Easy
$A$ series $LCR$ circuit containing a resistance $R$ has angular frequency $\omega$. At resonance,the voltage across the resistance and the inductor are $V_R$ and $V_L$ respectively. Then,the value of inductance $L$ will be:
- A$\frac{V_R R}{V_L \omega}$
- B$\frac{V_L}{V_R R \omega}$
- C$\frac{V_R \omega}{V_L R}$
- D$\frac{V_L R}{V_R \omega}$
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Similar Questions
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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In an $LCR$ circuit, at a particular angular frequency $\omega_0$, the capacitive reactance and inductive reactance are the same. If the angular frequency is doubled to $2\omega_0$, what will be the ratio of the reactance of the capacitor to that of the inductor?
DifficultMHT CET 2026
View SolutionObtain the equation of bandwidth for an $L-C-R$ series $AC$ circuit and deduce the equation of $Q$ factor.
Medium
View Solution$A$ resonance circuit having inductance and resistance $2 \times 10^{-4} \ H$ and $6.28 \ \Omega$ respectively oscillates at $10 \ MHz$ frequency. The value of the quality factor of this resonator is .........
$[\pi = 3.14]$
$[\pi = 3.14]$
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