Application of Probability to Mechanical Design
55
Figure 2.5 Parallel resistors for Example 2.11.
RT is
RT =- F(RI R2, RI + R2)
Substituting the mean values ~1 and ~2
#T ---- 6667 ohms
the standard deviation is Eq. (2.16)
Vi =1 /~D \2
-]1/2
Li=I\ ~:
J
Figure 2.5 Parallel Resistors for Example 2.11 taking partial derivatives of
RT with respect to R1 and R2 then substituting the mean values yields
ORT _
t~ 2
__
2
- 0.444
ORI (#1 + #2)
2
OR2 (#l + ~2) 2
substituting into the ~T = [(0-4444[300]) 2 + (0.1111 [600])2] I/2
~r = ±149.1 ohms
therefore
(#T,
~T) 6667 + 149.1 oh ms
This is considered to be an exact solution for the standard deviation. The
coefficient of variation is
Cr = ~r 100 = 2.24%