Question 484: AC fundamentals

Question

A condenser of capacity C is charged to a potential difference of V1. The plates of th condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is

Accepted Answer

q = q
0
cos$$\omega $$t
$$ \Rightarrow $$ cos$$\omega $$t = $${q \over {{q_0}}} = {{C{V_2}} \over {C{V_1}}} = {{{V_2}} \over {{V_1}}}$$
Current through the inductor
I = $${{dq} \over {dt}} = {d \over {dt}}\left( {{q_0}\cos \omega t} ight)$$ = - q
0
$$\omega $$ sin$$\omega $$t
$$ 	herefore $$ |I| = $${C{V_1}{1 \over {\sqrt {LC} }}{{\left[ {1 - {{\cos }^2}\omega t} ight]}^{1/2}}}$$
= $${V_1}\sqrt {{C \over L}} {\left[ {1 - \left( {{{{V_2}} \over {{V_1}}}} ight)} ight]^{1/2}}$$
= $${\left( {{{C\left( {V_1^2 - V_2^2} ight)} \over L}} ight)^{{1 \over 2}}}$$

Answer

$${\left( {{{C{{\left( {{V_1} - {V_2}} \right)}^2}} \over L}} \right)^{{1 \over 2}}}$$

Answer

$${{C\left( {V_1^2 - V_2^2} \right)} \over L}$$

Answer

$${{C\left( {V_1^2 + V_2^2} \right)} \over L}$$

Answer

$${\left( {{{C\left( {V_1^2 - V_2^2} \right)} \over L}} \right)^{{1 \over 2}}}$$

AC fundamentals

A condenser of capacity C is charged to a potential difference of V1. The plates of th condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is

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