| ⇦ | 
| ⇨ |
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 a double slit experiment, the two slits are 1mm apart and the screen is placed 1 m
- For the normal eye, the cornea of eye provides a converging power of 40D
- A photon of energy 10.2 eV collide inelastically with hydrogen atom in ground state
- A pellet of mass 1 g is moving with an angular velocity of 1 rad/s along a circle
- A short linear object of length L lies on the axis of a spherical mirror of focal
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In a double slit experiment, the two slits are 1mm apart and the screen is placed 1 m
- For the normal eye, the cornea of eye provides a converging power of 40D
- A photon of energy 10.2 eV collide inelastically with hydrogen atom in ground state
- A pellet of mass 1 g is moving with an angular velocity of 1 rad/s along a circle
- A short linear object of length L lies on the axis of a spherical mirror of focal
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😃