3-चरण संतुलित प्रणाली में शक्ति को मापने के लिए दो वाटमीटर का उपयोग किया जाता है। जब एक वाटमीटर दूसरे से दो बार पढ़ता है तो लोड का पावर फैक्टर क्या होता है?

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  1. 0
  2. 0.5
  3. 0.866
  4. 1

Answer (Detailed Solution Below)

Option 3 : 0.866
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Detailed Solution

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कॉन्सेप्ट:

दो वाटमीटर विधि में, पश्चगामी भार के लिए

पहले वाटमीटर का पठन (W1) = VL IL cos (30 – ϕ)

दूसरे वाटमीटर का पठन (W2) = VL IL cos (30 + ϕ)

परिपथ में कुल शक्ति (P) = W1 + W2

परिपथ में कुल प्रतिघाती शक्ति \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

शक्ति गुणक = cos ϕ

\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

गणना:

दिया गया है, W1 = 2W2

\(\tan \phi = \frac{{\sqrt 3 \left( {2{W_2} - {W_2}} \right)}}{{\left( {2{{\rm{W}}_2} + {W_2}} \right)}} = \frac{1}{{\sqrt 3 }}\)

⇒ ϕ = 30°

शक्ति गुणक = cos 30 = 0.866

महत्वपूर्ण बिंदु:

p.f. कोण

(ϕ)

p.f.(cos ϕ)

W1

[VLIL cos (30 + ϕ)]

W2

[VLIL cos (30 - ϕ)]

W = W1 + W2

[W = √3VLIL cos ϕ]

अवलोकन

1

\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)

\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)

√3 VLIL

W1 = W2

30°

0.866

\(\frac{{{V_L}I}}{2}\)

VLIL

1.5 VL IL

W= 2W1

60°

0.5

0

\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)

\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\)

W1 = 0

90°

0

\(\frac{{ - {V_L}{I_L}}}{2}\)

\(\frac{{{V_L}{I_L}}}{2}\)

0

W1 = -ve

W2 = +ve

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