| ⇦ | 
| ⇨ |
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: - If oxygen has root mean square velocity of C m/s, then root mean square velocity of H
- A mass m moves in a circle on a smooth horizontal plane with velocity v₀
- An ideal heat engine works between temperature T₁=500 K and T₂=375 K.
- An engine pumps water continuously through a hose. Water leaves the hose
- A sample of HCl gas is placed in an electric field of 5 x 10⁴ N/C
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If oxygen has root mean square velocity of C m/s, then root mean square velocity of H
- A mass m moves in a circle on a smooth horizontal plane with velocity v₀
- An ideal heat engine works between temperature T₁=500 K and T₂=375 K.
- An engine pumps water continuously through a hose. Water leaves the hose
- A sample of HCl gas is placed in an electric field of 5 x 10⁴ N/C
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😃