Expression for Relation Between E.M.F , Terminal Potential Difference and internal resistance
Derive relation between e.m.f. and terminal potential difference and thus find expression for internal resistance. Consider a cell of e.m.f. $\varepsilon$ and internal resistance $r$ connected to an external resistance $R$ through a key (K). Case 1 : When key (K) is closed , current is drawn from the cell by the circuit, which is given by $$I = \frac{\varepsilon}{R+r}$$ R and r are in series, so (R + r) is the equivalent resistance of the circuit. $\varepsilon = IR + Ir$ ...(i) According to Ohm's law : That is, $V = IR$ ...(ii) Hence equation (i) becomes $\varepsilon = V + Ir$ $V = \varepsilon - Ir$...(iii) This shows that the terminal potential difference of the cell is less than the e.m.f. of the cell. Now the voltmeter connected across the cell will read the value as $V$ which is less than the value of e.m.f. ($\varepsilon$). Case 2 : When key (K) is open, $I = 0$ Hence eqn. (iii) becomes $V = \varepsilon$ Thus, terminal potential difference between the electrodes of the cell ...