Multiple leds, and basic wiring advice?

[Leezygeezer]

Hi I am very new to electronics and experimenting with very basic timing circuits using a 3v supply, 2 resistors, a capacitor and an led (red).
I have managed to glean that the forward voltage on an led is approx 2v, so with a 3v supply and a 2000ohm resistor, a 1000uf cap and a 100ohm resistor to connect to ground I get a 2 second delay before the cap reaches 2/3rd voltage i.e. 2 volts, enough to light the led.
swapping the 3v supply for a 9v, how would I be able to set up the same circuit (different R1 and C1 values?) and get the timing delay?

Please forgive my ignorance in wiring matters but I have a question on this basic circuit, why does the led attach between R1 and the cap, and why is a resistor required to go back to ground, why not just a jumper wire?

Why will the resistor followed by the cap followed by the led in series not work?

If I wanted to use a 9v supply to power say 5 red led's, the forward voltage will be approx 10v and therefore some of the leds may not light. If i wire a resistor in series with each led but each led in parallel will all 5 light? Will this just pull 0.1amp of current from the battery (if each led is pulling 0.02a) will each led light up?
Sorry if this is basic but I need to get a firm grasp of the basics before I even attempt to move on to other stuff lol

 

[Grossel]

I'll give you some steps to follow. Assuming the circut is like this (see picture).

How much current goes through the forward resistor when the led lits?

When you applies a 9V battery, how much current should go through the forward resistor to get the led lit?

Hint: It's ohm's law all the way.


Probably not important for you at this time, but I let you read this anyway:
The charge curve for 3V and 9V voltage supply will be different (if you could see it on a scope), even if the delay for a given cap charge is calculated to be the same.
Because of the characteristic of a LED, you won't get it to light fully with excactly the same time delay. But it'l be close enough.