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 small candle,2.5cm in size is placed at 27cm in front of a concave mirror of radius of curvature 36cm.At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image.If the is moved closer to the mirror, how would the screen have to be moved?
A 4.5cm needle is placed 12cm away from a convex mirror of focal length 15cm. Give the location of the image and the magnification.Describe what happens as the needle is moved farther from the mirror.
A tank is filled with water to a height of 12.5cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

(a)and(b)show the refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface
A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge? The refractive index of water is 1.33. (Consider the bulb to be a point source.)
