Page 166 - Numerical Analysis Using MATLAB and Excel

Classification

Solution:

∂Z ∂Z
We will first find the partial derivatives ------- and ---------- ; then we compute the change in impedance
∂R ∂X C
from the total differential dZ . Thus,

X
∂Z --------------------------- and ---------- = ---------------------------
R
∂Z
C
------- =
∂R 2 2 ∂X 2 2
R + X C C R + X C
and
∂Z ∂Z R dR + X dX
C
C
dZ = ------- dR + ---------- dX = ----------------------------------------
∂R ∂X C C R + X C 2
2
and by substitution of the given values
(
(
)
–
–
----------------------------------------------------- =
-------------------------- =
dZ = 4 0.25 + 3 0.125 ) 1 0.375 0.125
2
4 + 3 2 5
Therefore, if increases by 6.25 % and X C decreases by 4.167% , the impedance increases by
R
Z
4.167% .
Example 5.5
A light bulb is rated at 120 volts and 75 watts. If the voltage decreases by volts and the resis-5
tance of the bulb is increased by 8 Ω , by how much will the power change?
Solution:

At V = 120 volts and P = 75 watts, the bulb resistance is

------------ =
------ =
R = V 2 120 2 192 Ω
P 75
and since

------
------ then ------- =
P = V 2 ∂P 2V ∂P – V 2
------- and ------- =
R ∂V R ∂R R 2
and the total differential is
(
∂P ∂P 2V V 2 2 120 ) 120 2
)
dP = ------- dV + ------- dR = -------dV ------– dR = ----------------- –( 5 – ----------- 8() = – 9.375
∂V ∂R R R 2 192 192 2
That is, the power will decrease by 9.375 watts.

Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−5

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