| ⇦ | 
| ⇨ |
An alternating voltage e=200s in100t is applied to a series combination of R= 30Ω and an inductor of 400 mH. The power factor of the circuit is
Options
(a) 0.01
(b) 0.05
(c) 0.042
(d) 0.6
Correct Answer:
0.6
Explanation:
Xʟ = Lω = 0.4 × 100 = 40
R = 30 Ω
.·. Power factor, cos ɸ = R / √(R² × Xʟ²) = 30 / √(R² × Xʟ²) = 30 / √(30² × 40²) = 0.6
Related Questions: - ₆C¹⁴ absorbs an energitic neutron and emits beta particles. The resulting nucleus is
- The threshold frequency for a photosensitive metal is 3.3 x 10¹⁴ Hz.
- The moment of inertia of a semicircular ring about the centre is
- The total energy of a particle executing SHM is 80 J
- The angle between the dipole moment and electric field at any point on the equatoria
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- ₆C¹⁴ absorbs an energitic neutron and emits beta particles. The resulting nucleus is
- The threshold frequency for a photosensitive metal is 3.3 x 10¹⁴ Hz.
- The moment of inertia of a semicircular ring about the centre is
- The total energy of a particle executing SHM is 80 J
- The angle between the dipole moment and electric field at any point on the equatoria
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