Assuming the potentiometer resistance {{R}}_{1} is proportional to x, derive the expression for the output voltage {{v}}_{o} as a function of x.
The length of the pot is L and its total resistance is {{R}}_{1} + {{R}}_{2}. Figure 6.2.5b shows the circuit diagram of the system. From the voltage-divider rule,
v_{o}={\frac{R_{1}}{R_{1}+R_{2}}}V (1)
Because the resistance {{R}}_{1} proportional to x,
R_{1}=(R_{1}+R_{2})\left(\frac{x}{L}\right)Substituting this into equation (1) gives
v_{o}={\frac{x}{L}}V=K xwhere K = V/L is the gain of the pot.