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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?

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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.

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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.

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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.

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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.

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