Difficult
In a series $RLC$ circuit,the resonance frequency is $12 \text{ kHz}$. If $R = 5 \text{ } \Omega$ and $X_L$ at resonance is $300 \text{ } \Omega$,the half-power frequencies will be:
- A$11900 \text{ Hz}, 12100 \text{ Hz}$
- B$9000 \text{ Hz}, 10000 \text{ Hz}$
- C$200 \text{ Hz}, 210 \text{ Hz}$
- DNone of the above
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Match List-$I$ with List-$II$ :
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 $emf$
Choose the correct answer from the options given below :
| 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 $emf$ |
Choose the correct answer from the options given below :
Resonance frequency of $L-C-R$ series $AC$ circuit is $f_0$. Now,if the inductance is reduced to $\frac{1}{4}$ times and the capacitance is increased to $16$ times,then the new resonance frequency becomes:
The resonant frequency of a series $LCR$ circuit is $f_R$. The circuit is connected to a sinusoidally alternating e.m.f. of frequency $2 f_R$. The inductive reactance becomes $X_{L_1}$ and capacitive reactance becomes $X_{C_1}$ after changing the frequency. $X_{C_1}$ is equal to:
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