 16

Does this mean I need a relay rated for 16 Amps or 21 amps listed on the power supply specs?

OR

Do I work back to Watts from the relay's specs?

Example relay is rated at 5A / 220V AC:

W = 5A * 220V = 1100

Though I've read choosing a 16A or above relay is a good idea since it shouldn't be the first thing that fails in a power surge.

More info: Here's the power supply I have: https://www.amazon.com/gp/product/B01IU8QBCO/ref=oh_aui_detailpage_o07_s00?ie=UTF8&psc=1

And will be purchasing one of these for 12V landscape lighting: https://www.amazon.com/gp/product/B06XR3ZLSG/ref=ox_sc_saved_title_8?smid=A3S4SVMWTDK1H2&psc=1

Looking at a relay like this to be on the safe side: https://www.amazon.com/AC100V-250V-2-Channel-High-low-Trigger-Arduino/dp/B077W1NVLM/ref=pd_ybh_a_15?_encoding=UTF8&psc=1&refRID=RAE8FR66XX0GN9YQANXR

EDIT: Suggested alternative: Though I'm confident wiring mains I like to avoid it. I'd suggest anyone not confident to go buy one of these (I'll be getting one for my project): https://www.adafruit.com/product/2935

https://www.amazon.com/Iot-Relay-Enclosed-High-Power-Raspberry/dp/B00WV7GMA2/ref=sr_1_1?s=wireless&ie=UTF8&qid=1542226953&sr=1-1&keywords=Controllable+Four+Outlet+Power+Relay#customerReviews

Answer 1

It depends on which side you are switching.

If you're switching the secondary then you got it right.

A = (4 * 96W) / 24V = 16

Choosing a 16 A relay here is fine but it might be a good idea to choose a higher rating to be on the safe side.

If you are switching the mains voltage, I suggest the following:

Looking at the information given we can see that the power supply has an efficiency of 82% and a maximum power output of 500W. With this information we can calculate the maximum power input by taking:

500 W / 0.82 = ~610 W

With this information we can now calculate the current on the primary side:

610 W / 120 VAC = ~5,1 A

So a relay rated for atleast 5 A would be enough in your case since you won't be using the full 500W.

Be careful if you choose to control the primary side!