Step 1: Understanding the Concept:
A full-wave rectifier is an electronic circuit that converts both halves (positive and negative) of an alternating current (AC) input into a pulsating direct current (DC) output. It uses two p-n junction diodes and a center-tapped transformer. The basic principle is that a p-n junction diode allows current to flow only when it is forward-biased.
Step 2: Circuit Diagram and Waveforms: 
During the positive half-cycle of the AC input: The end A of the secondary coil is positive with respect to the center tap C, and the end B is negative. This makes the diode \(D_1\) forward-biased and the diode \(D_2\) reverse-biased. As a result, current flows through diode \(D_1\) and the load resistor \(R_L\) from top to bottom. No current flows through \(D_2\).
During the negative half-cycle of the AC input: The polarity reverses. The end A becomes negative, and the end B becomes positive with respect to the center tap C. This makes diode \(D_2\) forward-biased and diode \(D_1\) reverse-biased. Current now flows through diode \(D_2\) and the load resistor \(R_L\). Importantly, the direction of current through \(R_L\) is again from top to bottom. \end{enumerate} In both half-cycles, the current flows through the load resistor in the same direction. Therefore, we get a continuous, pulsating DC voltage across the load, as shown in the output waveform diagram.
Step 4: Final Answer:
A full-wave rectifier uses a center-tapped transformer and two diodes to convert both halves of the AC cycle into a unidirectional (DC) output, making it more efficient than a half-wave rectifier.
Assuming in forward bias condition there is a voltage drop of \(0.7\) V across a silicon diode, the current through diode \(D_1\) in the circuit shown is ________ mA. (Assume all diodes in the given circuit are identical) 


For the given logic gate circuit, which of the following is the correct truth table ? 