Question

A series RLC circuit has a $200kHz$ resonance frequency. What is the resonance frequency if

a. The resistor value is doubled?

b. The capacitor value is doubled?

Short answer

(a) Resonance frequency when resistor value is doubled remains unchanged. So $\omega =200kHz$

(b) Resonance frequency when capacitor value is doubled is$\omega =142.85kHz$

Step-by-step solution

Part(a) Step 1: Given information 

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

we need to find resonance frequency when resistor value is doubled.

Part(a) Step 2: Solution

At resonance condition

$$\begin{gathered} X_L=X_C \\ \omega =1/\sqrt{LC} \end{gathered}$$

As value of resonance frequency do not depend on value of resistance.

Hence we can say that value of resonance frequency will remain same $i.e200kHz$on doubling value of resistor.

Part(b) step 1: Given information 

RLC circuit having resonance frequency(${\omega }_i$)=$200kHz$

we need to find resonance frequency when capacitor value is doubled.

Part(b) Step 2: Solution

At resonance condition

$$\begin{gathered} X_L=X_C \\ \omega =1/\sqrt{LC} \end{gathered}$$

When capacitor value is doubled

$$\begin{gathered} \omega =1/\sqrt{L\times \left(2C\right)} \\ {\omega }_f={\omega }_i/\sqrt{2} \\ {\omega }_f=200/\sqrt{2} \\ {\omega }_f=142.85kHz \end{gathered}$$

Hence on doubling value of capacitor resonance frequency becomes$142.85kHz$.