KVPY 2020-21 SB SX Exam Previous Papers

Question : 25 of 120

Marks: +1, -0

The output voltage (taken across the resistance) of a LCR series resonant circuit falls to half its peak value at a frequency of 200 Hz and again reaches the same value at 800 Hz. The bandwidth of this circuit is :

200 Hz
$200√ 3$ Hz
400 Hz
600 Hz

Validate

Solution:

{V}_{\text "Output "}={V}_{{R}}

={i}_{{\rms}} {R}

={{V}_{0} {R}}/{{Z}}={{V}_{0} {R}}/{\√{{R}^{2}+({X}_{{L}}-{X}_{{C}})^{2}}}

For peak

X_{L}=X_{C} \⇒V_{\text "peak "}=V_{0}

For

V_{\text "Output "}={V_{0}}/{2}

{V_{0}}/{2}={V_{0} R}/{\√{R^{2}+(X_{L}-X_{C})^{2}}}

R^{2}+(X_{L}-X_{c})^{2}=4 R^{2}

X_{L}-X_{C}=\±\√{3} R

\ωL-{1}/{\ωC}=\±\√{3} R

\ω^{2} L C \ ∓\√{3} R \ωC-1=0

\ω={\±\√{3} R C \±\√{3 R^{2} C^{2}+4 L C}}/{2 L C}

\ω_{1}={-\√{3} R C+\√{3 R^{2} C^{2}+4 L C}}/{2 L C}

=200 \×2 \π

\ω_{2}={+\√{3} R C+ √{3 R^{2} C^{2}+4 L C}}/{2 L C}

=800 \×2 \π

\ω_{2}-\ω_{1}=600 \×2 \π=\√{3} {{R}}/{{L}}

Bandwidth

={{R}}/{{L}}={2 \π\×600}/{\√{3}}

\Δ{f}={1}/{2 \π} {{R}}/{{L}}={600}/{\√{3}}=200 \√{3}

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