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Asked in GUJCET 2025 · Rectifiers and filters
Given: load resistance R = 200 Ω and filter capacitance C = 15 μF = 15×10⁻⁶ F.
Idea: between one rectified peak and the next the capacitor is left to discharge through the load, and that discharge is exponential with time constant τ = RC.
τ = 200 × 15×10⁻⁶
τ = 3×10⁻³ s = 3 ms
The larger this time constant is compared with the gap between peaks, the less charge the capacitor loses and the flatter the output.
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