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Alternating Current question

2010 · Q83
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Alternating Current question

2010 · Q83

NEETPhysicsAlternating CurrentMCQ+4 / −1
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
  1. A
    (C(V1−V2)2L)12{\left( {{{C{{\left( {{V_1} - {V_2}} \right)}^2}} \over L}} \right)^{{1 \over 2}}}(LC(V1​−V2​)2​)21​
  2. B
    C(V12−V22)L{{C\left( {V_1^2 - V_2^2} \right)} \over L}LC(V12​−V22​)​
  3. C
    C(V12+V22)L{{C\left( {V_1^2 + V_2^2} \right)} \over L}LC(V12​+V22​)​
  4. D
    (C(V12−V22)L)12{\left( {{{C\left( {V_1^2 - V_2^2} \right)} \over L}} \right)^{{1 \over 2}}}(LC(V12​−V22​)​)21​
View written solutionFree

Correct answer: D

q = q0 cosω\omega ωt

⇒\Rightarrow⇒ cosω\omega ωt = qq0=CV2CV1=V2V1{q \over {{q_0}}} = {{C{V_2}} \over {C{V_1}}} = {{{V_2}} \over {{V_1}}}q0​q​=CV1​CV2​​=V1​V2​​

Current through the inductor

I = dqdt=ddt(q0cos⁡ωt){{dq} \over {dt}} = {d \over {dt}}\left( {{q_0}\cos \omega t} \right)dtdq​=dtd​(q0​cosωt) = - q0ω\omega ω sinω\omega ωt

∴\therefore∴ |I| = CV11LC[1−cos⁡2ωt]1/2{C{V_1}{1 \over {\sqrt {LC} }}{{\left[ {1 - {{\cos }^2}\omega t} \right]}^{1/2}}}CV1​LC​1​[1−cos2ωt]1/2

= V1CL[1−(V2V1)]1/2{V_1}\sqrt {{C \over L}} {\left[ {1 - \left( {{{{V_2}} \over {{V_1}}}} \right)} \right]^{1/2}}V1​LC​​[1−(V1​V2​​)]1/2

= (C(V12−V22)L)12{\left( {{{C\left( {V_1^2 - V_2^2} \right)} \over L}} \right)^{{1 \over 2}}}(LC(V12​−V22​)​)21​

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