Study the circuits $(a)$ and $(b)$ shown in the figure and answer the following questions.
$(a)$ Under which conditions would the $rms$ currents in the two circuits be the same?
$(b)$ Can the $rms$ current in circuit $(b)$ be larger than that in $(a)$?

(N/A) Let the $rms$ current in circuit $(a)$ be $I_{a}$ and in circuit $(b)$ be $I_{b}$.
For circuit $(a)$,the impedance is $Z_{a} = R$. Thus,$I_{a} = \frac{V_{rms}}{R}$.
For circuit $(b)$,the impedance is $Z_{b} = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$. Thus,$I_{b} = \frac{V_{rms}}{\sqrt{R^{2} + (X_{L} - X_{C})^{2}}}$.
$(a)$ For $I_{a} = I_{b}$,we must have $Z_{a} = Z_{b}$.
$\therefore R = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.
Squaring both sides,$R^{2} = R^{2} + (X_{L} - X_{C})^{2}$,which implies $(X_{L} - X_{C})^{2} = 0$,or $X_{L} = X_{C}$.
This is the condition of electrical resonance.
$(b)$ Since $Z_{b} = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$,it is clear that $Z_{b} \geq R$ for all values of $X_{L}$ and $X_{C}$.
Since $I_{b} = \frac{V_{rms}}{Z_{b}}$ and $I_{a} = \frac{V_{rms}}{R}$,and $Z_{b} \geq R$,it follows that $I_{b} \leq I_{a}$.
Therefore,the $rms$ current in circuit $(b)$ can never be larger than that in circuit $(a)$.