A capacitor will be used on a 230 V, 50 Hz AC supply. What minimum peak-voltage rating should be specified for the capacitor to ensure safe operation?

Difficulty: Easy

Correct Answer: 325 V (approximately √2 × 230 V)

Explanation:


Introduction / Context:
Capacitors used on sinusoidal AC must be rated for the maximum instantaneous voltage, not merely the RMS value. Exceeding the dielectric strength can cause breakdown, heating, and catastrophic failure. Therefore, selecting an appropriate peak-voltage rating is a basic but critical design step.



Given Data / Assumptions:

  • Supply: 230 V RMS, 50 Hz sinusoid.
  • Ideal capacitor; no waveform distortion.
  • We seek the minimum peak (not RMS) voltage rating.


Concept / Approach:
For a pure sinusoid v(t) = V_max sin(ωt), the RMS value is V_rms = V_max / √2. Therefore V_max = √2 * V_rms. A device connected across the line must withstand at least this peak. In practice, designers add margin for transients (surges/spikes), but the theoretical minimum for a clean sinusoid is √2 times the RMS value.



Step-by-Step Solution:

Compute peak: V_max = √2 * 230 V.Evaluate: V_max ≈ 1.414 * 230 ≈ 325 V.Therefore, specify at least a 325 V peak rating (or higher with safety margin, e.g., 400 V DC rating for mains capacitors).


Verification / Alternative check:

Check with RMS–peak relation: 230 V = 325 / √2 V, consistent.


Why Other Options Are Wrong:

230 V is the RMS value and under-rates the peak.460 V corresponds to 2× RMS; it is conservative but not the minimum peak requirement.115 V or 160 V are below RMS or misapply √2 in the wrong direction.


Common Pitfalls:

Confusing RMS with peak; forgetting surge margins required in real mains applications.


Final Answer:

325 V (approximately √2 × 230 V)

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