Test: Diode Rectifiers - Electrical Engineering (EE) MCQ

For the series RLC diode circuit, the current i(t) is given by

Here 

The diode will stop conducting when ω_r t₁ = π.
∴ Conduction time of diode,

For α < 90° V_m = V_p
 For α > 90° V_m sinα = V_p
V_RMS = 200 V
⇒ V_m = 200 √2
200 √2 sin α = 200

α = 45°
α = 90 + 45 = 135°

The rms value of output voltage is:


Heater resistance

∴ Power absorbed by heater element

Given, I₀ = 10A
The rms value of diode current,

The average value of diode current,

If v_i is of very large negative value Iv_i| > |E|, then diode D₁ will be forward biased (ON) while diode D₂ will be reverse biased (OFF).

• The full-bridge rectifier delivers a constant load current of 20 A, which alternates in direction through the source, forming a pulsating AC waveform.
• For this waveform:
  • RMS value: Equal to the load current, I_rms=20 A.
  • Average value: Calculated from the magnitude, I_avg=20 A.
• Both RMS and average values match because the current magnitude remains constant.
• Correct answer: (a) 20 A, 20 A.

∴ Voltage regulation

A single-phase half-wave rectifier is popularly known as one-pulse converter because for one cycle of supply voltage, there is one half-cycle of output, or load voltage. Hence, both assertion and reason are true but reason is not the correct explanation of assertion.

The load current is mostly continuous in 3-phase converters. Hence reason is a false statement.