Fritzing as suggested in a couple of answers will help you draw a nice wiring diagram, but it does not explain what's wrong with yours.
This is what you've made. R1 and R2 should be obvious: they connect directly between the battery's plus and minus terminals. R3 is a different matter. To understand that you have to know how a solderless breadboard is constructed.
Each red, blue and black line indicates a series of connected pins. The red and black don't seem to be a problem; you got those right. Then you connected the positive supply line to A22. According to the drawing above everything from A22 to E22 is now connected to the plus. Another wire to F22. Now also F22 to J22 are connected to the plus. And then a final wire from J22 to the battery's minus. And then it goes wrong. You've created a path between plus and minus with no resistance, which is a short-circuit! Power supplies don't like them. Your third resistor is also short-circuited by the connections between F22 and J22, so it doesn't do anything. And since R1 and R2 are short-circuited too, they don't serve a function either.
I'm not so fond of the Fritzing drawing in Ignacio's answer, because the blue lines don't show wires, but electrical connections. Some are wires (though they share pins with the resistors :-( ), and one isn't. Here's how I would build it, no Fritzing:
The top and bottom right resistors are connected through the five pin row. And the blue wire makes that the bottom left resistor is parallel to the bottom right one.
I'm not a Fritzer myself, but I took the picture from Ignacio's answer and edited it in my graphics editor to what I think the Fritzing wire diagram should have looked like.
I believe Fritzing can be very useful to help you wiring up your breadboard, but like any other tool it must be used correctly.
edit re the update of your question
You're almost there, but the top capacitor isn't right yet. It might have been if that capacitor bridged a split in the supply rails, like Oli's breadboard has them. According to the photo you posted yours doesn't, so the battery's voltage runs all the way from left to right, and the capacitor is shorted out. A little rearranging fixes this:
(Note that this circuit won't do anything on a DC voltage supply. The capacitors and inductors are AC components; the coils are just 0 Ω resistors for DC and the capacitors will block DC, so there won't be any current. You could use the low voltage output from a transformer, but that's a very low frequency, and you almost need an oscilloscope to analyze your circuits. You can always start with only resistors in every possible combination of parallel and series to get a feel of circuit analysis. Learn about Kirchhoff's Laws to calculate currents and voltages.)