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.
$$\begin{aligned} &\text { The phase angle ( } \phi \text { ) by which voltage leads the current in L-C-R series circuit is given by }\\ &\begin{aligned} & \tan \phi=\frac{X_L-X_C}{R}=\frac{2 \pi v L-\frac{1}{2 \pi v C}}{R} \\ & \tan \phi < 0\left(\text { for } \nu< \nu_0\right) \\ & \tan \phi > 0\left(\text { for } \nu > \nu_0\right) \\ & \tan \phi=0 \quad\left(\text { for } \nu=\nu_0=\frac{1}{2 \pi \sqrt{2 C}}\right) \end{aligned} \end{aligned}$$
A device ' $X$ ' is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in figure.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device $X$.
(a) We know that
$$\text { Power }=P=V I$$
that is curve of power will be having maximum amplitude, equals to multiplication of amplitudes of voltage $(V)$ and current $(I)$ curve. So, the curve will be represented by $A$.
(b) As shown by shaded area in the diagram, the full cycle of the graph consists of one positive and one negative symmetrical area.
Hence, average power over a cycle is zero.
(c) As the average power is zero, hence the device may be inductor ( $L$ ) or capacitor (C) or the series combination of $L$ and $C$.