In the series L – C – R circuit, the voltage across resistance, capacitance and inductance are 30V each at frequency f = f₀ .

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It the inductor is short-circuited, the voltage across the capacitor will be $30 \sqrt{2} V$

If the capacitor is short-circuited, the voltage drop across the inductor will be $\frac{30}{\sqrt{2}} V$

If the frequency is changed to 2f₀, the ratio of reactance of the inductor to that of the capacitor is 4 : 1.

If the frequency is changed to 2f₀, the ratio of the reactance of the inductor to that of the capacitor is 1 : 4.

answer is B, C.

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Detailed Solution

$\begin{aligned} V_{L} = V_{C} = V_{R}; \\ \Rightarrow X_{L} = X_{C} = R \end{aligned}$
when inductor is short circuited
$Z = \sqrt{R^{2} + x_{c}^{2}} = \sqrt{2} R$
$\begin{array}{l} ∴ I = \frac{30}{Z} = \frac{30}{\sqrt{2} R} \\ ∴ V_{L} = {ix}_{L} = \frac{30}{\sqrt{2} R} \times R = \frac{30}{\sqrt{2}} \end{array}$
$∴ (A)$ is incorrect and with similar calculations (B) will be correct. Here f0 is the resonance frequency as v_L = v_C
$\begin{array}{l} \Rightarrow f_{0} = \frac{1}{2 π \sqrt{LC}} \\ and ω_{0} = 2 {πf}_{0} = \frac{1}{\sqrt{LC}} \\ \frac{x_{L}}{x_{C}} = \frac{ωL}{1 / ωC} = ω^{2} LC \\ Given f = 2 f_{0} \\ \Rightarrow ω = 2 ω_{0} \end{array}$
$∴ \frac{x_{L}}{x_{C}} = 4$
$∴ (C)$ is also correct.