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Question: An LC circuit contains a 40 mH inductor and a 25\(\mu F\)capacitor .The resistance of the circuit is...

An LC circuit contains a 40 mH inductor and a 25μF\mu Fcapacitor .The resistance of the circuit is negligible. The time is measured from the instant the circuit is closed The energy stored in the circuit is completely magnetic at time (in milliseconds)

A

0, 3.14, 6.28

B

0, 1.57, 4.71

C

1.57 4717.85

D

1.573.1,4 71

Answer

1.57 4717.85

Explanation

Solution

:Here, L = 40 mH = 40 × 10310^{- 3} Sh

C=25μF=25×106FC = 25\mu F = 25 \times 10^{- 6}F

υ=22πLC\upsilon = \frac{2}{2\pi\sqrt{LC}}

Substituting the given values, we get

υ=12π40×103×25×106=1032πHz\upsilon = \frac{1}{2\pi\sqrt{40 \times 10^{- 3} \times 25 \times 10^{- 6}}} = \frac{10^{3}}{2\pi}Hz

T=1υ=2π103s=2π×103s=2πms\therefore T = \frac{1}{\upsilon} = \frac{2\pi}{10^{3}}s = 2\pi \times 10^{- 3}s = 2\pi ms

Energy stored is completely electrical at times

t=0,T2,T,32T,......t = 0,\frac{T}{2},T,\frac{3}{2}T,......

Energy stored is completely magnetic at times

t=T4,34T,54T,......t = \frac{T}{4},\frac{3}{4}T,\frac{5}{4}T,......

Hence, t=π2ms,3π2ms,5π2mst = \frac{\pi}{2}ms,\frac{3\pi}{2}ms,\frac{5\pi}{2}ms

=1.57ms,4.71ms,7.85ms= 1.57ms,4.71ms,7.85ms