to short DC terminal by using 5 sec delay and when ac source apply

[prakashkovaidt]

Hi,

i'm doing project in electrical based. in my project stage 7th i'm struggling.. i will explan what is it? Could you please guide me to improve my the project

1st thing, When applying ac Voltage in one circuit it contains lamp(220VAC) at the same time already finished circuit should enable and turn off ( 5sec delay ) second circuit disable.

the first circuit ac lamp automatically disable after 50Sec..
this is my needs.

 

[dorke]

Welcome to EP,
I can not understand what your are saying:
"should enable and turn off", what do you mean by that?

 

[prakashkovaidt]

yes.. i want to enable the circuit( 2nd in pic DC circuit) when circuit 1 (AC circuit)
is enabled

enable means circuit start run.

 

[elektrroda]

you meant , when do you supplies circuit 1 , after some time do you want on switch in 2 circuit ? Are coupled Load from 1 circuit and switch from 2 circuit ? If yes , you buy timer when has delayed output.

 

[dorke]

Well,
Here is one simple solution,using a 220v ac relay.
I will leave the delay issue for you...

 

[cjdelphi]

A 240v relay to activate a low voltage circuit? (Usually it's low voltage dc allowing 240vac to flow)

So circuit 1 activates, 5 seconds later, circuit 2 activates?

Right or wrong?

 

[dorke]

1.Usually? who says?
there are many ways to skin a cat...
2.Wrong,there is no 5sec delay(yet).

 

[cjdelphi]

I was clarifying on what the op wanted....

A link to a relay triggered by 240ac?