1. ## Maximum Percentage error

The resonant frequency of an oscillation in electrical circuits is given by the formula f = 1/2pÖLC. If the error in measuring L is 4%, and that measuring C is 2%, calculate the maximum percentage error in calculating f?

Many thanks

2. Hello, Stazzer5!

I think I've solved it.
Someone please check my reasoning and my work.

The resonant frequency of an oscillation in electrical circuits
. . is given by the formula: .$\displaystyle f \:=\:\tfrac{\pi}{2}\sqrt{LC}$

If the error in measuring $\displaystyle L$ is 4%, and that measuring $\displaystyle C$ is 2%,
. . calculate the maximum percentage error in calculating $\displaystyle f$?
We are given: . $\displaystyle \begin{Bmatrix}\dfrac{dL}{L}\:=\: 0.04 \\ \\[-3mm]\dfrac{dC}{C} \:=\: 0.02 \end{Bmatrix}$ . [1]

We have: .$\displaystyle f \:=\:\tfrac{\pi}{2}(LC)^{\frac{1}{2}}$

Take differentials: .$\displaystyle df \;=\;\tfrac{\pi}{2}\cdot\tfrac{1}{2}(LC)^{-\frac{1}{2}}(L\,dC + C\,dL) \;=\;\frac{\pi(L\,dC + C\,dL)}{4\sqrt{LC}}$

Divide by $\displaystyle f\!:\;\;\frac{df}{f} \;=\;\frac{\frac{\pi(L\,dC + C\,dL)}{4\sqrt{LC}}}{\frac{\pi}{2}\sqrt{LC}} \;=\;\frac{L\,dC + C\,dL}{2LC} \;=\;\frac{L\,dC}{2LC} + \frac{C\,dL}{2LC}$

. . Hence: .$\displaystyle \frac{df}{f} \;=\;\frac{1}{2}\left(\frac{dC}{C} + \frac{dL}{L}\right)$

Substitute [1]: .$\displaystyle \frac{df}{f} \;=\;\tfrac{1}{2}(0.04 + 0.02) \;=\;0.03$

Therefore, the maximum percentage error in calculating $\displaystyle f$ is $\displaystyle \boxed{3\%}$

3. Hello everyone

I don't know enough about electrical circuits to say for sure, but, according to this Wikipedia article, it looks as though the formula should be:
$\displaystyle f= \frac{1}{2\pi\sqrt{LC}}$
It's easiest to deal with percentage errors with this type of equation if you take logs before differentiating:

So, writing $\displaystyle f$ as:
$\displaystyle f=(2\pi)^{-1}L^{-\tfrac12}C^{-\tfrac12}$
we get:
$\displaystyle \ln(f) = -\ln(2\pi)-\tfrac12\ln(L) - \tfrac12\ln(C)$
Then on differentiating we get:
$\displaystyle \frac{df}{f}=-\frac12\left(\frac{dL}{L}+\frac{dC}{C}\right)$
You can then substitute in the values for $\displaystyle \frac{dL}{L}$ and $\displaystyle \frac{dC}{C}$ as Soroban has done. (You get the same answer, 3%, but with a negative sign, indicating that if $\displaystyle L$ and $\displaystyle C$ are larger than the true value, $\displaystyle f$ will be smaller.)

Note that any formula where you have powers of a number of variables can be dealt with in this way:
$\displaystyle y = k\,a^p\,b^q\,c^r...$

$\displaystyle \Rightarrow \ln(y) = \ln(k) + p\ln(a)+q\ln(b)+r\ln(c) + ...$

$\displaystyle \Rightarrow \frac{dy}{y}= p\frac{da}{a}+q\frac{db}{b}+r\frac{dc}{c}+ ...$

4. hi there,

Thank you very much for your help.

stazzer5