Question
The output of the given circuit in Fig.

Answer

  1. Would be like a half wave rectifier with negative cycles in output.
Solution:
Key Concept:
  1. During positive half cycle,
Diode → Forward biased
Output signal → obtained
  1. During negative half cycle,
Diode → reverse biased
Output signal → not obtained
  1. Output voltage is obtained across the load resistance RL. It is not constant but pulsating (mixture of ac and dc) in nature.
  2. Average output in one cycle
  3. r.m.s. output: $\text{I}_\text{rms}=\frac{\text{I}_0}{2},\text{V}_\text{rms}=\frac{\text{V}_0}{2}$
When the diode is forward biased during positive half cycle of input AC voltage, the resistance of p-n junction is low. The current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in a series of circuit. This result into zero output voltage across p-n junction.
And when the diode is reverse biased during negative half cycle of AC voltage, the p-n junction is reverse biased. The resistance of p-n junction becomes high which will be more than resistance in series. That is why, there will be voltage across p-n junction with negative cycle in output, hence option (c) is correct.

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