Question:medium

An a. c. supply of 220V is applied to a half wave rectifier through a transformer of turn ratio 20:1. In this circuit the peak inverse voltage is:

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Always be careful whether the given AC voltage is an RMS value or a peak value. In standard practice (like mains supply), the value given (e.g., 220V, 110V) is always the RMS value unless specified otherwise. Remember PIV values for standard rectifiers:
Half-wave: PIV = \(V_{s,peak}\)
Full-wave (center-tapped): PIV = \(2V_{s,peak}\) (where \(V_{s,peak}\) is across half the secondary)
Full-wave (bridge): PIV = \(V_{s,peak}\)
Updated On: Feb 20, 2026
  • 11 V
  • 311.08 V
  • 1.5554 V
  • 15.554 V
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The Correct Option is D

Solution and Explanation

Step 1: Concept Definition:
Peak Inverse Voltage (PIV) is the maximum reverse voltage a diode in a rectifier must withstand without breakdown. For a half-wave rectifier, during the negative half-cycle, the diode is reverse-biased. The maximum voltage experienced is the secondary AC waveform's peak voltage.
Step 2: Methodology:
1. Determine the secondary RMS voltage (\(V_{s,rms}\)). 2. Calculate the secondary peak voltage (\(V_{s,peak}\)). 3. For a half-wave rectifier, PIV equals \(V_{s,peak}\).
Relevant formulas:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_{s,rms} = V_{p,rms} \times \frac{N_s}{N_p} \] \[ V_{peak} = V_{rms} \times \sqrt{2} \] Step 3: Detailed Calculation:
1. Secondary RMS Voltage (\(V_{s,rms}\)):
Given: Primary voltage \(V_{p,rms} = 220\) V. Turns ratio \( \frac{N_p}{N_s} = \frac{20}{1} \). Calculation: \[ V_{s,rms} = V_{p,rms} \times \frac{N_s}{N_p} = 220 \, V \times \frac{1}{20} = 11 \, V \] 2. Secondary Peak Voltage (\(V_{s,peak}\)):
Relationship for sinusoidal AC: \[ V_{s,peak} = V_{s,rms} \times \sqrt{2} \] Calculation: \[ V_{s,peak} = 11 \, V \times \sqrt{2} \approx 11 \times 1.414 \] \[ V_{s,peak} \approx 15.554 \, V \] 3. Peak Inverse Voltage (PIV):
In a half-wave rectifier, the full secondary peak voltage appears across the diode during the negative cycle. \[ \text{PIV} = V_{s,peak} \approx 15.554 \, V \] Step 4: Conclusion:
The peak inverse voltage for the diode is approximately 15.554 V, corresponding to option (D).
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