Midwest Puddle Pirate
Join Date: Feb 2005
Location: Gardner, KS
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I would guess that with the buzzer and bulb in parallel, what happens is that the bulb acts like a very low resistance (near zero ohms) so effectively it absorbs all the power and the buzzer gets none. If the bulb has a resistance of 1/2 ohm, and the buzzer is a maybe 600 ohms...the voltage divides proportionately and the buzzer won't sound.
The simplest most reliable solution will cost you about five bucks more. You disconnect the bulb, and replace it with an inexpensive 12-volt "dual pole" (DPDT or DPST, either will do) relay Wire up the relay coil to where the bulb was. The connect a 12-volt power lead to one side of both the relay's "NO" (normally open) contact pairs. Connect the blub to one of the paris, the buzzer to the other, and then connect them to ground. (If you buy a relay designed for car fog lights, there's usually a diagram included with it.)
The relay "goes on" instead of the originallight,and it isolates the light and buzzer so each one acts without interference from the other. That also will allow you to use a much more powerful buzzer.
While this explantion sounds good, it works a little differently. The current divides proportionally, not the voltage. With a parallel circuit, each leg of the circuit gets a full 12 volts and a full ground. Provided of course that the power supply and the ground are capable. The current on the circuit is increased by whatever the new item draws. In a series circuit, the total current flow is limited to whatever the lowest current flow item is, in this case the light bulb. While the light bulb's resistance is very low when read with an ohm meter, because of the inductive resistance of the filament when operating, the actual current flow is much lower than if it were just a resistor. You can prove this to yourself merely by applying ohms law to the 1/2 ohm light bulb. 12/.5 = 24amps in other words we would have a 288 watt warning light. In reality he likely has about a 3 watt bulb drawing .25 amps. A series circuit will only flow as much current as the LOWEST current user, in this case .25 amps. A quarter of an amp is not going to operate both the bulb and the buzzer, so the light comes on and the .25 amps passes through the buzzer to ground without sounding the buzzer.
I think that a relay would add unneeded complexity to it, unless the buzzer draws more amps than the switch will allow. If it does, he needs a new buzzer.
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John USYacht 27 "Cora Lee"