Ok on question 1. Why does the 2k have 50v. and the 1k have 25v.? ...

For the wizards out there: This is limited to DC and pure resistnace. (anyone else that wants to pitch in here won't hurt my feelings)

Okay, a little circuit theory. I think you are just missing what some of the terms mean.

We got Ohm E = I x R That gives the relation for voltage and current for a resistor. If you know the current through a resistor and the resistance, then you can calculate the voltage across the resistor.

So the current around the loop is: I = (75V)/(3000ohms) = .025A

So, if you put your meter probes across the 1000ohm resistor what is the voltage? E = I x R = .025 x 1000 = 25V

So, if you put your meter probes across the 2000ohm resistor what is the voltage? E = I x R = .025 x 2000 = 50V.

--- Shouldn't the more resistance lower the voltage? ...

I'm not sure what you are asking. More resistance give less voltage drop? Well, no.

Voltage across a resistor = I x R (right?)

In this case, both reisitors have the same current through them (right?)

So with the current the same, the highe the resistor, the higher voltage drop.

.025 x 1000 is less than .025 x 2000

Draw the circuit. Label all the compoents, values, current through them. Calculate the Voltages. See what you get.

cf