okay the diagram is still a little confusing, there are two different ground symbols, at least thats all I learned those two symbols as. And the white triangle on the right side is labeled as "D3" like a diode, but its not a diode...
I will assume that the Ground symbol on the left is 0V reference and that the white triangle on the right is +30V (this is probably the 'power source' you mentioned).
ericgibbs - theres a much quicker solution to this problem than calculating currents in each resistor. I think the point the problem is trying to drive home is the effect of diodes.
Each diode is noted with a voltage. This is probably the intrinsic voltage drop of the diode. The ideal diode model states that if a diode has 0V or less across it, it does not conduct, and that if there is a positive voltage across it, it will conduct any current. Instead of 0V, your diagram says this voltage is 5V. So the diagram seems to instruct that every diode IS conducting, and each one has a 5V drop across it. (decent ideal diode explanation here:
Diodes )
So you might first look at this and think: what do circuit elements in parallel have in common? Parallel circuit elements will have the same voltage. Series circuit elements will have the same current. Starting at the +30V reference (on the right) youll notice that there is a parallel combination of a resistor and diode. The diode is conducting, and it drops 5V. This means that the parallel combination of that diode (D1) and that resistor (R1) both have the same voltage across them: 5V.
Similarly, there is another, similar parallel combination to the left of that, again, with the 5V diode "clamping" the voltage across the resistor.
Now, you know the voltage across both R1 and R2. So now you know a voltage and a resistance, and are asked for a current (current in R2). You must know the circuit law to find that.
This is kind of interesting as the current in R2 is not the current in R3. If there are 2 5V diodes, that makes the voltage across R3 20V. 20V / 1kohm = 20 mA. So when you find the current in R2, you can also find the current in D2, and then the current in R1 and D1, and so on.
The point here is the "clamping" effect of the diodes.
Sorry if I solved too much, try another similar problem just to be sure youve grasped the concept.