2 added 196 characters in body edited Nov 19 '16 at 3:10 jonk 38.1k11 gold badge3131 silver badges8282 bronze badges A potential difference sets up an electric force which causes conduction electrons to move. A battery is a piece of complicated chemistry, with anions and cations, and I don't want to delve into the details there. Plenty of better sources for that, than me. But the fundamental problem with your imagination is illustrated where you write, "it's voltage is 0!" Up until that point, you were talking about potential differences (appropriate.) Suddenly, right there you skip tracks and then conflate (confuse) the idea of a voltage at a single point with the idea of potential differences. These aren't even close to the same thing. A voltage at aexactly one node is entirely arbitrary. I could look at your circuit and argue correctly that the node you identified as 0 V is really at 1,000,000 V. The number is completely arbitrary. You can make it up. So can I. And we are both right, so far as the idea goes. Such values depend only on your choice of reference. And you and I are allowed to make different choices, there. (Of course, once you've chosen a reference point and assigned it any arbitrary value, you can't do that again. You only get to chose a reference point and assign it a value, just once.) The only question here is whether or not there is a non-zero electric field intensity (which is a vector), $$\\vec{\varepsilon}=\left(\frac{\textrm{d}V}{\textrm{d}x}, \frac{\textrm{d}V}{\textrm{d}y}, \frac{\textrm{d}V}{\textrm{d}z}\right)\$$, which provides a motive force. The battery itself has a field intensity direction inside it, as well. So the point you called "0" has a different potential to one side and to the other side, so there is still a field intensity direction at that point, as well. Everywhere in the circuit, in fact, has an electric field gradient. So the electrons continue to be propelled by the force acting on them throughout the circuit. A potential difference sets up an electric force which causes conduction electrons to move. A battery is a piece of complicated chemistry, with anions and cations, and I don't want to delve into the details there. Plenty of better sources for that, than me. But the fundamental problem with your imagination is illustrated where you write, "it's voltage is 0!" Up until that point, you were talking about potential differences (appropriate.) Suddenly, right there you skip tracks and then conflate (confuse) the idea of a voltage at a single point with the idea of potential differences. These aren't even close to the same thing. A voltage at a node is entirely arbitrary. I could look at your circuit and argue correctly that the node you identified as 0 V is really at 1,000,000 V. The number is completely arbitrary. You can make it up. So can I. And we are both right, so far as the idea goes. Such values depend only on your choice of reference. And you and I are allowed to make different choices, there. The only question here is whether or not there is a non-zero electric field intensity (which is a vector), $$\\vec{\varepsilon}=\left(\frac{\textrm{d}V}{\textrm{d}x}, \frac{\textrm{d}V}{\textrm{d}y}, \frac{\textrm{d}V}{\textrm{d}z}\right)\$$, which provides a motive force. The battery itself has a field intensity direction inside it, as well. So the point you called "0" has a different potential to one side and to the other side, so there is still a field intensity direction at that point, as well. Everywhere in the circuit, in fact, has an electric field gradient. So the electrons continue to be propelled by the force acting on them throughout the circuit. A potential difference sets up an electric force which causes conduction electrons to move. A battery is a piece of complicated chemistry, with anions and cations, and I don't want to delve into the details there. Plenty of better sources for that, than me. But the fundamental problem with your imagination is illustrated where you write, "it's voltage is 0!" Up until that point, you were talking about potential differences (appropriate.) Suddenly, right there you skip tracks and then conflate (confuse) the idea of a voltage at a single point with the idea of potential differences. These aren't even close to the same thing. A voltage at exactly one node is entirely arbitrary. I could look at your circuit and argue correctly that the node you identified as 0 V is really at 1,000,000 V. The number is completely arbitrary. You can make it up. So can I. And we are both right, so far as the idea goes. Such values depend only on your choice of reference. And you and I are allowed to make different choices, there. (Of course, once you've chosen a reference point and assigned it any arbitrary value, you can't do that again. You only get to chose a reference point and assign it a value, just once.) The only question here is whether or not there is a non-zero electric field intensity (which is a vector), $$\\vec{\varepsilon}=\left(\frac{\textrm{d}V}{\textrm{d}x}, \frac{\textrm{d}V}{\textrm{d}y}, \frac{\textrm{d}V}{\textrm{d}z}\right)\$$, which provides a motive force. The battery itself has a field intensity direction inside it, as well. So the point you called "0" has a different potential to one side and to the other side, so there is still a field intensity direction at that point, as well. Everywhere in the circuit, in fact, has an electric field gradient. So the electrons continue to be propelled by the force acting on them throughout the circuit. 1 answered Nov 19 '16 at 0:00 jonk 38.1k11 gold badge3131 silver badges8282 bronze badges A potential difference sets up an electric force which causes conduction electrons to move. A battery is a piece of complicated chemistry, with anions and cations, and I don't want to delve into the details there. Plenty of better sources for that, than me. But the fundamental problem with your imagination is illustrated where you write, "it's voltage is 0!" Up until that point, you were talking about potential differences (appropriate.) Suddenly, right there you skip tracks and then conflate (confuse) the idea of a voltage at a single point with the idea of potential differences. These aren't even close to the same thing. A voltage at a node is entirely arbitrary. I could look at your circuit and argue correctly that the node you identified as 0 V is really at 1,000,000 V. The number is completely arbitrary. You can make it up. So can I. And we are both right, so far as the idea goes. Such values depend only on your choice of reference. And you and I are allowed to make different choices, there. The only question here is whether or not there is a non-zero electric field intensity (which is a vector), $$\\vec{\varepsilon}=\left(\frac{\textrm{d}V}{\textrm{d}x}, \frac{\textrm{d}V}{\textrm{d}y}, \frac{\textrm{d}V}{\textrm{d}z}\right)\$$, which provides a motive force. The battery itself has a field intensity direction inside it, as well. So the point you called "0" has a different potential to one side and to the other side, so there is still a field intensity direction at that point, as well. Everywhere in the circuit, in fact, has an electric field gradient. So the electrons continue to be propelled by the force acting on them throughout the circuit.