AC

Dunman High School (Senior High Physics) 9646 Physics (2014) Topic 17: Alternating Currents 17-1 H2 Topic 17 Alternating Currents A typical electrical transformer found at power sub-stations A typical power adaptor for an electrical device

Dunman High School (Senior High Physics) 9646 Physics (2014) Topic 17: Alternating Currents 17-2 Learning Objectives Content • Characteristics of alternating currents • The transformer • Rectification with a diode Learning Outcomes Candidates should be able to: (a) show an understanding and use the terms period, frequency, peak value and root -mean-square value as applied to an alternating current or voltage. (b) deduce that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current. (c) represent an alternating current or an alternating voltage by an equation of the form x = x 0 sin ωt. (d) distinguish between r.m.s. and peak values and recall and solve problems using the relationship Irms = I0 / 2 for the sinusoidal case. (e) show an understanding of the principle of operation of a simple iron-cored transformer and recall and solve problems using Ns /Np = Vs /Vp = Ip /Is for an ideal transformer. (f) explain the use of a single diode for the half -wave rectification of an alternating current.

Dunman High School (Senior High Physics) 9646 Physics (2014) Topic 17: Alternating Currents 17-3 17.0 Introduction It is important to understand alternating currents (a.c.) because they are so much a part of our everyday life. Every time we turn on a television set, computer or any of a multitude of other electric appliances, we are using a.c. to provide the power to operate them. At the microscopic level, a.c. can be considered as having the charge carriers oscillate about a fixed point. An alternating current (a.c.) is a current that varies periodically with time in magnitude and direction (Polarity of the voltage source constantly changes). 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. 17.1 Quantities Characterising an Alternating Current Quantity Symbol Description Period T Time taken by alternating current to make one complete alternation. Frequency f The number of times the current passes through its zero value in the same direction in unit time. Angular Frequency ω A way of expressing the frequency of the alternating current in terms of radians per second instead of cycles per second. Peak Value (Amplitude) Io Maximum value of the alternating current in either direction of zero value in a periodic cycle. Peak to Peak Value Difference between the positive peak value and the negative peak value of the a.c. within a cycle. Mean Value <I> The average value of an a.c. over a given time interval. Root Mean Square Value Irms The value of alternating current that is equal to the steady direct current which would dissipate heat at the same average rate in a given resistor. t I 0 t I 0 t I 0

Dunman High School (Senior High Physics) 9646 Physics (2014) Topic 17: Alternating Currents 17-4 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 current across the resistor can be expressed by the equation I = I o sin ωt Similarly, the potential difference across the resistor is given by V = Vo sin ωt 17.1.1 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 an a.c. f lows in a resistor, implying that the mean value of an a.c. does not represent the effective value of the a.c. 17.1.2 Root-mean-square value of a.c. Let us consider a simple circuit consisting of a resistor and an a.c. source as shown. As with d.c. , a.c. also causes heating effect in resistors. Recall that the electrical power P dissipated by a resistor is P = I2R In an a.c., the current keeps changing magnitude from zero to Io, hence the instantaneous power dissipated from a resistor is given by 𝑃𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 = 𝐼2𝑅 Mean power dissipated from a resistor over a time interval 𝑃𝑚𝑒𝑎𝑛 = mean value of 𝑃𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 = mean value of 𝐼2𝑅 = (mean value of 𝐼2) × 𝑅 = 𝐼𝑟𝑚𝑠2 𝑅 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. t I 0 I0 -I0 T/2 T Peak value 3T/2 Peak-to-peak value I > 0 I < 0 resistor

Dunman High School (Senior High Physics) 9646 Physics (2014) Topic 17: Alternating Currents 17-5 To find what value of current in a d.c. will dissipate the same power as the mean power dissipated by an a.c. with maximum current Io, we equate: Pdc = <Pac> Idc 2 R = <Iac 2 R> Idc 2 R = <Iac 2> R since R is constant Idc = >< 2 acI = Irms Hence, the r.m.s. value of an a.c. is the value of a steady d.c. which will dissipate energy at the same rate as the mean power dissipated by an a.c. in a given resistor. 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 an a.c. is the value of the steady direct current which would dissipate energy at the same rate as the a.c. 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 = >< 2 acI = T dtI T ∫0 2 = period one cycle curve for area under I 2 So to find the r.m.s. current, first we square the current, next find its average value and then take the square root of this average value. For a sinusoidal a.c., I = Io sin ωt To find the r.m.s. value: resistor Io P = Io 2 R resistor <P> 0 to Io I0 -I0 I t t I2 I0 2 t I0 2 I2 A B ½ I0 2 T T T I = I0 I2 = I0 2 sin2ωt I2 = I0 2 sin2ωt Squaring Mean = ½ I0 2 Square Root =

Dunman High School (Senior High Physics) 9646 Physics (2014) Topic 17: Alternating Currents 17-6 For a sinusoidal current, Irms = 2 oI refer to Annex A Similarly, Vrms = 2 oV Example 1 Calculate the mean value and r.m.s. value for each of the a.c. shown (a) (b) Mean value: 〈𝑰〉 = ∫ 𝑰𝒅𝒕 𝑻 𝟎 𝑻 = 𝟒(𝟎. 𝟎𝟏) − 𝟒(𝟎. 𝟎𝟏) 𝟎. 𝟎𝟐 = 𝟎 𝐀 r.m.s. value: 〈𝑰𝟐〉 = ∫ 𝑰𝟐𝒅𝒕𝑻 𝟎 𝑻 = 𝟏𝟔(𝟎.𝟎𝟐) 𝟎.𝟎𝟐 = 𝟏𝟔 𝐀𝟐 ∴ 𝑰𝒓𝒎𝒔 = �〈𝑰𝟐〉 = 𝟒 𝐀 Mean value: 〈𝑰〉 = ∫ 𝑰𝒅𝒕 𝑻 𝟎 𝑻 = 𝟐(𝟎. 𝟎𝟏) − 𝟒(𝟎. 𝟎𝟏) 𝟎. 𝟎𝟐 = −𝟏 𝐀 r.m.s. value: 〈𝑰𝟐〉 = ∫ 𝑰𝟐𝒅𝒕𝑻 𝟎 𝑻 = 𝟒(𝟎.𝟎𝟏)+𝟏𝟔(𝟎.𝟎𝟏) 𝟎.𝟎𝟐 = 𝟏𝟎 𝐀𝟐 ∴ 𝑰𝒓𝒎𝒔 = �〈𝑰𝟐〉 = 𝟑. 𝟏𝟔 𝐀 I 2 t 16 I 2 t 4 16

Dunman High School (Senior High Physics) 9646 Physics (2014) Topic 17: Alternating Currents 17-7 17.1.3 Mean power of of a.c. For a sinusoidal alternating current, I = Io sin ωt The instantaneous power dissipated in the resistor Pinstantaneous = I2 R = Io 2 R sin2 ωt Mean power, Pmean = (mean value of I2) R = Io 2 R <sin2 ωt > = 2 1 Pmax So the mean power P dissipated by a resistive load is half the maximum power available. The r.m.s. values of an a.c. will obey the many of the same equations as a d.c., for example, Vrms = Irms R Power dissipation in a resistor, Pmean = <I2 > R = Irms 2 R = Irms Vrms R Vrms 2 = These equations also apply if the maximum values of the current and voltage are used. Example 2 A tourist from U.S.A. brings an electric wat