NEET2026PhysicsChapterActual
Two ideal diodes D₁ and D₂ are connected in two parallel branches between nodes A and B as shown in the figure. An alternating voltage source V(t) = 200 (50 t) volts is applied across the nodes, where V(t) represents the potential difference V_A - V_B . At time t = 10 ms , which of the following statements is correct regarding the biasing of the diodes?
Options
- AD₁ is reverse biased, D₂ is forward biased
- BBoth D₁ and D₂ are forward biased
- CD₁ is forward biased, D₂ is reverse biased
- DBoth D₁ and D₂ are reverse biased
Correct answer
C. D₁ is forward biased, D₂ is reverse biased
Step-by-step solution
The potential difference between nodes A and B is given by V(t) = V_A - V_B = 200 (50 t) . At time t = 10 ms = 10 10⁻³ s = 0.01 s , the voltage is: V(0.01) = 200 (50 0.01) V(0.01) = 200 (0.5 ) = 200 ( 2 ) = 200 V Since V(0.01) > 0 , we have V_A - V_B > 0 , which implies V_A > V_B . Thus, node A is at a higher potential than node B. From the circuit diagram, the anode (p-side) of diode D₁ is connected to node A and its cathode (n-side) is connected towards node B. Since V_A > V_B , D₁ is forward biased. The cathode