3.16 POTENTIOMETER
A potentiometer. \( G \) is a galvanometer and \( R \) a variable resistance (rheostat). 1, 2, 3 are terminals of a two way key
(a) circuit for comparing emfs of two cells;
(b) circuit for determining internal resistance of a cell.
This is a versatile instrument. It is basically a long piece of uniform wire, sometimes a few meters in length across which a standard cell (\( B \)) is connected. In actual design, the wire is sometimes cut in several pieces placed side by side and connected at the ends by thick metal strip. (Fig. 3.28). In the figure, the wires run from \( A \) to \( C \). The small vertical portions are the thick metal strips connecting the various sections of the wire.
A current \( I \) flows through the wire which can be varied by a variable resistance (rheostat, \( R \)) in the circuit. Since the wire is uniform, the potential difference between \( A \) and any point at a distance \( l \) from \( A \) is
\[ \varepsilon(l)=\phi l \](3.89)
where \( \phi \) is the potential drop per unit length.Figure 3.28 (a) shows an application of the potentiometer to compare the emf of two cells of emf \( \varepsilon_1 \) and \( \varepsilon_2 \) . The points marked 1, 2, 3 form a two way key. Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to \( \varepsilon_1 \) . The jockey is moved along the wire till at a point \( N_1 \), at a distance \( l_1 \) from \( A \), there is no deflection in the galvanometer. We can apply Kirchhoff’s loop rule to the closed loop \( AN1G31A \) and get,
\[ \phi l_{1}+0-\varepsilon_{1}=0 \](3.90)
Similarly, if another emf \( \varepsilon_2 \) is balanced against \( l_2 (AN_2) \)
\[ \phi l_{2}+0-\varepsilon_{2}=0 \](3.91)
From the last two equations
\[ \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{l_{1}}{l_{2}} \](3.92)
This simple mechanism thus allows one to compare the emf’s of any two sources (\( \varepsilon_1\),\( \varepsilon_2 \) ). In practice one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then easily calculated from Eq. (3.92).
We can also use a potentiometer to measure internal resistance of a cell [Fig. 3.28 (b)]. For this the cell (emf \( \varepsilon \) ) whose internal resistance (\( r \)) is to be determined is connected across a resistance box through a key \( K_2 \), as shown in the figure. With key \( K_2 \) open, balance is obtained at length \( l_1 (AN_1) \). Then,
\[ \varepsilon=\phi l_{1} \](3.93(a))
When key \( K_2 \) is closed, the cell sends a current (\( I \)) through the resistance box (\( R \)). If \( V \) is the terminal potential difference of the cell and balance is obtained at length \( l_2 (AN_2) \),
\[ V=\phi l_{2} \](3.93(b))
So, we have
\[ \varepsilon / V=l_{1} / l_{2} \](3.94(a))
But, \( \varepsilon=I(r+R) \) and \( V=IR \), This gives
\[ \varepsilon / V=(r+R) / R \](3.94(b))
From Eq. [3.94(a)] and [3.94(b)] we have
\[ \begin{array}{l} (R+r) / R=l_{1} / l_{2} \\ r=R\left(\frac{l_{1}}{l_{2}}-1\right) \end{array} \](3.95)
Using Eq. (3.95) we can find the internal resistance of a given cell.The potentiometer has the advantage that it draws no current from the voltage source being measured. As such it is unaffected by the internal resistance of the source.
Example 3.10
A resistance of \( R\Omega \) draws current from a potentiometer. The potentiometer has a total resistance \( R_0\Omega \) (Fig. 3.29). A voltage \( V \) is supplied to the potentiometer. Derive an expression for the voltage across \( R \) when the sliding contact is in the middle of the potentiometer.
While the slide is in the middle of the potentiometer only half of its resistance (\( R_0/2 \)) will be between the points \( A \) and \( B \). Hence, the total resistance between \( A \) and \( B \), say, \( R_1 \), will be given by the following expression: \[ \begin{array}{l} \frac{1}{R_{1}}=\frac{1}{R}+\frac{1}{\left(R_{0} / 2\right)} \\ R_{1}=\frac{R_{0} R}{R_{0}+2 R} \end{array} \]
The total resistance between \( A \) and \( C \) will be sum of resistance between \( A \) and \( B \) and \( B \) and \( C \), i.e., \( R_1 + R_0/2 \)
\( \therefore \) The current flowing through the potentiometer will be \[ I=\frac{V}{R_{1}+R_{0} / 2}=\frac{2 V}{2 R_{1}+R_{0}} \]
The voltage \( V_1 \) taken from the potentiometer will be the product of current \( I \) and resistance \( R_1 \), \[ V_{1}=I R_{1}=\left(\frac{2 V}{2 R_{1}+R_{0}}\right) \times R_{1} \]
Substituting for \( R_1 \), we have a \[ V_{1}=\frac{2 V}{2\left(\frac{R_{0} \times R}{R_{0}+2 R}\right)+R_{0}} \times \frac{R_{0} \times R}{R_{0}+2 R} \] \[ \begin{aligned} & V_{1}=\frac{2 V R}{2 R+R_{0}+2 R} \\ \text { or } \quad V_{1} &=\frac{2 V R}{R_{0}+4 R} \end{aligned} \]