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Electrical Engineering QUESTION #2180
Question 1
A full-wave bridge rectifier with capacitor filter has $V_s = 220\,\text{V}$ (rms) at $50\,\text{Hz}$, $R_L = 1\,\text{k}\Omega$, and $C = 1000\,\mu\text{F}$. The approximate peak-to-peak ripple voltage is:
  • $3.11\,\text{V}$✔️
  • $6.22\,\text{V}$
  • $1.56\,\text{V}$
  • $12.4\,\text{V}$
Correct Answer Explanation
$V_{peak} = 220\sqrt{2} \approx 311\,\text{V}$. For a full-wave rectifier: $V_r = \frac{V_{peak}}{2fR_LC} = \frac{311}{2\times50\times1000\times10^{-3}} = \frac{311}{100} \approx 3.11\,\text{V}$.