If resonant frequency is f and capacitance become 4 times, then the resonant frequency will be:
\[
{\text{A}}{\text{. }}\dfrac{f}{2} \\
{\text{B}}{\text{. 2}}f \\
{\text{C}}{\text{. }}f \\
{\text{D}}{\text{. }}\dfrac{f}{4} \\
\]
Answer
657.3k+ views
Hint: A LC circuit is made up of a capacitance and an inductance. The resonant frequency of a LC circuit is inversely proportional to the square root of capacitance and inductance of the circuit so with increase in capacitance and inductance, the resonant frequency decreases.
Formula used:
The resonance frequency is obtained when reactance of capacitor given as ${X_C} = \dfrac{1}{{\omega C}}$ matches the reactance of inductance given as ${X_L} = \omega L$ where $\omega = 2\pi \nu $
The resonant frequency of LC circuit is given as
$f = \dfrac{1}{{2\pi \sqrt {LC} }}$
where f is the resonant frequency, L is the inductance of the coil and C is the capacitance.
Complete step-by-step answer:
We are given that initially a LC circuit has resonant frequency f. If initial capacitance is C and initial inductance is L then the expression for resonant frequency can be written as
$f = \dfrac{1}{{2\pi \sqrt {LC} }}{\text{ }}...{\text{(i)}}$
Now, the capacitance of the coil has increased 4 times. The new capacitance is given as
$C' = 4C$
The inductance is independent of capacitance and remains same, therefore,
$L' = L$
But the resonant frequency changes because it is dependent on capacitance of the coil. The expression for new resonant frequency can be written as
$f' = \dfrac{1}{{2\pi \sqrt {L'C'} }}{\text{ }}...{\text{(ii)}}$
Substituting the values of L’ and C’, we get
$
f' = \dfrac{1}{{2\pi \sqrt {L \times 4C} }} \\
{\text{ = }}\dfrac{1}{2} \times \dfrac{1}{{2\pi \sqrt {LC} }} \\
$
Using equation (i) here, we get
$f' = \dfrac{f}{2}$
Therefore, the resonant frequency becomes half on increasing the capacitance four times. Hence, the correct answer is option A.
Note: The condition of resonance is obtained in a circuit when reactance of the capacitance is equal to the reactance of inductance. The resonant frequency signifies the oscillations of electrical energy between inductance and capacitance.
Formula used:
The resonance frequency is obtained when reactance of capacitor given as ${X_C} = \dfrac{1}{{\omega C}}$ matches the reactance of inductance given as ${X_L} = \omega L$ where $\omega = 2\pi \nu $
The resonant frequency of LC circuit is given as
$f = \dfrac{1}{{2\pi \sqrt {LC} }}$
where f is the resonant frequency, L is the inductance of the coil and C is the capacitance.
Complete step-by-step answer:
We are given that initially a LC circuit has resonant frequency f. If initial capacitance is C and initial inductance is L then the expression for resonant frequency can be written as
$f = \dfrac{1}{{2\pi \sqrt {LC} }}{\text{ }}...{\text{(i)}}$
Now, the capacitance of the coil has increased 4 times. The new capacitance is given as
$C' = 4C$
The inductance is independent of capacitance and remains same, therefore,
$L' = L$
But the resonant frequency changes because it is dependent on capacitance of the coil. The expression for new resonant frequency can be written as
$f' = \dfrac{1}{{2\pi \sqrt {L'C'} }}{\text{ }}...{\text{(ii)}}$
Substituting the values of L’ and C’, we get
$
f' = \dfrac{1}{{2\pi \sqrt {L \times 4C} }} \\
{\text{ = }}\dfrac{1}{2} \times \dfrac{1}{{2\pi \sqrt {LC} }} \\
$
Using equation (i) here, we get
$f' = \dfrac{f}{2}$
Therefore, the resonant frequency becomes half on increasing the capacitance four times. Hence, the correct answer is option A.
Note: The condition of resonance is obtained in a circuit when reactance of the capacitance is equal to the reactance of inductance. The resonant frequency signifies the oscillations of electrical energy between inductance and capacitance.
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