Easy
$A$ $L-C-R$ series $AC$ circuit is tuned to resonance. The impedance of the circuit is now . . . . . . .
- A$R$
- B$\left[R^{2}+\left(\frac{1}{\omega C}-\omega L\right)^{2}\right]^{\frac{1}{2}}$
- C$\left[R^{2}+(\omega L)^{2}+\left(\frac{1}{\omega C}\right)^{2}\right]^{\frac{1}{2}}$
- D$\left[R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}\right]^{\frac{1}{2}}$
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$A$ series resonant circuit consists of an inductor $L$ and capacitor $C$ which produces resonant frequency $f$. If $L$ is increased by $2L$ (making the new inductance $L' = L + 2L = 3L$) and $C$ is changed to $9C$, the new resonant frequency will be:
DifficultMHT CET 2026
View SolutionThe resonant frequency of a series $LCR$ circuit is $f$. The circuit is now connected to a sinusoidally alternating e.m.f. of frequency $2f$. The new reactance $X_{L}^{\prime}$ and $X_{C}^{\prime}$ are related as:
An $LCR$ series circuit of capacitance $62.5 \, nF$ and resistance of $50 \, \Omega$ is connected to an $A.C.$ source of frequency $2.0 \, kHz$. For the maximum value of the amplitude of current in the circuit,the value of inductance is $.......... \, mH$. (take $\pi^2 = 10$)
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Easy
View SolutionWhat is the sharpness of resonance and obtain an equation for the $Q$ factor?
Easy
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