⇦ | ⇨ |
An electric bulb has a rated power of 50W at 100 V. if it used on an a.c. source 200V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of
Options
(a) 0.1 mH
(b) 1 mH
(c) 0.1 H
(d) 1.1 H
Correct Answer:
1.1 H
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - In adiabatic expansion, product of PV
- A circular disc of radius R is removed from a bigger circular disc of radius 2R
- When the wave of hydrogen atom comes from infinity into the first orbit,
- The mirrors are inclined at an angle of 50⁰. The number of images formed for an object
- Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube.
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In adiabatic expansion, product of PV
- A circular disc of radius R is removed from a bigger circular disc of radius 2R
- When the wave of hydrogen atom comes from infinity into the first orbit,
- The mirrors are inclined at an angle of 50⁰. The number of images formed for an object
- Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube.
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
maximum current that can flow =p/v = 50/100 =0.5
R= V²/p= 100*100/50 =200
vrms=Irma * Z
220 = 0.5 √(R²+X²)
X comes out to be 346.14
Now,
X = 2πfL
L = 346.4/(2*3.14*50)
Hope it helps😃