Question

At resonance the peak value of current in L-C-R series circuit is.

• A. $E_0/R$
• B. $\frac{E}{\sqrt{R^2+\left(\omega L - \frac{1}{\omega C}\right)^2}}$
• C. $\frac{E_0}{\sqrt2 \sqrt{R^2+\left(\omega^2 L - \frac{1}{\omega^2 C^2}\right)^2}}$
• D. $\frac{E_O}{\sqrt{2R}}$

Answer 1

Answer: (A)

At resonance.
$\omega L=\frac{1}{\omega C}\,\,\, (1)$
So, $C =\frac{ E _{0}}{\sqrt{ R ^{2}+\left(\omega L -\frac{1}{\omega C }\right)^{2}}}$
$=\frac{ E _{0}}{\sqrt{ R ^{2}+0}}$ By (1)
$=\frac{ E _{0}}{ R }$