Could someone tell me if I breadboard this circuit correctly? I am very new to breadboarding.

Note that

Green = Resistor (take all to have same resistance)

Red/Black = wires that hook up to battery

Blue = just regular wires.

I am trying to build the circuit below.

Did I do this right?

edit 2012-08-21

Is the above breadboard the same as the schematic circuit? Also I can't find any function generators or oscilloscopes on Fritzing. Does Fritzing not have it?

Furthermore, these configurations seem kinda repetitive. Do they become very complicated if I include other devices?

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I don't really have time to diagram it for you, but no. What you have is three resistors in parallel. Move the bottom of the two long ones to the + row above, and put the small one from the - to +. (slightly abuses that row, but it's quick and easy). –  Kevin Aug 18 '12 at 1:51
Actually, two resistors and a wire in parallel. Badness for the battery. –  DarenW Aug 18 '12 at 4:45
It might be a useful educational exercise to first stick a few parts and wires into the breadboard, DON'T HOOK IT TO A BATTERY, and draw the schematic. You might "get" it more effectively than going from schematic to breadboard, at least at first. –  DarenW Aug 18 '12 at 4:47
–  DarenW Aug 18 '12 at 4:49
@jak - In parallel means that the top of the wire is connected to the top of the resistors, and the bottom of the wire to the bottom of the resistors. So the wire actually short circuits the resistors. See the diagram in my answer. BTW, I would reconsider the accepted answer. It only says "no", and gives you an unclear (but nice looking) diagram instead of telling you what you did wrong. –  stevenvh Aug 18 '12 at 9:09
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## migrated from physics.stackexchange.comAug 18 '12 at 2:36

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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.

edit
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.)

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+1 - this answers the question properly. A minor point - in the edited Fritzing picture note that some breadboards have split power rails (mine does), so it's possible by moving the top resistor over you could disconnect it from the positive terminal (i.e. you need to move the red wire to a pad adjacent to the top resistor lead) –  Oli Glaser Aug 18 '12 at 13:28
@Oli - Thanks for the feedback. Fixed my artwork. Any idea why they split it? It's bound to cause confusion IMO. –  stevenvh Aug 18 '12 at 13:33
I think it's just so you can have more rails - my board has 6 banana/screw terminals with a couple of pads beside them, so you can wire from those to the various rails. I think this is the idea anyway (it's the way I use it) I agree it can could be confusing if not documented, although on my board they are marked V1, V2, V3, V4 etc. –  Oli Glaser Aug 18 '12 at 13:48
@Oli - but does the breadboard have a clear indication of the splits? Or are they like in the Fritzing diagram? Suppose I put one half of the rail at 5 V, and the other half at 15 V, could be painful for my 74HCT if I connect it to the wrong half. –  stevenvh Aug 18 '12 at 13:52

It's not correct.

I suggest you use the open source design software: Fritzing

You can draw the schematic and the breadboard and check how they match, I find this tool really useful for beginners.

This is their website: http://fritzing.org/

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How does the breadboard and the circuit diagram line up? –  sidht Aug 18 '12 at 3:23