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 $1.2 \mathrm{~rad} / \mathrm{s}$.
$$\Delta \omega=\text { Bandwidth }=1 \cdot 2-0.8=0.4 \mathrm{~rad} / \mathrm{s} $$
The alternating current in a circuit is described by the graph shown in figure. Show rms current in this graph.
$$\begin{aligned} I_{\mathrm{rms}} & =\mathrm{rms} \text { current } \\ & =\sqrt{\frac{1^2+2^2}{2}}=\sqrt{\frac{5}{2}}=1.58 \mathrm{~A} \approx 1.6 \mathrm{~A} \end{aligned}$$
The rms value of the current $\left(I_{\mathrm{rms}}\right)=1.6 \mathrm{~A}$ is indicated in the graph.
How does the sign of the phase angle $\phi$, by which the supply voltage leads the current in an L-C-R series circuit, change as the supply frequency is gradually increased from very low to very high values.
$$\begin{aligned} &\text { The phase angle ( } \phi \text { ) by which voltage leads the current in L-C-R series circuit is given by }\\ &\begin{aligned} & \tan \phi=\frac{X_L-X_C}{R}=\frac{2 \pi v L-\frac{1}{2 \pi v C}}{R} \\ & \tan \phi < 0\left(\text { for } \nu< \nu_0\right) \\ & \tan \phi > 0\left(\text { for } \nu > \nu_0\right) \\ & \tan \phi=0 \quad\left(\text { for } \nu=\nu_0=\frac{1}{2 \pi \sqrt{2 C}}\right) \end{aligned} \end{aligned}$$