Question
Download Solution PDFThe ripple voltage of a full-wave rectifier with a 100 μF filter capacitor connected to a load drawing 50 mA is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
The ripple voltage of a full-wave rectifier is given by:
\(V_R={2.4V_{dc}\over R_LC}\)
∵ \({V_{dc}\over R_L}=I_L\)
\(V_R={2.4I_{L}\over C}\)
where, VR = Ripple voltage
IL = Load current
C = Capacitor
Calculation
Given, C = 100 μF
IL = 50 mA
\(V_R={2.4\times 50\times 10^{-3}\over 100\times 10^{-6}}\)
VR = 1.2 kV
Last updated on May 29, 2025
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