| ⇦ | 
| ⇨ |
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: - A circular disc of radius R is removed from a bigger circular disc of radius 2R
- A steady current of 1.5 amp flows through a copper voltmeter for 10 minutes
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- In Young’s double slit experiment, when wavelength used is 6000Å and the screen
- A tuning fork gives 4 beats with 50 cm length of a sonometer
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A circular disc of radius R is removed from a bigger circular disc of radius 2R
- A steady current of 1.5 amp flows through a copper voltmeter for 10 minutes
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- In Young’s double slit experiment, when wavelength used is 6000Å and the screen
- A tuning fork gives 4 beats with 50 cm length of a sonometer
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