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Why does there flows no current through the common part of the circuit?
Why doesn't current flow through the common part of a circuit?How does current flow through a voltage source?How the Current Flows in a Car?How current flowsDirection of current through a circuit with multiple batteriesWhy does the circuits work with conventional current?Why doesn't current flow through the common part of a circuit?Why doesn't an electric charge use up all its potential energy travelling from positive to negative terminal?Why is there current flow in the sub-circuits?Switching positive and negative voltageNoob questions: I'm confused about conventional current and electron flow and how it runs through and powers a circuit. Can someone please explain?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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:
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?
batteries current circuit-analysis
asked May 29 at 21:46
user3302735user3302735
111
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
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:
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?
batteries current circuit-analysis
asked May 29 at 21:46
user3302735user3302735
111
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
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
add a comment |
$begingroup$
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:
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?
batteries current circuit-analysis
asked May 29 at 21:46
user3302735user3302735
111
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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:
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?
batteries current circuit-analysis
batteries current circuit-analysis
asked May 29 at 21:46
user3302735user3302735
111
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 29 at 21:46
user3302735user3302735
111
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 29 at 21:46
user3302735user3302735
111
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 29 at 21:46
user3302735user3302735
111
asked May 29 at 21:46
user3302735user3302735
111
111
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user3302735 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
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
add a comment |
2
$begingroup$
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
2
2
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
simulate this circuit – Schematic created using CircuitLab
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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).
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.
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
$endgroup$
add a comment |
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2 Answers
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oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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.
simulate this circuit – Schematic created using CircuitLab
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
simulate this circuit – Schematic created using CircuitLab
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
simulate this circuit – Schematic created using CircuitLab
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
$endgroup$
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.
simulate this circuit – Schematic created using CircuitLab
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
edited May 29 at 22:15
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
answered May 29 at 22:00
Dave Tweed♦Dave Tweed
128k10159275
128k10159275
$begingroup$
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
|
show 1 more comment
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
|
show 1 more comment
$begingroup$
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).
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.
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
$endgroup$
add a comment |
$begingroup$
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).
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.
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
$endgroup$
add a comment |
$begingroup$
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).
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.
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
$endgroup$
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).
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.
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
answered May 29 at 22:19
laptop2dlaptop2d
31.6k123898
31.6k123898
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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