Question

A resistive position transducer with a resistance of 5000 and a shaft stroke of 5.0 in. is used in the arrangement of Fig. (4). Potentiometer R 3 R 4 is also 5000, and V T = 5.0 V. The initial position to be used as a reference point is such that R 1 =R 2 (i.e.. the shaft is at midstroke). At the start of the test, potentiometer R 3 R 4 is adjusted so that the bridge is balanced (V E =0). Assuming that the object being monitored will move a maximum distance of 0.5 in. toward A, what will the new value of V E be?

Answer 1

The output voltage depends on the wiper position

When unloaded the output voltage is given by the formula

${V_O}=\frac{R_1}{R_1+R_2}\times V_T$

when the wiper moves 0.5 in toward A from mid stroke it will be 3.0 inches from B

therefore

$R_2=\frac{3.0}{5.0}\times5000=3000$

$V_o=V_{R2}-V_{R4}$

Since the bridge is balanced at the start of the test, $R_3=R_4=2500$ ohms

Therefore $V_o=(\frac{3000}{5000}\times5)-(\frac{2500}{5000}\times5)=0.5V$

Answer= 0.5 V