| ⇦ |
| ⇨ |
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: - A parallel plate condenser has a uniform electric field E(V/m) in the space between
- In a vaccum triode the function of grid voltage is
- A planar coil having 12 turns carries 15 A current. The coil is oriented with respect
- A calorimeter contains 0.2 kg of water at 30° C. 0.1kg of water at
- The range of voltmeter is 10V and its internal resistance is 50Ω.
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A parallel plate condenser has a uniform electric field E(V/m) in the space between
- In a vaccum triode the function of grid voltage is
- A planar coil having 12 turns carries 15 A current. The coil is oriented with respect
- A calorimeter contains 0.2 kg of water at 30° C. 0.1kg of water at
- The range of voltmeter is 10V and its internal resistance is 50Ω.
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😃