DHS 18 Alternating Current (Notes & Tutorial)

DUNMAN HIGH SCHOOL (SENIOR HIGH PHYSICS) YEAR 6 (2025) H2 Physics (2025) 18-1 Topic 18: Alternating Currents TOPIC 1 8: Alternating Current ___________________________________________________________________________________ GUIDING QUESTIONS  How is alternating current different from direct current?  How can we describe alternating current, e.g., mathematically, and graphically?  Why is alternating current used in the generation and transmission of electricity? 18.0 Introduction During the 1880s in the US there was a heated and acrimonious debate between two inventors over the best method of electric-power distribution. Thomas Edison (1847-1931) favoured direct current (d.c.) – i.e., steady current that does not vary with time. George Westinghouse (1846 - 1914) favoured alternating current (a.c.), with sinusoidally varying voltages and currents. He argued that transformers (which we will study in this chapter) can be used to step the voltage up and down with a.c. but not with d.c.; low voltages are safer for consumer us e, but high voltages and correspondingly low currents are best for long -distance power transmission to minimize I2R losses in the cables. Eventually, Westinghouse prevailed, and most present -day household and industrial power - distribution systems operate with a.c.. Every time you turn on a light, a television set, a stereo, or any of a multitude of other electrical appliances in a home, you are using alternating currents to provide the power to operate them.

DUNMAN HIGH SCHOOL (SENIOR HIGH PHYSICS) YEAR 6 (2025) H2 Physics (2025) 18-2 Topic 18: Alternating Currents In this chapter we will learn how resistors behave in circuits with sinusoidally varying voltages and currents, why transformers are useful and how they work, and how a single diode converts the a.c. to d.c. An alternating current is an electric current that periodically reverses its direction in a circuit with a frequency. Below are some examples of a.c.: The principles of direct current in resistors learned in previous topics can be applied to resistors in a.c. circuits. However, major differences in circuit analysis arise when inductors and capacitors are to be considered. 18.1 Quantities characterising an alternating current or voltage Quantity Symbol Description period T The period of an A .C. source refers to the time taken to complete one cycle. The SI units for time is seconds (s). frequency f The frequency of an A .C. source refers to the number of complete cycles per unit time. The SI units for frequency is Hertz (Hz) or s−1. Frequency is related to period by the equation 1f T . For a sinusoidal A .C. waveform, the concept of angular frequency 2 f  is useful. peak value I0 Maximum absolute value of the alternating current or voltage in either direction of zero value in a periodic cycle. For a sinusoidal waveform, this corresponds to the amplitude for simple harmonic motion. If the waveform is not symmetrical, it makes more sense to distinguish peak positive value and peak negative value. root mean square value Irms Root-mean-square current of alternating current is the value of the steady direct current which would dissipate heat at the same average rate in a given resistor. peak to peak value Difference between the positive peak value and the negative peak value of the a.c. within a cycle. t I 0 t I 0 t I 0

DUNMAN HIGH SCHOOL (SENIOR HIGH PHYSICS) YEAR 6 (2025) H2 Physics (2025) 18-3 Topic 18: Alternating Currents The most commonly encountered form of a.c. is the sinusoidal form, that is, it varies with time according to a sine or cosine function. The sinusoidal alternating current can be represented by the equations: 0 0 sin sin t V V t   I I ω ω In these expressions, I and V are the instantaneous current and voltage; I0 and V0 are the peak current and voltage (current and voltage amplitude); and ω is the angular frequency, equal to 2πf or 2π/T. Self-assessment: 1. For this AC source, the peak current is ……… ….. …… A. 2. The period is …………………….. s. 3. We can write the equation, with I measured in amperes and t measured in seconds, as I = … … …… sin … …. …… πt 4. The time when the current fir st reaches 4.8 A is t 1 and the next time the current has the same value of 4.8 A is t2. Using the equation to solve for these times, we have t1 = ……… ………. …….. ms and t 2 = … ………. ………….. ms . 18.1.1 Mean value of a.c. For the case of sinusoidal current, any positive value of current, there will be a corresponding negative value within a complete cycle, thus the mean value of current I is zero. However, heat is dissipated when it flows in a resistor, implying that the mean value of an a.c. does not represent the effective value of the a.c.

DUNMAN HIGH SCHOOL (SENIOR HIGH PHYSICS) YEAR 6 (2025) H2 Physics (2025) 18-4 Topic 18: Alternating Currents 18.1.2 Mean power of sinusoidal alternating current in a resistive load For a sinusoidal alternating current, I = I0 sin ωt The instantaneous power dissipated in the resistor 2 2 2 instantaneous 0 sinP R R t  I I Mean power,   2 2 2 0 2 2 0 2 0 2 0 max mean value of sin sin 1 2 1 2 1 2 P R t R t R R R P                I I I I I So, the mean power dissipated in a resistive load is ……………………………………………… . The mean power delivered by the source is converted to internal energy in the resistor, just as in the case of a d.c. circuit. 18.1.3 Root-mean-square value of a.c. Although the current is not in one direction only, power is converted in the resistor. This is because the power/heating depends on I2, so independent of current direction. Recall that the electrical power P dissipated in a resistor is P = I2R In an a.c., the instantaneous power dissipated in a resistor is given by P R  2 instantaneous I Mean power dissipated in a resistor over one cycle:

DUNMAN HIGH SCHOOL (SENIOR HIGH PHYSICS) YEAR 6 (2025) H2 Physics (2025) 18-5 Topic 18: Alternating Currents   P P R R R R      instantaneous 2 2 2 2 2 rms mean value of I I I I Irms is the square root of the mean value of I2 and hence is known as the root-mean-square (r.m.s.) current of the a.c. source. To find what value of current by a d.c. source will dissipate the same power in a resistor as the mean power dissipated by an a.c. when it flows through the resistor, we equate: P P R R R R       dc 2 2 dc 2 2 2 dc 2 dc rms since is constant I I I I I I I I In other words, a d.c. of magnitude Irms will produce the same heating effect as the a.c. Thus the r.m.s. value can be considered as the effective value of the a.c. The r.m.s. value of alternating current is the value of the steady direct current which would dissipate heat at the same average rate in a given resistor. Graphically, it is the square root of the mean value of the square of the instantaneous current over one cycle. Irms = 2I = T dt T  2 0 I = t 2area under - graph over one cycle period I any

DUNMAN HIGH SCHOOL (SENIOR HIGH PHYSICS) YEAR 6 (2025) H2 Physics (2025) 18-6 Topic 18: Alternating Currents Procedure for calculating the r.m.s. current: 1. Square the current at each point in time for the entire repeating pattern. 2. Find the average (mean value) of these squared values, over the entire period. 3. Take the square root of this average. In general, the r.m.s. current of an a.c. source is also known as the d.c. equivalent current. For a sinusoidal a.c., 0 rms 2  II (refer to Annex A) Note that in a similar way, we can also find the root-mean-square (r.m.s.) voltage for an a.c. circuit. So,