| ⇦ |
| ⇨ |
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: - A proton carrying 1 MeV kinetic energy is moving in a circular path of radius
- To demonstrate the phenomenon of interference, we require two sources which emit
- Mass of the nucleons together in a heavy nucleus is
- If the cold junction of a thermo-couple is kept at 0⁰C and the hot junction is kept
- An electron in a circular orbit of radius 0.05 nm performs 10¹⁶ rev/s. The magnetic
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A proton carrying 1 MeV kinetic energy is moving in a circular path of radius
- To demonstrate the phenomenon of interference, we require two sources which emit
- Mass of the nucleons together in a heavy nucleus is
- If the cold junction of a thermo-couple is kept at 0⁰C and the hot junction is kept
- An electron in a circular orbit of radius 0.05 nm performs 10¹⁶ rev/s. The magnetic
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