Expression for Relation Between E.M.F , Terminal Potential Difference and internal resistance

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

When key (K) is closed

$$I = \frac{\varepsilon}{R+r}$$

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, $I = 0$

When key (K) is open

Hence eqn. (iii) becomes $V = \varepsilon$

Thus, terminal potential difference between the electrodes of the cell is equal to the e.m.f. of the cell in an open circuit.

Determination of internal resistance : 

We know $I = \frac{\varepsilon}{R+r}$

Using this value in eqn. (iii), we get

 $V = \frac{\varepsilon R}{(R+r)}$

 $r= \left(\frac{\varepsilon}{V} - 1\right)R$  ...(iv)

Thus, knowing the values of $\varepsilon$, $V$ and $R$, 

we can determine the value of the internal resistance ($r$) of the cell.

Previous Post Next Post