Alternating current

Here, we discuss about the alternating current (A.C).

Alternating Current:

 "An alternating current is the flow of electric charge that periodically reverses."

Here's an illustration of A.C.

Graphically its as follows,

image credit: bbc

The graph above has the same shape as the graphs used to represent simple harmonic motion, and it can be interpreted in the same way. The electrons, in a wire carrying a.c thus move back and forth with SHM. The current varies like a sine wave and so it is described as sinusoidal .

Mathematically, 
$$I=I_{0}sin(\omega t)$$
Where,
$\omega$=angular frequency of the supply
$I_{0}$=The peak value of the alternating current (The amplitude of the graph)

And
$$V=V_{0}sin(\omega t)$$

Where:
$V_{0}$ = The peak of the voltage
$\omega$ = angular frequency

Calculating Power in A.C,

Before we define power, it is important that we define Root Mean Square value. $I_{rms}$ or root mean square current is the average value that is equivalent to the value of the steady direct current which dissipates energy in a given resistance at the same rate as the alternating current.

If  $I_{rms}$ value of an a.c. is the current , then the average power $<P>$ dissipated in a load of the resistance $R$ is given by,
$$<P>=I_{rms}^{2}R=\frac{V_{rms}^{2}}{R}$$

Mathematical relationship between the peak value $I_{0}$ of the alternating current and d.c current which delivers the same average electrical power is as follows,

$$I_{rms}=\frac{I_{0}}{\sqrt{2}}$$
Similarly for $V_{rms}$ ,
$$V_{rms}=\frac{V_{0}}{\sqrt{2}}$$

WORKED EXAMPLE

A sinusoidal p.d of peak value $25V$ is connected across $20\Omega$ resistor. What is the average power dissipated in the resistor.

STEP 1: Calculate the r.m.s value os the p.d, $V_{rms}=V_{0}/ \sqrt{2}$
$V_{rms}= 25/√2 $
$V_{rms}=17.7V$

STEP 2: Now calculate the average power dissipated using, $P=V^{2}/R$

P=17.72/20
P=15.6W

(Note: if we had used $V_{0}$ instead of Vrms we would’ve got P as $31.3W$, which is double the correct answer)

Note: In collaboration with Kartikay Buchar.

5.3 Current of Electricity

In this section we'll discuss the notion of electricity.

The electric current is the rate of flow charged particles. It's SI unit is ampere(A).
$$I=\frac{Q}{t}$$
where,$I=$ Current, in $'A'$
          $Q=$ total charge passing a point, in $'C'$
          $t=$ time for charge to flow across the point, in $'s'$

From this definition, we can also find the expressions in different forms for a current in the wire in terms of individual charges. We can almost approximately visualize the flowing electrons in the wire as the figure below,

 Then the expression for the current in the wire is given by,$$I=nqv$$
where,
          $I=$ Current, in $'A'$
          $q=$ charge of individual electrons, in $'C'$
          $v=$ drift velocity of the electrons, in '$ms^{-1}$'
          $n=$ Charge density (charge per unit length), in '$Cm^{-1}$'

Convention on the flow of the current

The case we are interested(in particular the flow of charge in metal) has the direction of current flow from negative terminal to the positive terminal, i.e. (negative charge flow). However, due to historical reasons, the convention of the the current is taken to be the flow of the positive charge(i.e. from Positive terminal to the Negative terminal). The direction of convention is as shown in the diagram below,


Here's an introduction to current,



5.2 Energy stored in Capacitors

So the question is now then, How much energy is stored in the capacitors?

We know from the previous section that the charge stored in the capacitors is given by,$$Q=CV$$
where, $Q=$ charge stored in the capacitors, in coulombs($Q$)
           $V=$ potential difference between the plates, measured in volt($V$)
           $C=$ capacitance, measured in farad($F$)

Now, we can plot the Potential difference vs Charge  (i.e. $V$ vs $Q$) to check that its a straight line as follows,

Then, the energy stored in the capacitor is the work done to store the charge on the capacitor at the applied p.d. and is given by,$$E=\frac{1}{2}VQ=\frac{1}{2}CV^{2}$$
where, $Q=$ charge stored in the capacitors, in coulombs($Q$)
           $V=$ potential difference between the plates, measured in volt($V$)
           $E=$ Energy stored, in Joules($J$)

The above relation is given by finding the area under the curve. So, Energy in the ($V$ vs $Q$) graph is the quantity of area under the graph.


Here, a wonderful video of the energy stored in the capacitors. The reason for $\frac{1}{2}$ in the formula is also explained intuitively in the video, enjoy!!