Question

A series RLC circuit has a $200kHz$ resonance frequency. What is the resonance frequency if the capacitor value is doubled and, at the same time, the inductor value is halved?

Short answer

Resonance condition is $\omega =1/\sqrt{LC}$where $L$is inductance and $C$is conductance.

Step-by-step solution

Given Information

We are given that a RLC circuit having resonance frequency,${\omega }_i=200kHz$

we need to find resonance frequency when

Final Capacitance($C_f$)=$2C_i$where localid="1649860894885" $C_i$is initial conductance

Final Inductance($L_f$)= $L_i/2$where $L_i$is initial inductance

Explanation

As we know at resonance $\omega =1/\sqrt{LC}$

Therefore,

$$\begin{gathered} {\omega }_f=1/\sqrt{L_f{C}_f} \\ {\omega }_f=1/\sqrt{2C_i\times L_i/2} \\ {\omega }_f={\omega }_i=200kHz \end{gathered}$$

Hence, our required answer is$200kHz$.