A full wave rectifier circuit with diodes (D1) and (D2) is shown in the figure. If input supply voltage V in = 220sin ( 100πt ) volt, then at t = 15 m sec

The correct answer is: D1 is reverse biased, D2 is forward biased

Step-by-step Explanation:

1. Identify the given parameters:
The input voltage to the primary of the transformer is given by:
$V_{\text{in}}=220sin\left(100\pi t\right)\text{V}$
We need to determine the state of the diodes D₁ and D₂ at time:
$t=15\text{ms}=15\times {10}^{-3}\text{s}=0.015\text{s}$

2. Calculate the phase angle at the given instant:
Substitute the value of t into the angular argument of the sine function:
$\theta =100\pi t=100\pi \times 0.015=1.5\pi =\frac{3\pi }{2}\text{rad}$

3. Determine the sine value and the input voltage:
Calculate the value of the sine function at this phase:
$sin\left(\frac{3\pi }{2}\right)=-1$
Now, find the value of the input voltage:
$V_{\text{in}}=220\times (-1)=-220\text{V}$
Since the input voltage is negative, the AC source is in its negative half-cycle.

4. Analyze the diode biasing:
The schematic shows a center-tapped transformer configuration:
- During the positive half-cycle (V_in > 0), the upper terminal of the secondary winding is at a positive potential and the lower terminal is at a negative potential relative to the center tap. This makes the anode of D₁ positive (forward biased) and the anode of D₂ negative (reverse biased).
- During the negative half-cycle (V_in < 0), which occurs at t = 15 ms, the polarity across the secondary winding reverses. The upper terminal of the secondary winding becomes negative relative to the center tap, and the lower terminal becomes positive.
Consequently:
- The anode of diode D₁ is connected to a negative potential, making it reverse biased.
- The anode of diode D₂ is connected to a positive potential, making it forward biased.