Alternating Current

We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter.

The phrases ac voltage and ac current are contradictory and redundant, * respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points.

Figure 7.1 shows a resistor connected to a source $\epsilon$ of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by

$$v=v_m\sin \omega t(7.1)$$ where $v_m$ is the amplitude of the oscillating potential difference and $\omega$ is its angular frequency.

To find the value of current through the resistor, we

3.12), to the circuit shown in Fig. 7.1 to get

$$v_m\sin \omega t=iR$$

$$\text{or}i=\frac{v_m}{R}\sin \omega t$$

Since $R$ is a constant, we can write this equation as

$$i=i_m\sin \omega t(7.2)$$ where the current amplitude $i_m$ is given by

$$i_m=\frac{v_m}{R}(7.3)$$

not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by $i^{2}R$ and depends on $i^2$

(which is always positive whether $i$ is positive or negative) and not on $i$. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is

$$p=i^{2}R=i_m^{2}R{\sin }^2\omega t(7.4)$$

The average value of $p$ over a cycle is*

$$\bar{p}=⟨i^{2}R⟩=⟨i_m^{2}R{\sin }^2\omega t⟩[7.5(a)]$$

where the bar over a letter (here, $p$) denotes its average value and $⟨......⟩$ denotes taking average of the quantity inside the bracket. Since, $i_m^2$ and $R$ are constants,

$$\bar{p}=i_m^{2}R⟨{\sin }^2\omega t⟩[7.5(b)]$$

Using the trigonometric identity, ${\sin }^2\omega t=\frac{1}{2}(1-\cos 2\omega t)$, we have $⟨{\sin }^2\omega t⟩=(1/2)(1-⟨\cos 2\omega t⟩)$ and since $⟨\cos 2\omega t⟩=0$**, we have,

$$⟨{\sin }^2\omega t⟩=\frac{1}{2}$$

Thus,

$$\bar{p}=\frac{1}{2}{i}_m^{2}R[7.5(c)]$$

To express ac power in the same form as dc power ($P=I^{2}R$), a special value of current is defined and used. It is called, root mean square (rms) or effective current (Fig. 7.3) and is denoted by $I_{rms}$ or $I$.

and is related to the peak current $i_m$ by $I=i_m/\sqrt{2}=0.707i_m$.

The average value of a function $F(t)$ over a period $T$ is given by $⟨F(t)⟩=\frac{1}{T}{\int }_0^{T}F(t)dt$

$$⟨\cos 2\omega t⟩=\frac{1}{T}{\int }_0^T\cos 2\omega tdt=\frac{1}{T}{\left[\frac{\sin 2\omega t}{2\omega }\right]}_0^T=\frac{1}{2\omega T}[\sin 2\omega t{]}_0^T=0$$

It is defined by

$$I=\sqrt{\overline{i^2}}=\sqrt{\frac{1}{2}{i}_m^2}=\frac{i_m}{\sqrt{2}}$$

$$V=\frac{v_m}{\sqrt{2}}$$

From Eq. (7.3), we have $v_m=i_{m}R$

$$\frac{v_m}{\sqrt{2}}=\frac{i_m}{\sqrt{2}}R$$

$$R=\frac{V^2}{P}=\frac{(220\mathrm{V}{)}^2}{100\mathrm{W}}=484\Omega$$

(b) The peak voltage of the source is

EXAMPLE 7.1

$$v_m=\sqrt{2}V=311\mathrm{V}$$

(c) Since, $P=IV$

$$I=\frac{P}{V}=\frac{100\mathrm{W}}{220\mathrm{V}}=0.454\mathrm{A}$$

$$I=\frac{i_m}{\sqrt{2}}=0.707i_m(7.6)$$

In terms of $I$, the average power, denoted by $P$ is

$$P=\bar{p}=\frac{1}{2}{i}_m^{2}R=I^{2}R(7.7)$$

Similarly, we define the rms voltage or effective voltage by

$$V=\frac{v_m}{\sqrt{2}}=0.707v_m(7.8)$$

$$\text{or,}V=IR(7.9)$$