All componets are 12VDC and an added circuit that turns on when the other turns off powers a buzzer of 10mA which I don't know the resistor for either so help would be appreciated.
2007-07-02 13:26:56 · 4 answers · asked by leolaushman 1 in Science & Mathematics ➔ Engineering
I see I errored in speaking earlier what you have guessed that the transformer is indeed outputting 12 VDC so it has a built in rectifier. The relay has a resistance of 160 ohms. What happens is the transformer powers the relay until a switch breaks the circuit thus transferring the power to the buzzer and activating it instead. You may ask what this is and to let you know it is a simple way of alerting me when mail has been delivered to my mailbox out on the street. I rarely get mail there and I thought this would be a good way to check it without going outside in bad weather. I also need a switch to reset the circuit when the mail has been picked up by me but for simplicity I left that out seeing I really wanted to know the resistors needed. This circuit will either be feeding the relay or the buzzer continuously thus the need for the protection of the relay and buzzer. The buzzer shows no value for resistance so I guess there is none there or it is to small to count.
2007-07-03 02:19:37 · update #1
If the relay is designed for 12v (which it is), it can be connected directly to the 12v supply and it will take 75mA when activated. No resistor is needed.
This is a trick question!!!!
Not enough information has been provided for the buzzer part of the question.
2007-07-02 14:09:00 · answer #1 · answered by Colin M 1 · 4⤊ 0⤋
It doesn't matter how much current might be able to come from the transformer. All you care about is the 75mA that the relay needs. You have not said what resistance the relay has. It might not need any additional series resistance to limit its current to 75mA if it is itself at 160 ohms (Ohms law of R=E/I for 12V/.075A).
You mention that the components are all DC but a transformer is an AC device that outputs an AC current. I am assuming that there is also a rectifier (and maybe a filter) to make that AC into DC for the relay.
2007-07-02 15:56:47 · answer #2 · answered by Rich Z 7 · 0⤊ 0⤋
use 200 ohm resistor (12000/75). this Will drive your relay smoothly, and don't forgot to add one Diode in parallel to Relay for blocking the reverse current.
2007-07-02 13:38:01 · answer #3 · answered by Sant 2 · 1⤊ 2⤋
V=IR
Voltage = Current * Resistance
That's all I remember from my Electrical Engineering class...
2007-07-02 13:33:10 · answer #4 · answered by Glock 32 2 · 0⤊ 2⤋