# Can EMF be lesser than terminal potential difference?

Can terminal potential difference be greater than EMF? If so, how?

Upto my knowledge and my research, during charging it can happen but the question remains same: how it happens. What is the basic principal behind this concept? By doing this, do we violate the principle of conservation of energy?

## 4 Answers

Yes it can, for example when a cell is being recharged the terminal pd is $V = {\mathcal E} + IR$ where $\mathcal E$ is the emf of the cell, $I$ is the current passing through the cell from positive to negative terminal and $R$ is the internal resistance of the cell.

Power is being supplied from an external source $VI$ and is producing heat in the internal resistance $I^2R$ and via a chemical reaction in the cell being converted to chemical energy $\mathcal E I$ as the cell is charged.

Yes, it can be.

We know that $V=E-IR$. When the current $I$ becomes negative, the equation becomes $V=E+IR$ i.e. $E$ is less than $V$

Now we have to know when $I$ becomes negative. When another cell with higher emf (electromotive force) is connected in opposite direction with the main battery the new battery produces a current in opposite direction, so the current in the previous battery becomes negative.

Yes it's possible. We know that V = E - IR, here current I is positive and E > V. Now if we make the current negative then the equation become V = E + IR . Now V> E If we connect a new cell with higher emf then the current flows through the opposite direction amd it's become negative.

We know that V =IR, Here current I is positive and E>V. Where E is the emf of the cell. I is the current passing through the cell from positive to negative terminal and R is the internal resistance of the cell. If we connect a new cell with higher emf then the current flows through the opposite direction and it's become negative. Then V >E and the equation become V =E+IR. And it is possible.