Question
Explain emf, potential difference and internal resistance of cell and derive the relation between them. (or derive the relation between emf, potential difference and internal resistance of the cell.)
✓
Answer
→The figure shows the electrolytic cell. Electrolytic cell is used to maintain a steady current in an electric circuit.
→As shown in the figure, two electrodes, positive (P) and negative (N) are immersed in an electrolytic solution.
→Because of being dipped in the solution, the electrodes exchange charges with the electrolyte.
→The potential difference between the positive electrode and the adjacent point A is $V _{+}$and the potential difference between negative electrode to the adjacent point B is $- V _{-}$.
→So, the potential difference between P and N is $V_{+}-\left(-V_{-}\right)=V_{+}+V_{-}$
→This potential difference is called the electromotive force (emf) of the cell and it is denoted by $\varepsilon$.
$\therefore \varepsilon= V _{+}+ V _{-}>0$
→Definition of emf : The potential difference between two electrodes when no current is flowing through the cell is called the emf of the cell.
→As shown in the figure, consider a resistor R connected across the cell and current I flows across R from C to D and current in electrolyte current flows from N to P . clearly, across the electrolyte →the same current flows through the electrolyte.
The electrolyte through which a current flows has a finite resistance $r \&$ it is called the internal resistance.
→Case I : $R =\infty$ (infinite)
According to ohm's law
$I =\frac{ V }{ R } \rightarrow I =0$
where V is the potential difference between P and N and it is called terminal voltage (or voltage) of the cell.
→The potential difference between P and N , V (terminal voltage or voltage)
$=$ Potential difference between P and A
+ Potential difference between A and B
+ Potential difference between B and N
$\begin{aligned}
V & = V _{+}+ V _{-} \\
\therefore V & =\varepsilon
\end{aligned}$
→Thus, emf $\varepsilon$ is the potential difference between the positive and negative electrodes in an open circuit i.e when no current is flowing through the cell.
→Case II : $R =$ Finite
when $R$ is finite, electric current flows through the cell. In that case the potential difference between P and N is
$\begin{aligned}
V & = V _{+}+ V _{-}- I r \\
\therefore \quad V & =\varepsilon- I r
\end{aligned}$
→The negative sign in the expression (Ir) for the potential difference between A and B . This is because the current I flows from B to A in the electrolyte.
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