संधारित्रासह अर्ध-वेव्ह रेक्टिफायरसाठी रिपल फॅक्टर याद्वारे दिला जातो:

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ALP CBT 2 Electrician Official Paper (Held On: 23 Jan 2019 Shift 3)
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  1. \(r=\frac{1}{{\sqrt 3 fC{R_L}}}\)
  2. \(r=\frac{1}{{4\sqrt 3 f{R_L}}}\)
  3. \(r=\frac{1}{{4\sqrt 3 fC{R_L}}}\)
  4. \(r=\frac{1}{{2\sqrt 3 fC{R_L}}}\)

Answer (Detailed Solution Below)

Option 4 : \(r=\frac{1}{{2\sqrt 3 fC{R_L}}}\)
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संकल्पना:

संधारित्र फिल्टरसह हाफ वेव्ह रेक्टिफायरसाठी रिपल फॅक्टर

\(Ripple\;factor,(r) = \frac{1}{{2\sqrt 3 fC{R_L}}}\)

संधारित्र फिल्टरसह फुल वेव्ह रेक्टिफायरसाठी रिपल फॅक्टर

\(Ripple\;factor,(r) = \frac{1}{{4\sqrt 3 fC{R_L}}}\)

कॅपेसिटर फिल्टरसह रेक्टिफायरची व्यवस्था खाली दर्शविली आहे:F3 S.B 6.5.20 Pallavi D1

 

XC आणि R ची मूल्ये अशी निवडली जातात जेणेकरून |XC| << RL

म्हणजे, \(\frac{1}{{{\omega _o}C}} < < {R_L}\)  हाफ वेव्ह रेक्टिफायरसाठी

      \(\frac{1}{{2{\omega _o}C}} < < {R_L}\) फुल वेव्ह रेक्टिफायरसाठी

  • संधारित्र (C) आणि RL समांतर असल्याने, ते प्रवाह विभाजक परिपथ तयार करतात.
  • उच्च वारंवारतेसाठी संधारित्र शॉर्ट परिपथ म्हणून काम करेल, म्हणजे बहुतेक प्रवाह Iac संधारित्रामधून वाहतो आणि एक अतिशय लहान AC प्रवाह Rमधून जातो.
  • त्याचप्रमाणे, संधारित्र DC साठी ओपन परिपथ म्हणून वर्तन करत असल्याने, IDC हा Rमधून प्रवाहित होईल.


खालील तक्ता FWR आणि HWR साठी भिन्न संबंध दाखवते.

 

HWR

FWR

रिपल व्होल्टेज(Vr)

\(\frac{{{I_{DC}}}}{{2{f_o}C}}\)

\(\frac{{{I_{DC}}}}{{{f_o}c}}\)

रिपल फॅक्टर (r)

\(\frac{1}{{2\;\sqrt 3 {f_o}C{R_L}}}\)

\(\frac{1}{{4\;\sqrt 3 {f_0}C{R_L}\;}}\)

DC आउटपुट व्होल्टेज (VDC)

\({V_m} - \frac{{{I_{DC}}}}{{2{f_o}c}}\)

\({V_m} - \frac{{{I_{DC}}}}{{4\;{f_o}c}}\)

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