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Q. In an $L C$ oscillator circuit, $L=10\, mH , C=40\, \mu F$. If initially at $t=0$, the capacitor is fully charged to $4\, \mu C$, then find the current (in $mA$ ) in the circuit when the capacitor and inductor share equal energies.

Alternating Current

Solution:

$U_{C}=\frac{Q^{2}}{2 C}=0.2\, \mu \,J ($ total $)$
$U_{L}'=U_{C}'=0.1 \,\mu\, J =\frac{1}{2} L i^{2} $
$\Rightarrow i=4.47 \,mA$