Question: Resonance frequency of a circuit is f. If the capacitance is made 9 times the initial value, then th...

Question

Resonance frequency of a circuit is f. If the capacitance is made 9 times the initial value, then the resonance frequency will become

Answer 1

The resonance frequency ( $f$ ) of an LC circuit is given by the formula:

$f = \frac{1}{2\pi\sqrt{LC}}$

From this formula, we can see that the resonance frequency is inversely proportional to the square root of the capacitance:

$f \propto \frac{1}{\sqrt{C}}$

Let the initial resonance frequency be $f_1$ and the initial capacitance be $C_1$ .

So, $f_1 = \frac{1}{2\pi\sqrt{LC_1}}$

The problem states that the capacitance is made 9 times the initial value. So, the new capacitance $C_2 = 9C_1$ .

Let the new resonance frequency be $f_2$ .

$f_2 = \frac{1}{2\pi\sqrt{LC_2}}$

Substitute $C_2 = 9C_1$ into the equation for $f_2$ :

$f_2 = \frac{1}{2\pi\sqrt{L(9C_1)}}$

$f_2 = \frac{1}{2\pi\sqrt{9}\sqrt{LC_1}}$

$f_2 = \frac{1}{2\pi \cdot 3\sqrt{LC_1}}$

$f_2 = \frac{1}{3} \left( \frac{1}{2\pi\sqrt{LC_1}} \right)$

Since $f_1 = \frac{1}{2\pi\sqrt{LC_1}}$ , we can substitute $f_1$ into the equation for $f_2$ :

$f_2 = \frac{1}{3} f_1$

Given that the initial resonance frequency is $f$ , so $f_1 = f$ .

Therefore, the new resonance frequency $f_2 = \frac{f}{3}$ .