How Does Adding Load Resistor RL Affect Vout in a Circuit? Real-Life Analogies?

Hi guys, this is basic question but yeah it's really need deep understanding how electric goes by and how to make a good analogy for !
lets assumecase 1: I have specific circuit .. and at the end of the circuit there's vout pin which it's in the air I mean vout is "cutout" there's no resistance RL !
case 2: Now lets assume that I put where's the pin Vout a resistance RL .. so the new Vout will be bigger or smaller than the previous case?!
what's confusing me ..how can I imagine if I load the circuit ..will that decrease the vout or increase it? and may anyone please give me a more senseable real life analogy if we "load" more then voltage will be lesser/bigger?!

Ryan, you have already answered your own question.Please go back to your post on maximum power point transfer and you will see how the load affects the output.And how are we supposed to understand what you are talking about if you don't post the diagram you are referring to?

I mean something in general !

OK, here's the table from before with the voltage on the load added:

RL (Ω)	VLOAD	I (amps)	P (watts)	Effi %
0	0.0	4	0	0.00
5	16.7	3.3	55	0.17
10	28.6	2.8	78	0.28
15	37.5	2.5	93	0.37
20	44.4	2.2	97	0.44
25	50.0	2	100	0.50
30	54.5	1.8	97	0.54
40	61.5	1.5	94	0.63
60	70.6	1.2	83	0.69
100	80.0	0.8	64	0.80
1000	97.6	0.098	9.5	0.98
OC	100.0	0	0	0.00

OC for RL = Open Circuit (no load) = ie your "air".I have added a column for voltage on the load - remember that this is a 100V DC source with a 25 ohm resistor in series.You should be able to calculate this yourself easily, for any circuit.  Lets take the case where RL - 100 ohms.So you have a simple voltage dividerVL - VBATT * 100 / (100+25) = 100 * 100/125 = 80VIf this makes no sense to you you need to go back to basics and should not be trying to do this sort of problem.