An amplitude modulated (AM) radio operates at $$550 kHz$$ to $$1650 kHz$$. If $$L$$ is fixed and $$C$$ is varied for tuning then minimum and maximum value of $$C$$ is
Explanation
Answer : A$$ switch\quad in\quad position\quad 2:\\ { I }_{ max }=\dfrac { 150 }{ \left| L\omega -\dfrac { 1 }{ C\omega } \right| } =3\quad \Rightarrow \left| L\omega -\dfrac { 1 }{ C\omega } \right| =50\\ open\quad switch\quad condition\quad :\\ \hat { Z } =R+j(L\omega -\dfrac { 1 }{ C\omega } )=Z\angle -60\quad \Rightarrow \dfrac { { R }^{ 2 } }{ { R }^{ 2 }+{ \left| L\omega -\dfrac { 1 }{ C\omega } \right| }^{ 2 } } ={ cos }^{ 2 }(-60)\quad \Rightarrow \dfrac { { R }^{ 2 } }{ { R }^{ 2 }+{ 50 }^{ 2 } } =\left( \dfrac { 1 }{ 2 } \right) ^{ 2 }\Rightarrow R=\dfrac { 50 }{ \sqrt { 3 } } A\\ $$
Let the same reading be $$x V $$,then let, $${ Z }_{ total }=R+j({ X }_{ L }-{ X }_{ C })=Z\angle \theta \\ I=\dfrac { { V }_{ S } }{ Z } \angle -\theta \\ therefore,\quad { V }_{ R }=\left| \dfrac { { V }_{ S } }{ Z } R\angle -\theta \right| \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ L }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ L }\angle 90-\theta \right| \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ C }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ C }\angle -90-\theta \right| \\ given,\quad { V }_{ R }={ V }_{ L }={ V }_{ V }=x\\ \quad \Rightarrow \quad R={ X }_{ L }={ X }_{ C }\\ \Rightarrow { Z }_{ total }=R=Z\angle 0=Z\quad \quad \\ \Rightarrow \quad \quad { V }_{ R }=\left| \dfrac { { V }_{ S } }{ Z } R\angle -\theta \right| ={ V }_{ S }\\ \quad \quad \quad \quad { V }_{ L }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ L }\angle 90-\theta \right| =\left| j{ V }_{ S } \right| ={ V }_{ S }\\ \quad \quad \quad \quad { V }_{ C }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ C }\angle -90-\theta \right|=\left| -j{ V }_{ S } \right| ={ V }_{ S }\\ \Rightarrow x={ V }_{ S }=100V $$
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