We know that inductive reactance $X_L=2 \pi f L$

and capacitive reactance $X_C=\frac{1}{2 \pi f C}$

For very high frequencies $(f \rightarrow \infty), X_L \rightarrow \infty$ and $X_C \rightarrow 0$

When reactance of a circuit is infinite it will be considered as open circuit. When reactance of a circuit is zero it will be considered as short circuited.

So, $C_1, C_2 \rightarrow$ shorted and $L_1, L_2 \rightarrow$ opened.

So, effective impedance $=R_{\text {eq }}=R_1+R_3$

Let, $$\begin{aligned} & \left(I_{\mathrm{ms}}\right) a=\mathrm{rms} \text { current in circuit (a) } \\ & \left(I_{\mathrm{rms}}\right) b=\mathrm{rms} \text { current in circuit (b) } \\ & \left(I_{\mathrm{rms}}\right) a=\frac{V_{\mathrm{rms}}}{R}=\frac{V}{R} \\ & \left(I_{\mathrm{rms}}\right) b=\frac{V_{\mathrm{rms}}}{Z}=\frac{V}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \end{aligned}$$

(a) When

$$\begin{aligned} \left(I_{\mathrm{rms}}\right) a & =\left(I_{\mathrm{rms}}\right) b \\ R & =\sqrt{R^2+\left(X_L-X_C\right)^2} \end{aligned}$$

$\Rightarrow \quad X_L=X_C$, resonance condition

(b) As $Z \geq R$

$$\begin{aligned} \Rightarrow \quad \frac{\left(I_{\mathrm{rms}}\right) a}{\left(I_{\mathrm{rms}}\right) b} & =\frac{\sqrt{R^2+\left(X_L-X_{\mathrm{C}}\right)^2}}{R} \\ & =\frac{Z}{R} \geq 1 \\ \Rightarrow \quad\left(I_{\mathrm{rms}}\right) a & \geq\left(I_{\mathrm{rms}}\right) b \end{aligned}$$

No, the rms current in circuit (b), cannot be larger than that in (a).

Let the applied emf

$$E=E_0 \sin (\omega t)$$

and current developed is

$$I=I_0 \sin (\omega t \pm \phi)$$

Instantaneous power output of the AC source

$$P=E I=\left(E_0 \sin \omega t\right) \quad\left[I_0 \sin (\omega t \pm \phi)\right]$$

$$\begin{aligned} & =E_0 I_0 \sin \omega t \cdot \sin (\omega t+\phi) \\ & =\frac{E_0 I_0}{2}[\cos \phi-\cos (2 \omega t+\phi)]\quad\text{... (i)} \end{aligned}$$

$$\begin{aligned} \text { Average power } \quad P_{\mathrm{av}} & =\frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}} \cos \phi \\ & =V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi\quad\text{.... (ii)} \end{aligned}$$

where $\phi$ is the phase difference.

Clearly, from Eq. (i)

when $$\begin{gathered} \cos \phi<\cos (2 \omega t+\phi) \\ P<0 \end{gathered}$$

Yes, the instantaneous power output of an $A C$ source can be negative

From Eq. (ii)

$$\begin{aligned} P_{\mathrm{av}} & >0 \\ \text{Because}\quad\cos \phi & =\frac{R}{Z}>0 \end{aligned}$$

No, the average power output of an AC source cannot be negative.

Consider the diagram .

Bandwidth $=\omega_2-\omega_1$

where $\omega_1$ and $\omega_2$ corresponds to frequencies at which magnitude of current is $\frac{1}{\sqrt{2}}$ times of maximum value.

$$I_{\mathrm{rms}}=\frac{I_{\mathrm{max}}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~A}$$

Clearly from the diagram, the corresponding frequencies are $0.8 \mathrm{~rad} / \mathrm{s}$ and $1.2 \mathrm{~rad} / \mathrm{s}$.

$$\Delta \omega=\text { Bandwidth }=1 \cdot 2-0.8=0.4 \mathrm{~rad} / \mathrm{s} $$

$$\begin{aligned} I_{\mathrm{rms}} & =\mathrm{rms} \text { current } \\ & =\sqrt{\frac{1^2+2^2}{2}}=\sqrt{\frac{5}{2}}=1.58 \mathrm{~A} \approx 1.6 \mathrm{~A} \end{aligned}$$

The rms value of the current $\left(I_{\mathrm{rms}}\right)=1.6 \mathrm{~A}$ is indicated in the graph.