Question 3.2: The resistors used in the voltage-divider circuit shown in F...

From Eq. 3.22,

v_{2}=i R_{2}=v_{s} \frac{R_{2}}{R_{1}+R_{2}}\quad (3.22).

the maximum value of v_{o} occurs when R_{2} is 10% high and R_{1} is 10% low, and the minimum value of  v_{o} occurs when R_{2} is 10% low and R_{1} is 10% high. Therefore

v_{o}\left(\max \right)=\frac{\left(100\right)\left(110\right) }{110+22.5}=83.02 V, .

v_{o}\left(\min \right)=\frac{\left(100\right)\left(90\right) }{90+27.5}=76.60 V, .

Thus, in making the decision to use 10% resistors in this voltage divider, we recognize that the no-load output voltage will lie between 76.60 and 83.02 V.