Question

In an LCR circuit having \[L=8H\], \[C=0.5\mu F\] and \[R=100\Omega \] in series, the resonance frequency in \[rad/s\] is?
A) \[600\]
B) \[200\]
C) \[\dfrac{250}{\pi }\]
D) \[500\]

Answer 1

Hint: Before finding the answer to the given question, we should be able to answer what is resonance frequency. The resonant frequency is the oscillation of a system at its natural or unforced resonance. Resonance occurs when a system can store and easily transfer energy between different storage modes, such as Kinetic energy or Potential energy as you would find with a simple pendulum. Resonance is witnessed in objects that are in equilibrium with acting forces and could keep vibrating for a long time under perfect conditions.

Formula Used:
 \[f=\dfrac{1}{2\pi \sqrt{LC}}\]

Complete step by step solution:
We have been given the value of the inductance of the inductor, the value of the capacitance of the capacitor and the value of the resistance in the circuit.
Capacitance of the Capacitor \[(C)=0.5\mu F=0.5\times {{10}^{-6}}F\] since \[1\mu F={{10}^{-6}}F\].
The value of inductance of the inductor \[(L)=8H\].
From the formula to calculate the resonance frequency of the circuit, we have \[f=\dfrac{1}{2\pi \sqrt{LC}}\] where \[f\] is the resonant frequency, \[L\] is the inductance value and \[C\] is the capacitance value.
Substituting the values, we get
\[\begin{align}
  & f=\dfrac{1}{2\pi \sqrt{8\times 0.5\times {{10}^{-6}}}} \\
 & \Rightarrow f=\dfrac{{{10}^{3}}}{4\pi } \\
 & \Rightarrow f=\dfrac{250}{\pi }rad/s \\
\end{align}\]

Hence we can now say that option (C) is the correct answer to the given question.

Note: We must always keep in mind that only the inductor and the capacitor are responsible for resonance in a circuit. Therefore, even though we were given the value of resistance present in the circuit, we did not make any use of it. Also, we had been asked the value of the resonant frequency in \[rad/s\], so we did not need to convert but if we had been asked the answer in any other unit, a conversion would be elemental. So keep an eye out for the unit in which the answer is to be given.