indicator light for switch

Question

I have a led in series with my toggle switch to be an indicator to power a 30/40amp relay. The power source is a 12v battery. When I close the switch the led comes on, but the relay stays open?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Given that it looks like an car project I'd guess that by LED you means an indicator lamp with built-in series resistor.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (Left) What you've done. (Right) what you need to do.

If there is a series resistor in the LED it will limit the current through the circuit to about 10 or 12 mA. This won't be enough to energise your relay and if you measure the voltage across the relay you probably only have a volt or so.

Instead, wire the indicator in parallel with the relay. Relays give an inductive reverse-voltage "kick" when switched off and this is enough to blow the LED. Unless you know that either the relay or the indicator has protection built in then add D3 to protect the indicator LED.

Finally, note the use of the ground or chassis symbols in my schematic and omitted from yours. You have no return path for the current so, as drawn, neither the relay or LED would work. The schematics above have a ground return and so complete the loop to allow current to flow.

Answer 2

You want to have your led in parallel with the coil of your relay and you want a resistor in series with your led to limit the current through it. You should always limit the current through an led with a resistor , else you will just end up blowing it up.

^{simulate this circuit – Schematic created using CircuitLab}