Question

An LC circuit contains a 40 mH inductor and a 25 $\mu$ F capacitor. 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 times (in milliseconds)

• A. 0, 3.14, 6.28
• B. 0, 1.57, 4.71
• C. 1.57, 4.71, 7.85
• D. 1.57, 3.14, 4.71

Answer 1

The correct option is C 1.57, 4.71, 7.85

Here, $L=40mH=40\times {10}^{-3}H$
$C=25\mu F=25\times {10}^{-6}F,\upsilon =\frac{1}{2\pi \sqrt{LC}}$

Substituting the given values, we get
$\nu =\frac{1}{2\pi \sqrt{40\times {10}^{-3}\times 25\times {10}^{-6}}}=\frac{{10}^3}{2\pi }Hz$

$\therefore T=\frac{1}{\nu }=\frac{2\pi }{{10}^3}=\frac{2\pi }{{10}^3}s=2\pi \times {10}^{-3}s=2\pi ms$

Energy stored is completely electrical at times
$t=0,\frac{T}{2},T,\frac{3}{2}T,.....$

Energy stored is completely magnetic at times
$t=\frac{T}{4},\frac{3}{4}T,\frac{5}{4}T,.....$

Hence, $t=\frac{\pi }{2}ms,\frac{3\pi }{2}ms,\frac{5\pi }{2}ms$, = 1.57 ms, 4.71 ms, 7.85 ms