Concept: A full-wave rectifier converts both halves of an AC input into pulsating DC output using p–n junction diodes. It provides higher efficiency than a half-wave rectifier.

Circuit Diagram (Centre-tapped full-wave rectifier): Components:
Centre-tapped transformer
Two diodes \( D_1, D_2 \)
Load resistor \( R_L \)
Connections:
Anodes of diodes connected to the ends of secondary winding
Cathodes joined together and connected to load
Centre tap connected to other end of load
Working: Positive half cycle:
Upper end of secondary becomes positive
Diode \( D_1 \) forward biased → conducts
Diode \( D_2 \) reverse biased → off
Current flows through \( R_L \) in one direction
Negative half cycle:
Lower end of secondary becomes positive
\( D_2 \) conducts, \( D_1 \) off
Current again flows through load in same direction
Thus, both halves of AC are rectified → full-wave rectification. Input–Output Waveforms: Input waveform:
Sinusoidal AC wave
Positive and negative halves symmetric
Output waveform:
Both halves appear positive
Pulsating DC with double frequency of input
Graph description:
Input: sine wave about zero axis
Output: series of positive humps (no negative portion)
Key Features:
Output frequency = \( 2f \)
Higher efficiency than half-wave rectifier
Less ripple

A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).