how to find the trick plzz help

Thread Starter

muchy

Joined Dec 14, 2009
3
i got a homework is that:
we have one capasitor one resistor one DC source (9V) and one lamp.

the flasher circuit is required
The lamp must be lighted for 4 sec. (when it has 4volts) and for 4 secs isnt lighted.

notice:we just have one resistor one capasitor one DC source(9v) and a lamp.

pls guys help me im stuck.
 

ELECTRONERD

Joined May 26, 2009
1,147
You can't just make a flasher with a resistor, cap, lamp, and DC source. You need something that will switch between the two LEDs. You may do that by saturating two transistors, using the 555 timer, and many other techniques. One important equation to remember is times constant:

T = RC

Where:
T = Time
R = Resistor value
C = Capacitor value

Austin
 

KL7AJ

Joined Nov 4, 2008
2,229
You can't just make a flasher with a resistor, cap, lamp, and DC source. You need something that will switch between the two LEDs. You may do that by saturating two transistors, using the 555 timer, and many other techniques. One important equation to remember is times constant:

T = RC

Where:
T = Time
R = Resistor value
C = Capacitor value

Austin

Actually, you can, if that lamp happens to be a NEON lamp. (See relaxation oscillator). But I don't think that's what the poster had in mind. :)

eric
 

Ron H

Joined Apr 14, 2005
7,063
You can't just make a flasher with a resistor, cap, lamp, and DC source. You need something that will switch between the two LEDs. You may do that by saturating two transistors, using the 555 timer, and many other techniques. One important equation to remember is times constant:

T = RC

Where:
T = Time
R = Resistor value
C = Capacitor value

Austin
I didn't see any mention of LEDs by the OP.
 

hobbyist

Joined Aug 10, 2008
892
Here is a guess at how it might be done.

I can't give the exact answer for this is homework,

but think of the scenario given, what would happen if you put a cap across (parrallel) with a load,
the capacitor would act as a short circuit across the load, after the capacitor gains enough charge the load would begin to conduct current, but wait, if the load starts conducting current the capacitor is affected by a dis..chr,...,.,., hmmm seems like a see saw affectr where the cap. is getting charge than losing charge........ that's all I can give without given away the answwer to a homework question.

Then again this is just a guess at how I would try it, may not be the rightr answer afterall.

Only experimenting woulod tell..
 
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