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JEE Mains · Physics · STD 12 -7. Alternating current

यहाँ दर्शाये गये \(LCR\) परिपथ में, विद्युतधारा, आरोपित वोल्टता से अग्रगामी (आगे) रहती है। परिपथ में जुड़े संधारित्र \(C\) के साथ एक अतिरिक्त संधारित्र \(C^{\prime}\) जोड़ने से, इस परिपथ का शक्ति-गुणक इकाई (एकक) हो जाता है। तो, संधारित्र \(C ^{\prime}\) को अवश्य ही जोड़ा गया होगा :

  1. A series with \(C\) and has a magnitude \(\frac{C}{{\left( {{\omega ^2}LC - 1} \right)}}\)
  2. B series with \(C\) and has a magnitude \(\,\frac{{\left( {1 - {\omega ^2}LC} \right)}}{{{\omega ^2}L}}\)
  3. C parallel with \(C\) and has a magnitude \(\,\frac{{\left( {1 - {\omega ^2}LC} \right)}}{{{\omega ^2}L}}\)
  4. D parallel with \(C\) and has a magnitude \(\frac{C}{{\left( {{\omega ^2}LC - 1} \right)}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) parallel with \(C\) and has a magnitude \(\,\frac{{\left( {1 - {\omega ^2}LC} \right)}}{{{\omega ^2}L}}\)

Step-by-step Solution

Detailed explanation

Power factor \(\cos \phi=\frac{R}{\sqrt{R^{2}+\left[\omega L-\frac{1}{\omega\left(C+C^{\prime}\right)}\right]^{2}}}=1\) On solving we get, \(\omega L=\frac{1}{\omega\left(C+C^{\prime}\right)}\) \(C' = \frac{{1 - {\omega ^2}LC}}{{{\omega ^2}L}}\) Hence option \((c)\) is the…
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