Question

Please answer this following question regarding the procedure below:

-Explain any errors greater than 1.0% in Table 8. Show your schematics and calculations.

Procedure: 1MΩ Resistor Voltage Divider

1. Wire Figure 1 with the following values. Figure one is two resistors in series and can be seen in the pictures below
2. Complete Table 8 below.
3. Figure 1 is seen below
   w R1 w R2
   

Answer 1

The circuit is shown below. Ri E R Vref the source voltage Here, output voltage Voep is related to the relation, by Vref= V. R2 R+R2 » Vref v (R+R2 Ą - Vref (It ha)= v R2 > Vnef Ri R = - V- Vres Here, the difference between the measurud value and the calculated value arise due to the talerance of the resistors The measured value of a resistor is equal to the average value plus the absolute error. Thus, R, = (R)+OR, and R2 = <R22+ ORL Thus, KR,?+0R)-V)+ov - Vref Vref (R2+ OR2 Vref <R) Re> OV- TERE <R)+OR, (R) +AR Vrer (R) KR) =) OV= [KB (R) + T.CRI) (R2) - TCR2) 2T ov [T = Tolerance (3- van x200% [using equ". M 1-T now, putting values, ov 4.767 % <v) considering Tsmall, 1-7 ~ 1, Vref = 5.714, <V) = 12 V, We get, T= 0.04548 = 4.51. Thus the tolerance of the resistons are 4.5%