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In the following circuit , the AC source is an ideal voltage source$(V\sin \omega t)$ . I/ my book it is written that the amplitude of steady state current through the inductor at resonance is $V\sqrt{C/L}$

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But it should be either V/R as at steady state the resistance of inductor is zero .

or

at resonance capaciitive reactance and inductive reactance is equal so half the current will flow through inductor .

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At resonance, the $LC$ resonator in that circuit behaves as an open circuit, not a short one (it's a parallel resonator, not a series one).

In fact, the admittance of the $LC$ resonator is

$$Y(\mathrm{j}\omega) = \mathrm{j}\omega C-\mathrm{j}\frac{1}{\omega L}.$$

When $\omega = \omega_0 = 1/\sqrt{LC}$, we get

$$Y(\mathrm{j}\omega_0) = \mathrm{j}\frac{C}{\sqrt{LC}}-\mathrm{j}\frac{\sqrt{LC}}{L} = 0.$$

The impedance is thus infinite, no current crosses the resistor and the voltage across the $LC$ resonator is $V$. At resonance, the magnitude of the current through the inductor is then $V/(\omega_0L) = V\sqrt{C/L}$.

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  • $\begingroup$ If Y=0 then impedance should be zero . $\endgroup$ – search May 1 '17 at 10:54
  • $\begingroup$ @search The impedance is the reciprocal of the admittance, $Z = 1/Y$: if $Y=0$, $Z=\infty$. $\endgroup$ – Massimo Ortolano May 1 '17 at 10:58
  • $\begingroup$ We can directly calculate Z as $Z=1/(C\omega ) - \omega L =0$ $\endgroup$ – search May 1 '17 at 11:10
  • $\begingroup$ @search No, that is wrong. I think you have to study a bit more the basics ;-) $\endgroup$ – Massimo Ortolano May 1 '17 at 11:10
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    $\begingroup$ @search, the inductor and capacitor are parallel connected and so, the combined impedance is $$Z = \frac{Z_LZ_C}{Z_L + Z_C}=\frac{j\omega L}{1 - \omega^2LC}$$ Note that the denominator vanishes when $\omega = 1/\sqrt{LC}$ $\endgroup$ – Alfred Centauri May 1 '17 at 12:51
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I would like to address two misconceptions in your question:

But it should be either V/R as at steady state the resistance of inductor is zero .

The ideal inductor always has zero resistance so I assume you meant zero reactance but that's only true for DC steady state. In AC steady state, the inductor has non-zero reactance.

at resonance capaciitive reactance and inductive reactance is equal so half the current will flow through inductor .

First, this isn't true; the reactance of the capacitor at the resonance frequency is equal to the negative of the reactance of the inductor.

Second, on resonance, the capacitor and inductor have the same (AC) current which circulates around the loop formed by the capacitor and inductor; charge flows back and forth through the inductor and capacitor but not through the resistor.

The energy stored in the LC circuit 'sloshes' back and forth between all electrostatic energy (zero current, maximum voltage) and all magnetic energy (maximum current, zero voltage).

The mechanical analogy is a damped mass-spring system driven on resonance. There, the energy stored in the mass-spring system 'sloshes' back and forth between all potential energy (zero velocity, maximum displacement) and all kinetic energy (maximum velocity, zero displacement).

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In order to have the resonance we should have w=1/(LC)^(0,5) because we need to have the smallest value for the impedance Z.

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  • $\begingroup$ Sorry , see my edited question. $\endgroup$ – search May 1 '17 at 10:29

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