| ⇦ | 
| ⇨ |
In a circuit, L,C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45⁰. The value of C is
Options
(a) 1/πf(2πfL+R)
(b) 1/πf(2πfL-R)
(c) 1/2πf(2πfL-R)
(d) 1/2πf(2πfL+R)
Correct Answer:
1/2πf(2πfL+R)
Explanation:
The phase difference ɸ between current and voltage is given by
tan ɸ = (Xʟ – Xc) / R
(Xʟ – Xc) / R = tan 45° = 1 ⇒ Xc = X⇒ + R
1 / 2πfC = 2πfL + R
Therefore, C = 1 / 2πf(2πfL + R)
Related Questions: - The sodium nucleus ₁₁Na²³ contains
- A coil of resistance 10 Ω and an inductance 5 H is connected to a 100 volt
- In Young’s double slit experiment with sodium vapour lamp of wavelength
- Two equal and opposite charges of masses m₁ and m₂ are accelerated
- Two poles of same strength attract each other with a force of magnitude F
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The sodium nucleus ₁₁Na²³ contains
- A coil of resistance 10 Ω and an inductance 5 H is connected to a 100 volt
- In Young’s double slit experiment with sodium vapour lamp of wavelength
- Two equal and opposite charges of masses m₁ and m₂ are accelerated
- Two poles of same strength attract each other with a force of magnitude F
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