Question 326: AC fundamentals

Question

A series resonant circuit consists of inductor ' $L$ ' of negligible resistance and a capacitor ' $C$ ' which produces resonant frequency ' $f$ '. If $L$ is changed to 3 L and ' C ' is changed to 6 C , the resonant frequency will become.

Accepted Answer

In a series resonant circuit, the resonant frequency is given by the formula:
$$ f = \frac{1}{2\pi \sqrt{LC}} $$
When the inductor $L$ is changed to $3L$, and the capacitor $C$ is changed to $6C$, the new resonant frequency $f'$ becomes:
$$ f' = \frac{1}{2\pi \sqrt{(3L)(6C)}} $$
Simplifying the expression inside the square root:
$$ f' = \frac{1}{2\pi \sqrt{18LC}} $$
This can be further simplified to:
$$ f' = \frac{1}{2\pi \sqrt{(9 	imes 2)LC}} = \frac{1}{2\pi \cdot 3 \cdot \sqrt{2LC}} $$
This simplifies to:
$$ f' = \frac{1}{3 \sqrt{2}} \cdot \frac{1}{2\pi \sqrt{LC}} $$
Given that the original resonant frequency $f$ is:
$$ f = \frac{1}{2\pi \sqrt{LC}} $$
The new frequency $f'$ can be expressed in terms of $f$ as:
$$ f' = \frac{f}{3 \sqrt{2}} $$
Thus, the resonant frequency will become:
Option D: $\frac{\mathrm{f}}{3 \sqrt{2}}$

Answer

$\frac{\mathrm{f}}{6}$

Answer

$\frac{\mathrm{f}}{3}$

Answer

$\frac{\mathrm{f}}{2 \sqrt{2}}$

Answer

$\frac{\mathrm{f}}{3 \sqrt{2}}$

AC fundamentals

A series resonant circuit consists of inductor ' L ' of negligible resistance and a capacitor ' C ' which produces resonant frequency ' f '. If L is changed to 3 L and ' C ' is changed to 6 C , the resonant frequency will become.

Choose the correct answer: