How do I monitor voltage on a circuit even when switched off?

Question

Here's the situation: I have a baby. Soon, I'll have two. So I don't really like spending half an hour trying to get him to sleep and then somebody rings the doorbell, waking him up.

I've already installed a switch on the doorbell circuit which disables it. I'd still like to know if somebody is trying to ring it though. For simplicity sake, lets say it's a simple light bulb that I'm always looking at, so it doesn't need to stay on. (Just a momentary flash will do).

The diagram on the right is the simple doorbell circuit. On the left, I have my indicator circuit with the 12v relay I assume I'm going to be needing. I believe the relay coil needs to go in parallel with my switch, but I'm not sure if this will work. Do I need a resister between the two?

What am I missing to complete this solution?

****EDIT:** Maybe I did myself a disservice by oversimplifying. The indicator isn't a lamp, it's actually my alarm panel which can monitor for either an open or closed circuit (but not a good idea to put the actual bell on that same circuit). This will allow me to be notified of a doorbell signal by email/text message, or whatever method I choose, even if I'm away from home.

Additionally, I have a dog that barks regardless of whether the doorbell is disabled or not. This is probably outside the scope of the question, however.

Answer 1

The simplest solution is to teach babies to just sleep in a noisy environment. That might be easier said than done though.

As an alternative you could make this which saves some components.

^{simulate this circuit – Schematic created using CircuitLab}

Note that the indicator lamp should not use too much power as it adds to the power needed by the bell. I would suggest using one or more 12 V LED lights, the ones that replace 12 V halogen lamps for example. These are low power and very bright.