# pls help me with this simple computation

I HAVE THE SETUP BELOW, AND I WANT TO THEORETICALLY COMPUTE (W/O THE USE OF A MULTIMETER) THE
TOTAL CURRENT DRAW AND WATTAGE (POWER CONSUMPTION) OF THE SYSTEM AS A WHOLE …
I’M REALLY CONFUSED WITH THE DATASHEET OF THE RELAY
so i can’t really theoretically measure what i’m looking for …

i want to know the current draw and wattage whenever the switch is triggered on (the relay being energized)
tnx for the help

Well the coil side's easy: 12V @ 75mA is (12 x 0.075) Watts.

The other side is more difficult because as you have it it's going to blow those lamps through the roof: you really going to power 28V lamps on the mains?

NO SIR, ACTUALLY THE BOX WHERE THE 28V 7W RATING IS PLACED IS CALLED THE DRIVER, THE DRIVERS SUPPLIED WITH
A 220VAC INPUT AND THEN CONVERTS IT TO DC POWER TO SUPPLY THE LAMPS

So don't show the 28V lamps across the mains in the picture then.

So here's the numbers:

On the lamps: Power = 2x 7W = 14W.

And assuming the driver box thing is 100% efficient (it won't be), and also then consuming 14W, you have mains current of 14W/220V Amps.

oh im very sorry sir to the caps thing , i dont mean anything bad . sorry sir

so sir u are saying that all in all, the total wattage would be :

from the relay = 12v x 75mA
from the lamps = 14W

and the total current draw would be:
from the relay = 75mA
from the lamps = 14W/220 <----- may i ask sir why not use the 28v instead of the 220 to compute the current here ?

and also sir, do we need not to consider the contact ratings if we are getting the total current draw and power c?

from the lamps = 14W/220 <----- may i ask sir why not use the 28v instead of the 220 to compute the current here ?

On the DC side of the converter box thing, the curent is 14W/28VDC = DC amps, and 14/220 is the AC mains current.

ok sir so ur saying that if i am computing for the total and general power consumption of the system i must use the ac mains one sir and not the dc ? is that right sir?

and also sir, do we need to consider also the Contact Rating of the relay if we are computing the power and current loading ?

urbanmiles:
ok sir so ur saying that if i am computing for the total and general power consumption of the system i must use the ac mains one sir and not the dc ? is that right sir?

In theory the power is the same on both sides; the current differs because the voltage isn't the same.

Power = Volts x Current, and Current = Power / Volts

So on the DC side:

We know power is 14, so the current is 14/28

On the AC side:

Assuming no energy is lost in the conversion, the power is still 14, and the current is 14/220

But it's the AC you're switching by the relay (I think, but your ndrawing shows 220V on the 28V lamps so you really ought to make a better schematic) so it's the AC voltage, current, power you need be concerned about for the relay's specification.

ok sir everything is clear to me now except for the “losses thing” when it comes to the AC side…

tnx sir for the advices !

Well all I mean by losses is that the 220VAC to 28VDC conversion will not be 100% efficient. So although the DC side needs 14W, you might actually consume 20W (or more, I actually have no idea) on the AC side because of heat lost in the conversion.

Maybe on the AC side you should double up and go with say 30W to be safe.

Make it clear though: are you switching the 220VAC or the 28VDC with the relay, ie is the relay before or after your converter box? If you switch the AC you will need to be very careful and also make sure you are not doing anything illegal.

edit… and if you switch the DC, then your relay doesn’t handle 28V, it says only 24.

actually i'm using the ac line to connect with to the relay , not the 28vdc .. and it works just fine sir .
since i'm getting only 14w/220v= 63.6mA (from led lamps) to flow through the relay contacts which are 5Amax,
but i'm not actually sure if i can use that 5Amax contact knowing that im using only 12V 1A adapter

The 12V is across the relay coil: the spec says 75mA there I think, so your 1A supply is ok. The fact that the supply CAN give 1A doesn't mean it WILL. Current is not forced by the supply, it's drawn by the device, so the 75mA is well within the 1A that the supply can give.

But you're confused: the 64mA is on the "other side" of the relay, the output, and that's got nothing to do with the 75 on the input (coil) side.

yes sir hehe i think i really am CONFUSED ...
actually that's what i am asking to a different forum,
if the current running through the coil (75mA) has something to do with the relay's contact power (5Amax)
meaning if the relays contact power also changes if the current supplied to the coil is diminished to a value lower than the rated voltage for the coil to operate ?

well since we are here sir, may i ask sir the meaning of the contact rating
5A 220VAC
5A 24VDC

is this right as i can get a maximum contact 5A to loads having 220V Ac supply and
5Amax contact to loads having 24V DC supply ???

well since we are here sir, may i ask sir the meaning of the contact rating
5A 220VAC
5A 24VDC

is this right as i can get a maximum contact 5A to loads having 220V Ac supply and
5Amax contact to loads having 24V DC supply ???

Yes, that's on the output side.

if the current running through the coil (75mA) has something to do with the relay's contact power (5Amax)

The coil and the contacts are independent: that's the whole idea of a relay. A relatively small voltage at a low current energises the coil.... the coil is an electromagnet.... the magnet pulls the contacts closed and they in turn are connected to the load.

ok sir got it , so if i'm using the relay in the diagram which is a 3PDT relay,

does this mean that i can use 5A for 220VAC and 28VDC for each of the 3 pole
OR
i can only use a total of 5A for the 3 poles together ?

and also sir, i am planning on connecting a siren supplied by a 12V dc adapter , to the 2nd pole of the relay (the 1st pole being the LED LAMPS),
is it ok sir if i connect an AC supply (LAMP) on the 1st pole and a DC supply on the 2nd pole (siren)
on the same relay ??

I would say no.

Look att the isolation voltage between contacts, is it 500 volts=no, is it 4000 volts = OK.