Question 471: AC fundamentals

Question

A condenser of capacity C is charged to a potential difference of V
1
. 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 V
2
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 V 1 . 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 V 2 is

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