Question 316: AC fundamentals

Question

A $1 \mu \mathrm{~F}$ capacitor is charged to 50 V and is then discharged through 10 mH inductor of negligible resistance. The maximum current in the inductor is

Accepted Answer

To find the maximum current in the inductor, we need to equate the energy stored in the capacitor to the energy stored in the inductor when the current is at its peak.
Energy stored in the capacitor
:
The energy $ E $ stored in a capacitor is given by:
$    E = \frac{1}{2} \mathrm{CV}^2    $
where:
$ C = 1 \, \mu \mathrm{F} = 1 	imes 10^{-6} \, \mathrm{F} $
$ V = 50 \, \mathrm{V} $
Energy stored in the inductor
:
Since all the energy from the capacitor is transferred to the inductor, the energy in the inductor is:
$    \frac{1}{2} \mathrm{LI}_0^2    $
where:
$ L = 10 \, \mathrm{mH} = 10 	imes 10^{-3} \, \mathrm{H} $
$ I_0 $ is the maximum current
Setting the two energies equal
:
$    \frac{1}{2} \mathrm{CV}^2 = \frac{1}{2} \mathrm{LI}_0^2    $
Solving for $ I_0 $:
$    \mathrm{I}_0^2 = \frac{\mathrm{CV}^2}{\mathrm{L}} = \frac{1 	imes 10^{-6} 	imes 50^2}{10 	imes 10^{-3}}    $
$    \mathrm{I}_0^2 = \frac{2500 	imes 10^{-6}}{10 	imes 10^{-3}}    $
$    \mathrm{I}_0^2 = 25 	imes 10^{-2}    $
Calculating the maximum current $ I_0 $
:
$    \mathrm{I}_0 = \sqrt{25 	imes 10^{-2}} = 0.5 \, \mathrm{A}    $
Therefore, the maximum current in the inductor is $ 0.5 \, \mathrm{A} $.

Answer

0.5 A

Answer

1.5 A

Answer

1 A

Answer

0.15 A

AC fundamentals

A 1 \mu \mathrm{~F} capacitor is charged to 50 V and is then discharged through 10 mH inductor of negligible resistance. The maximum current in the inductor is

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