Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.

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

Inductive reactance XL= 2 $\pi \vartheta$ L

And capacitive reactance Xc= $\frac{1}{2\pi \vartheta \mathrm{C}}$

For very high frequencies XL= infinity and Xc=0

When reactance of the circuit is infinite it will be considered as open circuit . when reactance of a circuit is zero it will be considered as short circuited.

So C1, C2 = shorted and L1, L2=opened

So R_eff= R₁+R₂

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p><span data-teams="true">I</span>nductive reactance XL= 2&nbsp;<math><mi>π</mi><mi>ϑ</mi></math>&nbsp;L</p><p>And capacitive reactance Xc=&nbsp;<math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn><mi>π</mi><mi>ϑ</mi><mi mathvariant="normal">C</mi></mrow></mrow></mfrac></math></p><p>For very high frequencies XL= infinity and Xc=0</p><p>When reactance of the circuit is infinite it will be considered as open circuit . when reactance of a circuit is zero it will be considered as short circuited.</p><p>So C1, C2 = shorted and L1, L2=opened</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743055955php7AM6Se_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743055955php7AM6Se.jpeg" alt="" width="341" height="129"></picture></div><div>So R_eff= R₁+R₂</div></div>

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