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Power dissipated in an LCR series circuit connected to an a.c source of emf ? is
Options
(a) ?ￂﾲ √ [ R² + ( L? – 1 / C?)² ] / R
(b) ?ￂﾲ [ R² + ( L? – 1 / C?)² ] / R
(c) ?ￂﾲR / √ [ R² + ( L? – 1 / C?)² ]
(d) ?ￂﾲR / [ R² + ( L? – 1 / C?)² ]
Correct Answer:
?ￂﾲR / [ R² + ( L? – 1 / C?)² ]
Explanation:
Power dissipated in series LCR;
P = IR = / (Z) R
= ?ￂﾲR / [ R² + ( L? – 1 / C?)² ]
where Z = √ R² + ( L? – 1 / C?)²
is called the impedance of the circuit.
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Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If radius of the ₁₃Al²⁷ nucleus is taken to be Rᴀı, then the radius of ₅₃Te¹²⁵ nucleus
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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