resistance of relay?

An Ohm meter. 

Also it isn't just the relay coil load, it's the load up to 40 amps it's switching. At that load average battery fully charged to flat about 90 minutes. I remember some years ago installing a system in a VW Sirocco, 3 x Soundstream or Nak amps, 2 x subs and satellites, 19 minutes with engine off and I needed to jump start it.

I just read your previous post again. A starter kill WON'T give you a flat battery if wired correctly. You can read my article in the "Car Security Hot Topics" section on this site or look at this diagram. 6C4_starter_cut_dei.bmp

tinecidy627 wrote: BTW, I use the relay as part of an all-time switch, which I want to be working all the time regardless of engine on/off, so I guess I have to sacrifice the battery?

No. That's called connecting the load full time to the battery, and disconnecting the battery when you don't use it. Otherwise you will need another battery to start the vehicle, and you will probably be buying new batteries every month or few. Besides which, the relay de-energises with the flat battery so the circuit is pointless anyhow (unless it retriggers). Such circuits should NEVER require a relay to be "normally" energised - ie, on when left idle and not running. But as Howard wrote, yours doesn't anyhow, so why the concern? If it does, just connect it and see since you are willing to sacrifice the battery...

Thank all of you for your time, and the patience for my dumbness. I figured out my goal can be achieved by a diode, so no need for a "constant" relay. However, since I learned some electric stuff in college, I want to discuss it with you guys and make sure I have the right picture: For this relay, the coil resistance is about 80-90 Ohm. So when the relay is on, the voltage on it is 12 Volt, current will be ~0.14 Amp, so the power would be ~2 Watt. The 30/40A stuff is the maximum allowed current on the load side, in which 87 and 30 connect, right? If this is all true, then let's consider the case howie mentioned, with a lot of high-power equipment on and the battery exhausted at ~90 min. In this case, current 30A, voltage 12V, power 360W. So the load power is about 200 fold higher than the relay itself. If I use this relay to drive something low current, say a LED, say it ~1W. So the battery won't be depleted for couple of months under such a low power, right? I think that's why a starter kill relay won't flatten the battery so much, since it only consumes about ~2 W on the relay it self (85-86), and it cuts the starter circuit so there is no power on the load side (87-30). So if all my understandings are correct, then I'm happy with all these. I totally agree that a relay is designed to use a low current signal to control/switch a high current circuit. Actually I was trying to install a indicator LED, which turns on when the alarm system is on. So a single diode should work, no need to use a relay since the circuit is low enough. Please point out if anything is illogical here so that I could learn more about the electric stuff. Thank you for reading this and helping me. My gratefulness~

Your first 2 paras are correct. If you were to drive a LED, you would not use a relay. Wire the LED direct. Why use a 0.14A load to drive a 20mA (0.02A) load? You shouldn't even need a diode for the LED since the LED itself is a diode. (That can depend on other circuits, but rarely.) You just need to connect the "on" signal to the LED and a resistor. Whether the other side of the LED & its resistor go to +12V or GND depends on whether the "on" signal is GND or +12V (or +5V etc) respectively.

A starter cut relay DOESN'T draw any power if wired correctly. This is the THIRDtime I've posted this drawing since Saturday:- starter_cut.bmp And if you know so much what made you think you'd need a relay to power an LED? By the way your calculations don't take into account diode induced (minor) voltage drops and circuit inefficiency adding to more draw. It's STILL bad practice to have a relay constantly on.

In practice about 800ohms for a 2.5 volt LED