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

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

Step-by-step Explanation:

1. Analyze the Input Voltage Equation:
The input supply voltage to the full-wave rectifier circuit is given by:
$V_{\mathrm{in}}=220sin\left(100\pi t\right)\text{V}$

2. Calculate the Input Voltage at the Given Time:
We need to determine the state of the circuit at time $t=15\text{msec}=15\times {10}^{-3}\text{s}$.
Substituting this value of $t$ into the voltage equation:
$V_{\mathrm{in}}=220sin\left(100\pi \times 15\times {10}^{-3}\right)$
$V_{\mathrm{in}}=220sin\left(1.5\pi \right)$
$V_{\mathrm{in}}=220sin\left(\frac{3\pi }{2}\right)$
Since $sin\left(\frac{3\pi }{2}\right)=-1$:
$V_{\mathrm{in}}=220\times (-1)=-220\text{V}$

3. Determine the Biasing of Diodes D1 and D2:
Since the input voltage is negative at this instant, the circuit is operating in the negative half cycle:
- The upper terminal of the secondary winding of the center-tapped transformer becomes negative relative to the center tap, which puts a negative potential at the anode of diode D₁. Therefore, D₁ is reverse biased.
- The lower terminal of the secondary winding becomes positive relative to the center tap, which puts a positive potential at the anode of diode D₂. Therefore, D₂ is forward biased.

Thus, at $t=15\text{msec}$, diode D1 is reverse biased and diode D2 is forward biased.