Question

The circuit that follows is for a laboratory potentiometer for measuring unknown voltages V_x.

Assume the resistor AB is a slide wire whose resistance is directly proportional to its length. With the standard Weston cell (1.018 V) in the circuit, a null point was observed when contact C was moved to a position 84.3 cm from point A. When the Weston cell was replaced with a cell of an unknown voltage V_x, null was observed at 44.3 cm. Find the value of V_x.

Short answer

The voltage of the cell is V_x=1.937V

Step-by-step solution

Given information

Given the diagram below:

Point C is at 84.3 cm which is the null point of Weston cell of voltage 1.018V. A null point is observed at 44.3 cm for unknown cell of voltage V_x

Explanation:

The expression for standard voltage will be:

$V=V_b\frac{AC}{AB}$

where V_c is the voltage of the battery.

For Weston cell.

$$\begin{gathered} 1.018=V_b\frac{84.3}{AB}\dots \dots \dots \left(1\right) \\ Forcellofunknownvoltage \\ {V}_x=V_b\frac{44.3}{AB}\dots \dots \dots \dots \left(2\right) \\ Dividing\left(1\right)by\left(2\right)weget: \\ {V}_x=1.018V\times \frac{84.3}{44.3} \\ {V}_x=1.937V \end{gathered}$$