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I saw there was a post asking the same question here but I still don't understand it.

I am reading the book Code: The hidden language of computer hardware and software and the book shows the following circuit:

enter image description here

where the red wires show the flow of electricity in the circuit. I don't understand why there flows no current through the common. Could someone please explain? Why doesn't the electricy flow like this:

So from the the negative terminal of battery 1, through the right bulb, and into the positive terminal of battery 1? Is it because the currents (the one coming out of the negative terminal of battery 1 and the one coming out of the negative terminal of battery 2) are exactly equal and opposite so they cancel eachother out?

If that is the case why don´t they cancel out in this image? enter image description here

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    \$\begingroup\$ NET current flow is zero because current is flowing in opposite directions for each lamp. If the lamps are identical and therefore draw the exact same current then these currents would cancel out ….net zero current flow for that segment. \$\endgroup\$ – Jack Creasey May 29 at 21:56
  • \$\begingroup\$ Thanks a lot that clears up a lot doubts I had! But why don't they cancel out in the last image? The current that comes out of the negative terminals is opposite as well no? \$\endgroup\$ – user3302735 May 29 at 22:11
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Basically, there are two currents flowing in the common line, one for each bulb, but since they are equal and they flow in opposite directions, the effectively cancel each other out. If you draw the current for each bulb lit individually, and then superimpose them, you can see this.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks a lot! Indeed, superimposing makes it very clear! Almost like an XOR. However, why don't they cancel out in the last image? The current that comes out of the negative terminals is opposite as well no? \$\endgroup\$ – user3302735 May 29 at 22:11
  • \$\begingroup\$ No. And that isn't the way to think about it. When both lamps are on, it's the lamp current that cancels out the battery current on that side of the circuit. You can think of it as the lamp current replacing the current that would otherwise flow through the common wire. \$\endgroup\$ – Dave Tweed May 29 at 22:21
  • \$\begingroup\$ Sorry but what do you mean with lamp current versus battery current? And I was referring to the last image, the blue one in my original post. Maybe we mixed things up ;), I only saw you added pictures later. Thanks! \$\endgroup\$ – user3302735 May 29 at 22:39
  • \$\begingroup\$ In my last diagram, the green arrow pointing down next to LAMP5 is the same amount of current represented by the red arrow pointing up next to BAT5 -- there is nothing left over to flow through the common wire. \$\endgroup\$ – Dave Tweed May 29 at 22:43
  • \$\begingroup\$ Ah I see. But still why don't they cancel out in this image i.stack.imgur.com/ICecB.png (taken from youtu.be/w82aSjLuD_8?t=195, here we see two opposite currents as well right? \$\endgroup\$ – user3302735 May 29 at 23:09
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Its easier to see the symmetry if you redraw the circuit.

If the circuit is symmetrical, then the voltages of the batteries will be at the same potential because the voltage will be equal due to symmetry of the currents. An equal voltage means no current flow. However, in the real world this would not be possible, there would be some mismatch (it's really hard to match resistances), the batteries would also need to be identical. and all the wires the same length because wires are resistors (just really small ones).

schematic

simulate this circuit – Schematic created using CircuitLab

So from the the negative terminal of battery 1, through the right bulb, and into the positive terminal of battery 1?

Yes another thing to consider is if one of the switches are off, both of the switches are off, and if both are on. If only one switch is on, then the current will travel through the 'center wire' through the battery switch and light bulb.

Is it because the currents (the one coming out of the negative terminal of battery 1 and the one coming out of the negative terminal of battery 2) are exactly equal and opposite so they cancel eachother out?

If all things are equal (which can happen in the ideal world of circuit diagrams) then the currents will cancel out, no current will flow and the voltage will be the same on both of the negative battery terminals.

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Your second picture is of a different situation. The two negative terminals of the batteries are at the “same” voltage, aka the voltage between them is 0, so no current flows from one to the other. Same in quotes to show we are in the ideal world.

This is the exact reason no current flows in the first image. However if you measured the voltage between the negative side of your lamp and your batteries negative terminal there would be an ever so small but quantifiable voltage across the length of wire. It is small because the current has little resistance along the length of wire, therefore little voltage drop occurs along it.

Think V=IR for the wire. The current is the same flowing through your lamp and the resistance is from the internal resistance of the wire, very low for normal copper wiring, therefore the voltage across the wire would also be very low. And in this ideal example the current would split equally at the junction of two wires and flow to each battery, with the length of wire there showing half the voltage drop the same length of wire would have shown if you had only 1 wire and battery. Remember that without voltage difference no current flows.

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