Analog dinosaur needs simple buzzer when timed power source ceases
Find a low-power 12 V buzzer and hook it up to a 12 V source and measure the current required using your multimeter. Let's say it's 12 mA.
From Ohm's law work out the equivalent resistance of the buzzer. \$ R = \frac {V}{I} \$. For 12 V and 12 mA that would be 1 kΩ.
Let's get a rough idea of the capacitance required from the RC time constant formula \$ \tau = RC \$. (This tells us the time it takes to charge or discharge a capacitor by 63%.) Setting \$ \tau \$ = 20 s we get \$ C = \frac {20}{1k} = 20 \ \mathrm {mF} \$ (20,000 µF)








simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Suggested circuit.

How it works:


When the relay is energised C1 charges up to 12 V. R1 limits the charge current to 0.12 A to prevent a sudden load on the supply on power-up. It will dissipate \$ \frac {V^2}{R} = \frac {12^2}{100} = 1.44 \ \mathrm W \$ for a moment on power-up but a 1/2 W resistor should be OK.
When the relay switches off the capacitor will discharge through the buzzer.


See Must Calculate to assist with RC calculations.