Question 145: AC fundamentals

Question

In series LCR circuit, the capacitance is changed from $C$ to $4 C$. To keep the resonance frequency unchanged, the new inductance should be:

Accepted Answer

The resonance frequency $$ f_0 $$ of an LCR circuit is given by:
$$ f_0 = \frac{1}{2\pi\sqrt{LC}} $$
where:
$$ L $$ is the inductance.
$$ C $$ is the capacitance.
To keep the resonance frequency unchanged when the capacitance is changed from $$ C $$ to $$ 4C $$, we must adjust the inductance $$ L $$ to a new value $$ L' $$ such that:
$$ \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{L' \cdot 4C}} $$
Now solving for $$ L' $$:
$$ \sqrt{LC} = \sqrt{4CL'} $$
Squaring both sides, we have:
$$ LC = 4CL' $$
Dividing both sides by $$ 4C $$, we get:
$$ \frac{L}{4} = L' $$
Thus, the new inductance $$ L' $$ is one fourth of the original inductance $$ L $$. Therefore, to achieve the same resonance frequency with the capacitance increased to $$ 4C $$, the inductance should be reduced to a quarter of its initial value:
$$ L' = \frac{L}{4} $$
This means we have reduced the inductance by $$ \frac{3}{4}L $$, so the correct answer is:
Option C: reduced by $$ \frac{3}{4}L $$

Answer

increased by $2 \mathrm{~L}$

Answer

reduced by $\frac{1}{4} \mathrm{~L}$

Answer

reduced by $\frac{3}{4} \mathrm{~L}$

Answer

increased to $4 \mathrm{~L}$

AC fundamentals

In series LCR circuit, the capacitance is changed from C to 4 C. To keep the resonance frequency unchanged, the new inductance should be:

Choose the correct answer: