Question

=> Derivation of Bandwidth
$\rm \: \implies2 \Delta \omega = \omega_2 - \omega_1 = \frac{R}{L}$
​

Answer 1

EXPLANATION.

$\sf : \implies \: 2 \Delta \omega \: = \omega_{2} \: - \: \omega_{1} = \dfrac{ R }{L} \\ \\\sf : \implies \: i_{0} \: = \frac{ \epsilon_{0}}{z} \\ \\ \sf : \implies \: at \: \: \: \omega_{1} \: \: \: and \: \: \: \omega_{2} \: \implies \: resonance \: \: occur \: \: (X) \\ \\ \sf : \implies \: z \: = \sqrt{R {}^{2} + ( x_{l} \: - x_{c}) {}^{2} } \\ \\ \sf : \implies \: at \: \omega_{1} \: \: \: and \: \: \: \omega_{2} \: = \: half \: power \: frequency$

$\sf : \implies \: i_{0} = \dfrac{ \epsilon_{0} }{z} \\ \\ \sf : \implies \: i_{0} \: = \frac{ \epsilon_{0} }{ \sqrt{R {}^{2} + ( x_{l} \: - \: x_{c}) {}^{2} } } \\ \\ \sf : \implies \: i_{0} \: = \frac{ \epsilon_{0}}{ \sqrt{R {}^{2} + ( \omega_{l} - \dfrac{1}{ \omega_{c} } ) {}^{2} } } \\ \\ \sf : \implies \: at \: these \: frequency \: \implies \: p \: = \frac{ p_{max} }{2} \\ \\ \sf : \implies \: i_{0} \: = \frac{ l_{0} {max}}{2} \implies \: = \frac{ \epsilon_{0} }{R \sqrt{2} }$

$\sf : \implies \: \dfrac{ \epsilon_{0} }{R \sqrt{2} } = \dfrac{ \epsilon_{0} }{ \sqrt{R {}^{2} + ( \omega_{l} \: - \dfrac{1}{ \omega_{c} }) {}^{2} } } \\ \\ \sf : \implies \: = R {}^{2} + ( \omega_{l} \: - \frac{1}{ \omega_{c} }) {}^{2} = 2R {}^{2} \\ \\ \sf : \implies \: ( \omega \: l \: - \: \frac{1}{ \omega \: c} ) {}^{2} = R {}^{2}$

$\sf : \implies \: ( \omega \: l \: - \dfrac{1}{ \omega \: c} ) = \pm R \\ \\ \sf : \implies \: \omega_{1}l \: - \frac{1}{ \omega_{1}c } = - R \: \: \: .....(1) \\ \\ \sf : \implies \: \omega_{2}l \: - \frac{1}{ \omega_{2}c } = + R \: \: \: .....(2) \\ \\ \sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get$

$\\ \\ \sf : \implies adding \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get \: \: \\ \\ \sf : \implies \: ( \omega_{1} \: + \omega_{2})l \: - \dfrac{1}{c} ( \dfrac{1}{ \omega_{1}} + \dfrac{1}{ \omega_{2}} ) = 0 \\ \\ \sf : \implies \: ( \omega_{1} \: + \omega_{2})l \: - \: \frac{1}{c} ( \frac{ \omega_{2} \: + \: \omega_{1} }{ \omega_{1} \omega_{2} } ) = 0 \\ \\ \sf : \implies \: l \: = \frac{1}{c \omega_{1} \omega_{2} }$

$\sf : \implies \: subtract \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1})l \: - \frac{1}{ \omega_{2}c \: } + \frac{1}{ \omega_{1}c } = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1})l \: \: + \: \: \frac{1}{c}( \frac{1}{ \omega_{1} \: } - \frac{1}{ \omega_{2} } ) = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1})l \: \: + \: \: \frac{1}{c} ( \frac{ \omega_{2} \: - \: \omega_{1} }{ \omega_{1} \omega_{2} } ) = 2R$

$\sf : \implies \: ( \omega_{2} \: - \: \omega_{1}) \: \: (l + \dfrac{1}{c \omega_{1} \omega_{2}} ) = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1} \: ) \: (l \: + \: l) = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1}) \: (2l) = 2R \\ \\ \sf : \implies \: \orange{ \underline{ \underline{( \omega_{2} \: - \: \omega_{1}) = \frac{R}{l} \: \: \implies \: hence \: \: proved }}}$

Answer 2

EXPLANATION.

$\sf : \implies \: 2 \Delta \omega \: = \omega_{2} \: - \: \omega_{1} = \dfrac{ R }{L} \\ \\\sf : \implies \: i_{0} \: = \frac{ \epsilon_{0}}{z} \\ \\ \sf : \implies \: at \: \: \: \omega_{1} \: \: \: and \: \: \: \omega_{2} \: \implies \: resonance \: \: occur \: \: (X) \\ \\ \sf : \implies \: z \: = \sqrt{R {}^{2} + ( x_{l} \: - x_{c}) {}^{2} } \\ \\ \sf : \implies \: at \: \omega_{1} \: \: \: and \: \: \: \omega_{2} \: = \: half \: power \: frequency$

$\sf : \implies \: i_{0} = \dfrac{ \epsilon_{0} }{z} \\ \\ \sf : \implies \: i_{0} \: = \frac{ \epsilon_{0} }{ \sqrt{R {}^{2} + ( x_{l} \: - \: x_{c}) {}^{2} } } \\ \\ \sf : \implies \: i_{0} \: = \frac{ \epsilon_{0}}{ \sqrt{R {}^{2} + ( \omega_{l} - \dfrac{1}{ \omega_{c} } ) {}^{2} } } \\ \\ \sf : \implies \: at \: these \: frequency \: \implies \: p \: = \frac{ p_{max} }{2} \\ \\ \sf : \implies \: i_{0} \: = \frac{ l_{0} {max}}{2} \implies \: = \frac{ \epsilon_{0} }{R \sqrt{2} }$

$\sf : \implies \: \dfrac{ \epsilon_{0} }{R \sqrt{2} } = \dfrac{ \epsilon_{0} }{ \sqrt{R {}^{2} + ( \omega_{l} \: - \dfrac{1}{ \omega_{c} }) {}^{2} } } \\ \\ \sf : \implies \: = R {}^{2} + ( \omega_{l} \: - \frac{1}{ \omega_{c} }) {}^{2} = 2R {}^{2} \\ \\ \sf : \implies \: ( \omega \: l \: - \: \frac{1}{ \omega \: c} ) {}^{2} = R {}^{2}$

$\sf : \implies \: ( \omega \: l \: - \dfrac{1}{ \omega \: c} ) = \pm R \\ \\ \sf : \implies \: \omega_{1}l \: - \frac{1}{ \omega_{1}c } = - R \: \: \: .(1) \\ \\ \sf : \implies \: \omega_{2}l \: - \frac{1}{ \omega_{2}c } = + R \: \: \: .(2) \\ \\ \sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get$

$\\ \\ \sf : \implies adding \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get \: \: \\ \\ \sf : \implies \: ( \omega_{1} \: + \omega_{2})l \: - \dfrac{1}{c} ( \dfrac{1}{ \omega_{1}} + \dfrac{1}{ \omega_{2}} ) = 0 \\ \\ \sf : \implies \: ( \omega_{1} \: + \omega_{2})l \: - \: \frac{1}{c} ( \frac{ \omega_{2} \: + \: \omega_{1} }{ \omega_{1} \omega_{2} } ) = 0 \\ \\ \sf : \implies \: l \: = \frac{1}{c \omega_{1} \omega_{2} }$

$\sf : \implies \: subtract \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1})l \: - \frac{1}{ \omega_{2}c \: } + \frac{1}{ \omega_{1}c } = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1})l \: \: + \: \: \frac{1}{c}( \frac{1}{ \omega_{1} \: } - \frac{1}{ \omega_{2} } ) = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1})l \: \: + \: \: \frac{1}{c} ( \frac{ \omega_{2} \: - \: \omega_{1} }{ \omega_{1} \omega_{2} } ) = 2R$

$\sf : \implies \: ( \omega_{2} \: - \: \omega_{1}) \: \: (l + \dfrac{1}{c \omega_{1} \omega_{2}} ) = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1} \: ) \: (l \: + \: l) = 2R \\ \\ \sf : \implies \: ( \omega_{2} \: - \omega_{1}) \: (2l) = 2R \\ \\ \sf : \implies \: \orange{ \underline{ \underline{( \omega_{2} \: - \: \omega_{1}) = \frac{R}{l} \: \: \implies \: hence \: \: proved }}}$

Hope it helps

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