In a series LCR circuit, the plot of I_maxversus co is shown in figure. Find the bandwidth and mark in the figure.

This is a short answer type question as classified in NCERT Exemplar

In the figure

Band width = w₂-w₁ where these two corresponds to frequencies at which magnitude of current is 1/ $\sqrt[]{2}$ times of maximum value

I_rms= $\frac{I_{max}}{\sqrt[]{2}}$ = 1/ $\sqrt[]{2}$ = 0.7A

From the diagram the corresponding frequencies are 0.8 and 1.2rad/s

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>In the figure</p><p>Band width = w₂-w₁ where these two corresponds to frequencies at which magnitude of current is 1/<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span> times of maximum value</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743056409phpgRHGzx_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743056409phpgRHGzx.jpeg" alt="" width="163" height="119"></picture></div></div><p>I_rms=&nbsp;<math><mfrac><mrow><mrow><msub><mrow><mrow><mi>I</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></mrow></mrow><mrow><mrow><mroot><mrow><mrow><mn>2</mn></mrow></mrow><mrow></mrow></mroot></mrow></mrow></mfrac></math>&nbsp;= 1/&nbsp;<math><mroot><mrow><mrow><mn>2</mn></mrow></mrow><mrow></mrow></mroot></math>&nbsp;= 0.7A</p><p>From the diagram the corresponding frequencies are 0.8 and 1.2rad/s</p>

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