A series LCR circuit is connected across a so | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The resonant frequency of a series $LCR$ circuit is given by the formula: $f = \frac{1}{2\pi\sqrt{LC}}$.From this formula,we can see that the reso

Vedclass · Accepted Answer

$\left(\frac{1}{3}\right)^{1/4} f_0$

Vedclass · Answer

(D) The resonant frequency of a series $LCR$ circuit is given by the formula: $f = \frac{1}{2\pi\sqrt{LC}}$.From this formula,we can see that the resonant frequency $f$ is inversely proportional to the square root of the inductance $L$,provided the capacitance $C$ remains constant: $f \propto \frac{1}{\sqrt{L}}$.Let the initial resonant frequency be $f_0$ with inductance $L$. Thus,$f_0 = \frac{1}{2\pi\sqrt{LC}}$.When the inductance is increased by $\sqrt{3}$ times,the new inductance $L' = \sqrt{3}L$. The resistance $R$ does not affect the resonant frequency.The new resonant frequency $f'$ is given by: $f' = \frac{1}{2\pi\sqrt{L'C}} = \frac{1}{2\pi\sqrt{(\sqrt{3}L)C}}$.Dividing $f'$ by $f_0$: $\frac{f'}{f_0} = \frac{\frac{1}{2\pi\sqrt{\sqrt{3}LC}}}{\frac{1}{2\pi\sqrt{LC}}} = \frac{1}{\sqrt{\sqrt{3}}} = \left(\frac{1}{3^{1/2}}\right)^{1/2} = \left(\frac{1}{3}\right)^{1/4}$.Therefore,the new resonant frequency is $f' = \left(\frac{1}{3}\right)^{1/4} f_0$.

List-$I$	List-$II$
$(a)$ $\omega L > \frac{1}{\omega C}$	$(i)$ Current is in phase with $emf$
$(b)$ $\omega L = \frac{1}{\omega C}$	$(ii)$ Current lags behind the applied $emf$
$(c)$ $\omega L < \frac{1}{\omega C}$	$(iii)$ Maximum current occurs
$(d)$ Resonant frequency	$(iv)$ Current leads the $emf$

$A$ series $LCR$ circuit is connected across a source of alternating emf of changing frequency and resonates at frequency $f_0$. Keeping capacitance constant,if the inductance $(L)$ is increased by $\sqrt{3}$ times and resistance is increased $(R)$ by $1.4$ times,the resonant frequency now is

Similar Questions

The reading of the ammeter in the following circuit is (in $A$)

In an $LCR$ series circuit,if the frequency is increased,the impedance of the circuit

In an $LCR$ series resonant circuit,at resonance,the voltage across $L$ and $C$ will cancel each other because they are:

The total impedance of a series $L-C-R$ circuit varies with the angular frequency of the $AC$ source connected to it,as shown in the graph. The quality factor $Q$ of the series $L-C-R$ circuit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module