Question 397: AC fundamentals

Question

In series LCR circuit, at resonance the peak value of current will be [$$\mathrm{E_0}$$ is peak emf, R is resistance, $$\omega \mathrm{L}$$ is inductive reactance and $$\omega \mathrm{C}$$ is capacitive]

Accepted Answer

At resonance, the net reactance of the circuit is zero and the impedance is equal to the resistance.
$$	herefore I_0=\frac{E_0}{R}$$

Answer

$$\frac{E_0}{R}$$

Answer

$$\frac{E_0}{\sqrt{2} R}$$

Answer

$$\frac{E_0}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$$

Answer

$$\frac{E_0}{\sqrt{2} \sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$$

AC fundamentals

In series LCR circuit, at resonance the peak value of current will be [$\mathrm{E_0} is peak emf, R is resistance, \omega \mathrm{L} is inductive reactance and \omega \mathrm{C}$ is capacitive]

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