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(B) The initial reactance of the circuit is $X_i = |X_{Ci} - X_{Li}| = |\frac{1}{\omega C_i} - \omega L_i|$.Initially,the capacitor is filled with liq

Vedclass · Accepted Answer

$\frac{(1 - K)}{K(1 - \mu_r)}$

Vedclass · Answer

(B) The initial reactance of the circuit is $X_i = |X_{Ci} - X_{Li}| = |\frac{1}{\omega C_i} - \omega L_i|$.Initially,the capacitor is filled with liquid $(C_i = \frac{K \varepsilon_0 A}{\ell})$ and the inductor is empty $(L_i = \mu_0 n^2 A \ell)$.So,$X_i = |\frac{\ell}{\omega K \varepsilon_0 A} - \omega \mu_0 n^2 A \ell|$.After the liquid fills the inductor,the capacitor becomes empty $(C_f = \frac{\varepsilon_0 A}{\ell})$ and the inductor is filled with liquid $(L_f = \mu_0 \mu_r n^2 A \ell)$.So,$X_f = |\frac{\ell}{\omega \varepsilon_0 A} - \omega \mu_0 \mu_r n^2 A \ell|$.Taking the ratio $\frac{X_i}{X_f} = \left| \frac{\frac{\ell}{\omega K \varepsilon_0 A} - \omega \mu_0 n^2 A \ell}{\frac{\ell}{\omega \varepsilon_0 A} - \omega \mu_0 \mu_r n^2 A \ell} \right|$.Multiplying numerator and denominator by $\omega \varepsilon_0 A / \ell$,we get $\frac{X_i}{X_f} = \left| \frac{\frac{1}{K} - \omega^2 \mu_0 \varepsilon_0 n^2 A^2}{1 - \omega^2 \mu_0 \varepsilon_0 \mu_r n^2 A^2} \right|$.Using $\mu_0 \varepsilon_0 = \frac{1}{c^2}$ and $\omega^2 A^2 n^2 = c^2$,the term $\omega^2 \mu_0 \varepsilon_0 n^2 A^2 = \frac{c^2}{c^2} = 1$.Thus,$\frac{X_i}{X_f} = \left| \frac{\frac{1}{K} - 1}{1 - \mu_r} \right| = \left| \frac{1 - K}{K(1 - \mu_r)} \right| = \frac{1 - K}{K(1 - \mu_r)}$ (assuming $K > 1$ and $\mu_r > 1$ or vice versa to maintain magnitude).

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