bulb on

enter image description here

bulb off

enter image description here

I am not changing the orientation of the bulb, I understand that makes no difference anyway since it's an incandescent bulb.

As you can see, the difference in the pics is that where the light went on, I connected two wires manually.

But where the light was off (and I expected it on), I plugged the ends of those two wires into the breadboard with the rest of the wires, rather than connecting those two manually.


I do not understand why this would be shortening the bulb(what an answers says is happening). It looks like a circuit to me, that includes the bulb and the battery.

The current would go from -ve to +ve, A->B->C->Bulb->D->E->F->A

And if I unplug C and B and connect them manually instead of via the breadboard, I don't see why that should make any difference. The circuit would still be A->B->C->Bulb->D->E->F->A

enter image description here

  • \$\begingroup\$ I see - D and C are connected to each other twice. Once through the bulb and once through the breadboard \$\endgroup\$ – barlop Jan 20 '16 at 14:42

The bottom row is all connected. The cable marked D and C are physically connected, which means there is no voltage difference through the bulb. It's been shorted out. The path of least resistance is a ~ 0 Ohm wire, so the bulb isn't even in the circuit any more.


simulate this circuit – Schematic created using CircuitLab

Just because there is a gap between the holes, does not mean that row isn't connected.

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  • \$\begingroup\$ Your top paragraph is clear and hits the problem I had. As for your second paragraph, I know that, I know that the top row is all connected, and the bottom row is all connected. That's how I knew that F would connect to A. That's how I expected the bulb to light up when it did. But yeah your top paragraph explains it somewhat. \$\endgroup\$ – barlop Jan 20 '16 at 4:30
  • \$\begingroup\$ so the two wires were connected twice and travelling the low resistance path, skipping the bulb \$\endgroup\$ – barlop Jan 20 '16 at 4:36
  • \$\begingroup\$ @barlop yes. A dead short is the perfect low resistance path (in practice), so yes. A small wire being such a low resistance that it's practically 0 Ω. No current would flow through the bulb. \$\endgroup\$ – Passerby Jan 20 '16 at 4:51

Because you've shorted out the bulb.

The lines that run the length of the breadboard are all connected.

The following is an example of how the connections in the breadboard are wired up:


Image source

As a side note, this should be quite obvious if you put your hand on the battery. You will feel it getting very warm very quickly as you have, through a few wires and the breadboard, connected both of its terminals directly together.

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  • \$\begingroup\$ I know the breadboard is arranged like that but I don't see why one is shortening it. I have edited my question to clarify why I don't see how the bulb is being shorted. \$\endgroup\$ – barlop Jan 20 '16 at 4:19
  • \$\begingroup\$ ah I think I see.. have I made a parallel circuit for the bulb but withn no battery and the electricity is looping round? \$\endgroup\$ – barlop Jan 20 '16 at 4:23
  • \$\begingroup\$ Do one experiment. remove lamp wires from breadboard, then check continuity between two points where you are connecting lamp as per 2nd image. Buzzer (of multimeter) will beep showing short circuited path. That is the reason lamp is not glowing. Tom has given nice clarification. \$\endgroup\$ – Electroholic Jan 20 '16 at 4:25
  • \$\begingroup\$ @barlop C and D are both connected - notice how in the diagram above everything on row 'Y' is all connected? \$\endgroup\$ – Tom Carpenter Jan 20 '16 at 4:25
  • \$\begingroup\$ @TomCarpenter I know everything on row y, that top row, is connected.. that's how I expected the bulb to light up when it did. If it wasn't connected i'd have had no circuit at all.. I know that. And the bulb lighting up in the first picture was no surprise. \$\endgroup\$ – barlop Jan 20 '16 at 4:27

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