menu search
brightness_auto
more_vert
A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike

1 Answer

more_vert
 
verified
Best answer

Data: C = 100 µF = 100 × 10-6 F = 10-4 F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= \(\frac{1}{2}\)CV2
The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li2
Here, \(\frac{1}{2}\)CV2 = \(\frac{1}{2}\)Li2
∴ L = C\(\frac{V^{2}}{i^{2}}\)
∴ L = C\(\left(\frac{V}{i}\right)^{2}=10^{-4}\left(\frac{50}{5}\right)^{2}\) = 10-4 × 102
= 10-2H
This is the value of the inductance.

thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike

Related questions

thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 500 like thumb_down_off_alt 0 dislike
1 answer

5.2k questions

4.7k answers

75 comments

393 users

...