A current element X is connected across an AC source of emf \(V = V_0\ sin\ 2πνt\). It is found that the voltage leads the current in phase by \(\frac{π}{ 2}\) radian. If element X was replaced by element Y, the voltage lags behind the current in phase by \(\frac{π}{ 2}\) radian.
(I) Identify elements X and Y by drawing phasor diagrams.
(II) Obtain the condition of resonance when both elements X and Y are connected in series to the source and obtain expression for resonant frequency. What is the impedance value in this case?
In AC circuits, the relationship between voltage and current can be represented using phasor diagrams:

When a capacitor and an inductor are connected in series with a resistor (forming an RLC circuit), resonance occurs under specific conditions. The condition of resonance is when the inductive reactance (\( X_L \)) equals the capacitive reactance (\( X_C \)).
The resonance condition is given by:
\[ X_L = X_C \Rightarrow \omega L = \frac{1}{\omega C} \]
Solving for \( \omega \), we get:
\[ \omega^2 = \frac{1}{LC} \Rightarrow \omega = \frac{1}{\sqrt{LC}} \]
The resonant frequency \( f \) is given by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi \sqrt{LC}} \]
At resonance, the net reactance becomes zero because \( X_L = X_C \). Therefore, the total impedance \( Z \) in the circuit is purely resistive:
\[ Z = R \]
Resonance in an RLC circuit occurs when the inductive reactance equals the capacitive reactance, leading to maximum current and a purely resistive impedance. The resonant frequency is given by:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]
A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?