A series LCR circuit is connected to an AC source. At a certain angular frequency ω 1 , the current in the circuit is maximum. When the angular frequency is changed to ω 2 = 2 ω 1 , the current

Question

A series LCR circuit is connected to an AC source. At a certain angular frequency ω 1 ​ , the current in the circuit is maximum. When the angular frequency is changed to ω 2 ​ = 2 ω 1 ​ , the current becomes half of the maximum current. If the resistance of the circuit is R , what is the value of the inductance L in terms of R and ω 1 ​ ?

TestHub · Accepted Answer

At resonance, the current is maximum, so $ω_{1} = \frac{1}{L C}$ . The maximum current is $I_{ma x} = \frac{V}{R}$ . When the frequency changes to $ω_{2} = 2 ω_{1}$ , the current becomes $I = \frac{I _{ma x}}{2} = \frac{V}{2 R}$ . This implies the impedance $Z = \frac{V}{I} = 2 R$ .

For an LCR circuit, $Z = R^{2} + (X_{L} - X_{C})^{2}$ . Substituting $Z = 2 R$ , we get $4 R^{2} = R^{2} + (X_{L} - X_{C})^{2}$ , which simplifies to $(X_{L} - X_{C})^{2} = 3 R^{2}$ , so $X_{L} - X_{C} = \pm 3 R$ . At $ω_{2} = 2 ω_{1}$ , $X_{L} = L ω_{2} = L (2 ω_{1})$ and $X_{C} = \frac{1}{C ω _{2}} = \frac{1}{C ( 2 ω _{1} )}$ . Since $C = \frac{1}{L ω _{1}^{2}}$ from resonance, $X_{C} = \frac{1}{( 1/ ( L ω _{1}^{2} )) ( 2 ω _{1} )} = \frac{L ω _{1}}{2}$ .

Substituting these into the impedance equation: $L (2 ω_{1}) - \frac{L ω _{1}}{2} = \pm 3 R$ . This gives $\frac{3 L ω _{1}}{2} = 3 R$ (taking the positive root as $L ω_{1} > 0$ ). Solving for $L$ : $L = \frac{2 3 R}{3 ω _{1}} = \frac{2 R}{3 ω _{1}}$ .

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