| ⇦ | 
| ⇨ |
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.
Related Questions: - The heart of a man pumps 5 litres of blood through the arteries per minute at
- Ohm’s law is not applicable except to
- In Bohr model of hydrogen atom, the force on the electron depends on the principal
- When a wire of uniform cross-section a, length l and resistance R
- If the band gap between valence band and conduction band in a material is 5.0 eV,
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The heart of a man pumps 5 litres of blood through the arteries per minute at
- Ohm’s law is not applicable except to
- In Bohr model of hydrogen atom, the force on the electron depends on the principal
- When a wire of uniform cross-section a, length l and resistance R
- If the band gap between valence band and conduction band in a material is 5.0 eV,
Topics: Alternating Current (96)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply