Question: 40. A full wave rectifier circuit with diodes ( \mathrm{D}_{1} ) and ( \mathrm{D}_{2} ) is shown in the figure. If input supply voltage Vin =220 \sin (100 \pi t) volt, then at t=15 m sec
(1) \mathrm{D}_{1} is forward biased, \mathrm{D}_{2} is reverse biased
(2) \mathrm{D}_{1} is reverse biased, \mathrm{D}_{2} is forward biased
(3) \mathrm{D}_{1} and \mathrm{D}_{2} both are forward biased
(4) \mathrm{D}_{1} and \mathrm{D}_{2} both are reverse biased
Answer: Option (2)
Explanation:
The input supply voltage is given as:
V_{in} = 220 \sin (100\pi t)At time t = 15 ms = 15 \times 10^{-3} s,
substitute the value in the equation.
100\pi t = 100\pi \times 15 \times 10^{-3} = \frac{3\pi}{2} \sin \left(\frac{3\pi}{2}\right) = -1Therefore, the instantaneous input voltage is:
V_{in} = 220 \times (-1) = -220\ \text{V}This shows that at t = 15 ms,
the input voltage is negative.
In a centre-tapped full wave rectifier, when the input voltage is negative,
the polarity across the secondary winding reverses.
As a result, diode \mathrm{D}_{2} becomes forward biased and conducts current,
while diode \mathrm{D}_{1} becomes reverse biased and does not conduct.
Hence, at t = 15 ms,
\mathrm{D}_{2} is forward biased and \mathrm{D}_{1} is reverse biased.
Therefore, the correct answer is Option (2).