Introduction to Alternating Currents and Circuits

ALTERNATING CURRENTS AND circuits

The basic principle of the ac generator is a direct consequence of Faraday's law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency $\omega$ a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is,

V = V₀ sin $\omega$t (i)

The usual circuit diagram symbol for an ac source is shown in Figure.

In Equation (i) V₀ is the maximum output voltage of the ac generator, or the voltage amplitude and $\omega$ is the angular frequency, equal to 2$\pi$ times the frequency f.

$\omega$ = 2$\pi$f

The frequency of ac in India is 50 Hz, i.e.,

f = 50 Hz

So, $\omega$ = 2$\pi$f $\approx$ 314 rad/s

The time of one cycle is known as time period T, the number of cycles per second the frequency f.

$T=\frac{1}{f}orT=\frac{2\pi }{\omega }$

A sinusoidal current might be described as,

i = i₀ sin $\omega$t

If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is,

${⟨i⟩}_{onecycle}={(i_{av})}_T=\frac{{\int }_0^{T}idt}{{\int }_0^{T}dt}=\frac{{\int }_0^{2\pi /\omega }(i_0\sin \omega t)dt}{{\int }_0^{2\pi /\omega }dt}=0$

Thus,

${⟨i⟩}_{onecycle}=0$

Similarly, the average value of the voltage (or emf) for one cycle is zero.

${⟨V⟩}_{onecycle}=0$

Since, these averages for the whole cycle are zero, the dc instrument will indicate zero deflection. In ac, the average value of current is defined as its average taken over half the cycle. Hence,

${⟨i⟩}_{Halfcycle}={(i_{av})}_{T/2}=\frac{{\int }_0^{T/2}idt}{{\int }_0^{T/2}dt}=\frac{{\int }_0^{\pi /\omega }(i_0\sin \omega t)dt}{{\int }_0^{\pi /\omega }dt}=\frac{2}{\pi }{i}_0$

This is sometimes simply written as, i_av. Hence,

$i_{av}={⟨i⟩}_{Halfcycle}=\frac{2}{\pi }{i}_0\approx 0.637i_0$

Similarly,

$V_{av}=\frac{2}{\pi }{V}_0\approx 0.637V_0$

A dc meter can be used in an ac circuit if it is connected in the full wave rectifier circuit. The average value of the rectified current is the same as the average current in any half cycle, i.e., $\frac{2}{\pi }$ time the maximum current i_{0 .} A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i² and finally take the square root of that average. This procedure defines the root-mean-square current denoted as i_rms. Even when i is negative, i² is always positive so i_rms is never zero (unless i is zero at every instant). Hence,

${⟨i^2⟩}_{onecycle}=\frac{{\int }_0^T{i}^{2}dt}{{\int }_0^{T}dt}=\frac{{\int }_0^{2\pi /\omega }(i_0^2{\sin }^2\omega t)dt}{{\int }_0^{2\pi /\omega }dt}=\frac{i_0^2}{2}$

$\therefore i_{rms}=\sqrt{{⟨i^2⟩}_{onecycle}}=\frac{i_0^{}}{\sqrt{2}}\approx 0.707i_0^{}$

Thus, $i_{rms}=\frac{i_0^{}}{\sqrt{2}}\approx 0.707i_0^{}$

Similarly, we get

$V_{rms}=\frac{V_0^{}}{\sqrt{2}}\approx 0.707V_0^{}$

The square root of the mean square value is called the virtual value and is the value give by ac instruments.

Thus, when we speak of our house hold power supply as 220 volts ac, this means that the rms voltage is 220 volts and its voltage amplitude is,

V₀ = $\sqrt{2}$V_rms = 311 volt

Form Factor

The ratio, $\frac{rmsvalue}{averagevalue}=\frac{V_0/\sqrt{2}}{2V_0/\pi }=\frac{\pi }{2\sqrt{2}}=1.11$

is known as or factor,

Note:

1. The average value of sin $\omega$t, cos $\omega$t, sin2$\omega$t, cos2$\omega$t, etc. is zero because it is positive half of the time and negative rest half of the time. Thus,

$⟨\sin \omega t⟩=⟨\cos \omega t⟩=⟨\sin 2\omega t⟩=⟨\cos 2\omega t⟩=0$

$\therefore$ If i = i₀ sin $\omega$t

then $⟨i⟩=⟨i_0\sin \omega t⟩=i_0⟨\sin \omega t⟩=0$

2. The average value of sin² $\omega$t and cos² $\omega$t is $\frac{1}{2}$

or $⟨{\sin }^2\omega t⟩=⟨{\cos }^2\omega t⟩=\frac{1}{2}$

$\therefore$If $i^2=i_0^2{\sin }^2\omega t$

then $⟨i^2⟩=⟨i_0^2{\sin }^2\omega t⟩=i_0^2⟨{\sin }^2\omega t⟩=\frac{i_0^2}{2}$

3. Like SHM, general expressions of current/voltage in an sinusoidal ac are,

i = i₀ sin ($\omega$t $\pm$ $\varphi$), V = V₀ sin ($\omega$t $\pm$ $\varphi$)

or i = i₀ cos ($\omega$t $\pm$ $\varphi$), and V = V₀ cos ($\omega$t $\pm$ $\varphi$)

Illustration 1: If the current in an ac circuit is represented by the equation,

i = 5 sin (300t - $\pi$/4)

Here, t is in second and i in ampere. Calculate,

(a) peak and rms value of current.

(b) frequency of ac

(c) average current

Solution:

(a) An in case of ac,

i = i₀ sin ($\omega$t $\pm$ $\varphi$)

$\therefore$The peak value i₀ = 5A

and $i_{rms}=\frac{i_0^2}{\sqrt{2}}=\frac{5}{\sqrt{2}}=3.535A$

(b) Angular frequency $\omega$ = 300 rad/s

$\therefore$ $f=\frac{\omega }{2\pi }=\frac{300}{2\pi }\approx 47.75HZ$ (c)

$i_{av}=\left(\frac{2}{\pi }\right)i_0^2=\left(\frac{2}{\pi }\right)(5)=3.18A$

Capacitor in an AC Circuit

If a capacitor of capacitance C is connected across the alternating source, the instantaneous charge on the capacitor,

q = CV_C = CV₀ sin $\omega$t and the instantaneous current i passing through it, is given by:

$i=\frac{dq}{dt}=C{V}_0\omega \cos \omega t$

$=\frac{V_0}{1/\omega C}\sin (\omega t+\pi /2)$

or i = i0 sin ($\omega$t + $\pi$/2)

Here, $V_0=\frac{i_0}{\omega C}$

This relation shows that the quantity $\frac{1}{\omega C}$ is the effective ac resistance or the capacitive reactance of the capacitor and is represented as X_C. It has unit as ohm. Thus,

$X_C=\frac{1}{\omega C}$

It is clear that the current leads the voltage by 90° or the potential drop across the capacitor lags the current passing it by 90°.

Figure shows V and i as functions of time t.

Inductor in an AC Circuit

Consider a pure inductor of self inductance L and zero resistance connected to an alternation source. Again we assume that an instantaneous current i = i₀ sin $\omega$t flows through the inductor. Although there is no resistance, there is a potential difference V_L between the inductor terminals a and b because the current varies with time, giving rise to self induced emf.

V_L = V_ab = - (induced emg) = $-\left(-L\frac{di}{dt}\right)$

Or V_L = $L\cdot \frac{di}{dt}=L{i}_0\omega \cos \omega t$

Or V_L= V₀ sin $\left(\omega t+\frac{\pi }{2}\right)$ (i)

Here V₀ = i₀($\omega$L) (ii)

Or i₀ = $\frac{V_0}{\omega L}$

$\therefore$ $i=\frac{V_0}{\omega L}\sin (\omega t)$ (iii)

Equation (iii) shows that effective ac resistance, i.e., inductive reactance of inductor is,

X_L = $\omega$L

And the maximum current,

$i_0=\frac{V_0}{XL}$

The unit of X_L is also ohm.

From Equations. (i) and (iii) we see that the voltage across the inductor leads the current passing through it by 90°.

Figure shows V_L and i as functions of time.

Illustration 2: A 100$\Omega$ resistance is connected in series with a 4H inductor. The voltage across the resistor is, V_R = (2.0V) sin (10³ rad/s)t:

(a) Find the expression of circuit current

(b) Find the inductive reactance

(c) Derive an expression for the voltage across the inductor.

Solution: (a) $i=\frac{V_R}{R}=\frac{(2.0V)\sin ({10}^{3}rad/s)t}{100}$

(2.0 x 10^-2 A) sin (10³ rad/s)t

(b) X_L = $\omega$L = (10³ rad/s) (4H)

= 4.0 ´ 10³ ohm

(c) The amplitude of voltage across inductor,

V₀ = i₀X_L = (2.0 x 10^-2 A) (4.0 x 10³ ohm)

= 80 volt

In an ac voltage across the inductor leads the current by 90° or $\pi$/2 rad. Hence,

V_L = V₀ sin ($\omega$t + $\pi$/2)= (80 volt) sin $\left\{({10}^{3}rad/s)t+\frac{\pi }{2}rad\right\}$

Note:

That the amplitude of voltage across the resistor (=2.0 volt) is not same as the amplitude of the voltage across the inductor (=80 volt), even though the amplitude of the current through both devices is the same.