Question

In an LCR series circuit, the resonance frequency is \( f \). If the capacitance is made 4 times, what will be the new resonance frequency?

• A. \(4f\)
• B. \(2f\)
• C. \(f/2\)
• D. \(f/4\)

Hint:

Recall the resonance frequency formula for a series LCR circuit and note that only the capacitance is changing here. Instead of resubstituting numbers into the full formula, try expressing the new frequency as a ratio of the old one using how $f$ depends on $C$.

Answer 1

The Correct Option is C

Concept: For an LCR series circuit, the resonance frequency is given by \[ f = \frac{1}{2\pi\sqrt{LC}} \] where \(L\) is the inductance and \(C\) is the capacitance. Thus, resonance frequency is inversely proportional to the square root of capacitance.

Step 1: Write the formula for resonance frequency. \[ f = \frac{1}{2\pi\sqrt{LC}} \]

Step 2: Substitute the new capacitance. Given that the capacitance becomes \[ C' = 4C \] The new frequency becomes \[ f' = \frac{1}{2\pi\sqrt{L(4C)}} \]

Step 3: Simplify the expression. \[ f' = \frac{1}{2\pi\sqrt{4LC}} \] \[ f' = \frac{1}{2\pi \cdot 2\sqrt{LC}} \] \[ f' = \frac{1}{2}\left(\frac{1}{2\pi\sqrt{LC}}\right) \] \[ f' = \frac{f}{2} \] \[ \boxed{f' = \frac{f}{2}} \]

Answer 2

Concept:

• When two quantities are connected by a power-law relationship, such as $f \propto C^{-1/2}$, a known change in one quantity can be translated directly into the change in the other using a ratio, without resubstituting into the full formula.
• For the resonance frequency of a series LCR circuit, $f = \dfrac{1}{2\pi\sqrt{LC}}$, so with $L$ held fixed, $f\sqrt{C}$ stays the same constant value before and after $C$ changes.

Step 1: Write the proportionality relation with $L$ fixed.
$f \propto \dfrac{1}{\sqrt{C}}$, which means $f\sqrt{C}$ is constant for fixed $L$.

Step 2: Set up the ratio between the new and old frequency using this constant.
$f_1\sqrt{C_1} = f_2\sqrt{C_2}$
$\dfrac{f_2}{f_1} = \sqrt{\dfrac{C_1}{C_2}}$

Step 3: Substitute the given change in capacitance.
$C_2 = 4C_1$
$\dfrac{f_2}{f_1} = \sqrt{\dfrac{C_1}{4C_1}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}$

Step 4: Solve for the new frequency.
$f_2 = f_1 \times \dfrac{1}{2}$

Final Answer: $f_2 = \dfrac{f_1}{2}$, i.e. the new resonance frequency is $\dfrac{f}{2}$