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Q. An LCR circuit has L = 10 mH, $R = 3\Omega$ , and $C = 1\,\mu F$ connected in series to a source of $15\,cos\,\omega t$ volt . The current amplitude at a frequency that is 10% lower than the resonant frequency is

Alternating Current

Solution:

$c_v=\frac{90}{100}c_{v_0}=\frac{90}{100}x\frac{1}{\sqrt{LC}}=9000\,rad/s$
$i_0=\frac{E_0}{\sqrt{R^2\left(\omega\,L-\frac{1}{\omega\,C}\right)^2}}$