In series LCR circuit, at resonance the peak value of current will be $\left[E_0\right.$ is peak emf, $R$ is…

In series LCR circuit, at resonance the peak value of current will be $\left[E_0\right.$ is peak emf, $R$ is resistance, $\omega L$ is inductive reactance, and $1 / \omega \mathrm{C}$ is capacitive reactance]
  1. $\frac{E_0}{R}$
  2. $\frac{E_0}{\sqrt{2} R}$
  3. $\frac{E_0}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$
  4. $\frac{E_0}{\sqrt{2} \sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$

Solution

At resonance, the net reactance of the circuit is zero and the impedance is equal to the resistance. $\therefore \mathrm{I}_0=\frac{\mathrm{E}_0}{\mathrm{R}}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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