what will happen when i do this?

Question

heres a concept drawing
http://img389.imageshack.us/my.php?image=save0240ty1.jpg

and what will happen when i disconnect the circuit from the switch in the drawing
(it's in front of the battery)

i also want to know how much volts will go into the bulb when i shut off the battery

Answer 1

This is a relatively simple circuit. You might as well draw all of the capacitors as just one, because they are all hooked up the same. When you disconnect the battery, you will have a capacitor discharge, which means you will have a logarithmic decline in voltage and current going through the light bulb. This is a first order decline, meaning the rate of loss is proportional to what remains. The bulb will grow dimmer and dimmer as the capacitors discharge, until they are fully discharged, which will depend on the resistance of the bulb filament and the total capacitance of the capacitors. In the field of Electrical Engineering, this is referred to as an RC circuit. LC circuits are more fun, but I will leave that for another time.

Answer 2

Now you're cookin' !

The light will remain lit for the length of time it takes the capacitors to discharge - a functioning RC (resistance-capacitance) circuit.

Be SURE you insert the capacitors with the polarity correct (+ to + and - to -)
If the capacitors are connected in reverse, they can overheat very quickly and BOOM!


Added in response to your additional question.

The initial voltage discharge from the capacitors will be the same as the battery that charged them - the voltage will diminish as the charge is drawn from the capacitors - similar to your battery, but much quicker. The length of time your light remains on depends on the value of the capacitors - the higher the uF, the longer the light.

Answer 3

Hi. The bulb will dim until the capacitors discharge then go out when the voltage no longer can push enough current through to keep the bulb lit.