लहरी घटक  कॅपेसिटर फिल्टरसह पूर्ण तरंग दुरुस्त करणासाठी दिलेले आहे:

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

Answer (Detailed Solution Below)

Option 3 : \( \frac{1}{{4\sqrt 3 fC{R_L}}}\)
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Detailed Solution

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संकल्पना: 

लहरी घटक कॅपेसिटर फिल्टरसह पूर्ण तरंग दुरुस्त करणासाठी

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

कॅपेसिटर फिल्टरसह दुरुस्त करणासाठी व्यवस्था खाली दर्शविली आहे.

F3 S.B 6.5.20 Pallavi D1

Xआणि R ची मूल्ये |X| अशी निवडली आहेत < < RL

म्हणजे,  \(\frac{1}{{{\omega _o}C}} < < {R_L}\)  for अर्ध  तरंग दुरुस्त करणासाठी

            \(\frac{1}{{2{\omega _o}C}} < < {R_L}\) for पूर्ण तरंग दुरुस्त करणासाठी

  • कॅपेसिटर (C) आणिRसमांतर असल्याने, ते धाराविद्युत विभाजक परिपथ तयार करतात.
  • उच्च वारंवारतासाठी कॅपॅसिटर लहान परिपथ  म्हणून काम करेल, म्हणजे बहुतेक धाराविद्युत Iac  कॅपेसिटरमधून वाहते आणि खूप लहान AC धाराविद्युत RL मधून जातो.
  • त्याचप्रमाणे, कॅपेसिटर DC साठी उघडा परिपथ म्हणून वर्तन करत असल्याने, IDC RL मधून प्रवाहित होईल.

 

खालील सारणी FWR आणि HWR साठी भिन्न संबंध दर्शवते,

  HWR FWR
लहरी विद्यु्युत दाब(Vr) \(\frac{{{I_{DC}}}}{{2{f_o}C}}\) \(\frac{{{I_{DC}}}}{{{f_o}c}}\)
लहरी घटक (r) \(\frac{1}{{2\;\sqrt 3 {f_o}C{R_L}}}\) \(\frac{1}{{4\;\sqrt 3 {f_0}C{R_L}\;}}\)
DC उत्पादन विद्यु्युत दाब(VDC) \({V_m} - \frac{{{I_{DC}}}}{{2{f_o}c}}\) \({V_m} - \frac{{{I_{DC}}}}{{4\;{f_o}c}}\)

 

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