| ⇦ | 
| ⇨ |
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 particle A suffers on oblique elastic collision with a particle B that
- A slab consists of two parallel layers of two different materials of same thickness
- The point of suspension λ of a simple pendulum with normal time period T₁ is moving
- Two spheres A and B of masses m and m respectively collide. A is at rest initially
- If a body travels half the distance with velocity v1 and the next half with velocity v2
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle A suffers on oblique elastic collision with a particle B that
- A slab consists of two parallel layers of two different materials of same thickness
- The point of suspension λ of a simple pendulum with normal time period T₁ is moving
- Two spheres A and B of masses m and m respectively collide. A is at rest initially
- If a body travels half the distance with velocity v1 and the next half with velocity v2
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😃