Question 100: AC fundamentals

Question

A capacitor of capacitance 500 $$\mu$$F is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _______ A.

Accepted Answer

At steady state charge stored on the capacitor,
$${q_{\max }} = CV$$
$$ = 500 	imes {10^{ - 6}} 	imes 100$$
$$ = 5 	imes {10^{ - 2}}\,C$$
Energy stored in the capacitor,
$${U_{\max }} = {{q_{\max }^2} \over {2C}}$$
Now, when electrostatic energy of capacitor converted to magnetic field energy then all energy of capacitor is transferrd to the inductor.
$$	herefore$$ Maximum energy stored in the inductor
$${U_{L\,\max }} = {1 \over 2}L\,I_{\max }^2$$
$$	herefore$$ $${1 \over 2}L\,I_{\max }^2 = {{q_{\max }^2} \over {2C}}$$
$$ \Rightarrow {I_{\max }} = {{{q_{\max }}} \over {\sqrt {LC} }}$$
$$ = {{5 	imes {{10}^{ - 2}}} \over {\sqrt {50 	imes {{10}^{ - 3}} 	imes 500 	imes {{10}^{ - 6}}} }}$$
$$ = {{5 	imes {{10}^{ - 2}}} \over {5 	imes {{10}^{ - 3}}}}$$
$$ = 10\,A$$

AC fundamentals

A capacitor of capacitance 500 $\mu$F is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _______ A.

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