If a L-C circuit is considered analogous to a harmonically oscillating springblock system, which energy of the L-C circuit would be analogous to potential energy and which one analogous to kinetic energy?
If we consider a L-C circuit analogous to a harmonically oscillating springblock system. The electrostatic energy $\frac{1}{2} C V^2$ is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\left(\frac{1}{2} L I^2\right)$ is analogous to kinetic energy.
Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.
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 $=R_{\text {eq }}=R_1+R_3$
Study the circuits (a) and (b) shown in 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)?
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).
Can the instantaneous power output of an AC source ever be negative? Can the average power output be negative?
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.
In series $L C R$ circuit, the plot of $I_{\max }$ versus $\omega$ is shown in figure. Find the bandwidth and mark in the figure.
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 $1.2 \mathrm{~rad} / \mathrm{s}$.
$$\Delta \omega=\text { Bandwidth }=1 \cdot 2-0.8=0.4 \mathrm{~rad} / \mathrm{s} $$