Shouldn't the electromotive force of all cells be zero?

Question

The electric energy supplied by the cell for a unit charge to move from any point in the circuit and come back to the same point by traversing the entire circuit along with the cell is called the electromotive force of the cell.

Now, if I move a unit charge from A to A, by traversing the entire circuit, shouldn't the energy provided be zero, in other words, should any energy be required for the unit charge to move from A to A?

We can see it like this as well: let the potential at point A be $V_A$. Now, the potential difference between A & A is $V_A-V_A=0$.

So, should any energy be required for a unit charge to traverse from A to A? Shouldn't the emf of all cells be zero?

Answer 1

Starting at the bottom right-hand corner and moving clockwise,

the potential rises when moving through the battery with work being done by an external source (a chemical reaction) to move charges between the terminals of the battery with the work done per unit charge being the emf of the battery, and

the potential drops when moving through the resistor with an equal amount of work being done driving the charges through the resistors (with the production of heat) when arriving back at the bottom right hand corner.

So in terms of energy, the chemical energy used in the battery is equal to the heat dissipated in the resistor.

Answer 2

$V_a - V_a = 0$, as you said. But that is the potential difference across the cell and the resistor, not just the cell.

It is like walking in a circular path that takes you up a hill and down the other side. You wind up at the same altitude with the same potential energy as you started with. Along the way you had more potential energy at the top of the hill and less at the bottom.

The battery contains chemicals that push electrons onto the negative terminal. Electrons repel each other. Normally a wire contains equal numbers of electrons and protons. If more electrons are pushed onto the wire, they are closer together than normal. They gain potential energy.

They try to spread out. They can travel easily in the wire. Electrons in the negative terminal push their neighbors away. They push their neighbors, and so on down to the resistor.

Electrons do not flow so easily through the resistor. So the wire between the negative terminal and the resistor has a (very slight) excess of electrons.

The other side of the resistor is connected to the positive terminal. The battery contains chemicals that attract electrons in the positive terminal and push them onto the negative terminal. This leaves the wire slightly depleted of electrons. Electrons have lower potential energy in this region.

Electrons are attracted to this wire through the resistor. So electrons flow from the negative terminal through the resistor to the positive terminal.

Inside the battery, chemicals push electrons from the positive terminal to the negative terminal. The electrons gain potential energy. This energy comes from the energy in the chemicals.

So if B is a point in the wire on the other side of the battery, $V_B - V_A > 0$.

Answer 3

$\int E \cdot dl = 0$ For a closed path,

$\int E \cdot dl $ Is $ V_{0} $when the path is chosen to be the ends of the terminal