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Automatic ON and OFF BULB for specific time only.

Following are considered to achieve the project...

  1. As 12 v is supplied, relay R1 is slowly activated and bulb B1 will glow for specific time T which is dependent on capacitor C. As C is fully charged after sometime, say 5v, it blocks current and hence B1 is off.

  2. In the meantime, R2 is always activated as long as 12v is supplied. Now, as 12v is disconnected, the energy stored in capacitor C is discharged through 87a of relay R2 and B2 now glows for specific time T again.

  3. In short, B1 glows for time T and off when capacitor C is full, and B2 glows for time T only when 12v source is disconnected.

Would like to hear any comments or suggestions to rectify my project in achieving the on and off bulb for specific time only. Thanks.enter image description here

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  • \$\begingroup\$ A quick google search gives, among many, arrow.com/en/research-and-events/articles/… \$\endgroup\$
    – Solar Mike
    Oct 6, 2019 at 12:46
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    \$\begingroup\$ What time are you hoping to achieve? If it's more than a few mS, likely not long enough for the relay to fully close, this this is an incorrect approach. Why? The time-constant of the circuit with the capacitor, which determines how long current will flow, is going to be quite LOW since the R (resistance) of the relay coil is low compared to the C (capacitance) value of the capacitor. There are much better ways to do this, perhaps using a 555 timer (or similar) IC. \$\endgroup\$
    – jwh20
    Oct 6, 2019 at 13:21
  • \$\begingroup\$ 3 to 5 minutes of time.. \$\endgroup\$
    – Thang Tons
    Oct 6, 2019 at 13:28
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    \$\begingroup\$ 3-5 minutes of delay is NOT achievable with any practical C value that I can imagine using this circuit. Again, I think your best bet is a simple 555 timer circuit which can easily achieve 3-5 minutes with readily available components. \$\endgroup\$
    – jwh20
    Oct 6, 2019 at 13:40
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    \$\begingroup\$ If you have a dynamo running off the wheels and your circuit works as you want, the doors will lock when the car reaches a speed where enough current is generated to power the relay. The doors will then unlock when the car slows down for a junction, as you speed up the doors will lock again. However the way you have used the capacitor will mean it won’t work at all. You likely need a microcontroller monitoring the speed of the car and the ignition switch. \$\endgroup\$
    – HandyHowie
    Oct 6, 2019 at 14:42

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