2018 Alternating currents Lecture notes (tutors)

9749 H2 Physics Lecture Notes Nanyang Junior College 1 Chapter 18 ALTERNATING CURRENT Content • Characteristics of alternating currents • The transformer • Rectification with a diode Learning Outcomes Candidates should be able to (a) show an understanding of and use the terms period, frequency, peak value and root-mean-square (r.m.s.) value as applied to an alternating current or voltage. (b) deduce that the mean power in a resistive load is half the maximum (peak) power for a sinusoidal alternating current. (c) represent an alternating current or an alternating voltage by an equation in the from sinoxx t ω= (d) distinguish between r.m.s. and peak values and recall and solve problems using the relationship o rms II= 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 ss P PPS NV I==NVI for an ideal transformer. (f) explain the use of a single diode for the half -wave rectification of an alternating current.

9749 H2 Physics Lecture Notes Nanyang Junior College 2 18.1 Introduction An alternating current (a.c.) is an electric current which has a flow direction that reverses periodically with time. An a.c. at the microscopic level can be considered as having the charge carriers oscillate about a fixed point. Examples of Direct Current (d.c.) Examples of Alternating Current (a.c.) It is important to understand a.c. because they are so much a part of our everyday life. Each time a television set, computer or any other electric appliances which are connected t o the wall socket are turned on, an a.c. provides the power to operate them . 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. 18.2 Quantities Characterising an Alternating Current Quantity Symbol Description Period T The time taken for the a.c. to complete one cycle. Frequency f The number of complete cycles undergone by the a.c. in one second. Angular Frequency ω A way of expressing the frequency of the a.c. in terms of radians per second instead of cycles per second. Peak Value (Amplitude) Io The maximum value of the a.c. in either direction within a cycle. Peak to Peak Value The 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 a steady d.c. that will dissipate energy at the same rate as the a.c. in a given resistor. t I 0 t I 0 t I 0 t I 0 t I 0

9749 H2 Physics Lecture Notes Nanyang Junior College 3 18.3 Sinusoidal Alternating Current The most commonly encountered a.c. takes a sinusoidal form as shown and the waveform can be expressed by the equation 𝐼𝐼 = 𝐼𝐼𝑜𝑜𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 As voltage is related to current by the equation =V IR , ( ) ( ) 0 0 0 sin sin sin ω ω ω = = = V I tR IR t Vt Hence, current and voltage across a resistor are in phase. For any positive value of current, there will be a corresponding negativ e value within a complete cycle, thus the mean value of current <I> is zero. However heat is dissipated when an a.c. flows in a resistor, effectively implying that the mean value of an a.c. does not represent the effective value of the a.c. 18.4 Mean Power The instantaneous power, P, is given by the following equations, ( )( )00 2 00 2 0 sin sin sin sin ωω ω ω = = = = P IV I tV t IV t Pt or ( ) ( ) 2 2 0 22 0 2 0 sin sin sin ω ω ω = = = = P IR I tR IR t Pt or ( ) 2 2 0 2 20 2 0 sin sin sin ω ω ω = = =  = VP R Vt R V tR Pt t I 0 I0 -I0 T/2 T Peak value 3T/2 Peak-to-peak value

9749 H2 Physics Lecture Notes Nanyang Junior College 4 Graphically, The mean power, P , can be found by considering the graph of P − t, Notice that the shaded regions under the curve and above the dashed line (A) have the same area as the shaded regions above the curve and below the dashed line (B). Thus, the mean power in a resistive load is half the maximum power for a sinusoidal alternating current i.e. 0 2= PP Note: 1. The period of power is half that of the sinusoidal a.c. and therefore its frequency is twice the frequency of a.c. 2. Mean power is half the maximum power i.e. 0 2= PP , for sinusoidal a.c. only. P t 0 P0 V0 I0 T 2T P P0 T 2T 0 t A B

9749 H2 Physics Lecture Notes Nanyang Junior College 5 18.5 Root Mean Square Value Since energy is dissipated at the same rate then I dc = Irms. 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. 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. Thus the r.m.s. value can be considered as the effective value of the a.c. I R R 22 2 ( 2 2 2 2 2 IP = I R = )R I R P= P I =I II II o ac rms dc rms o rms o rms 〈〉〈 〉 = 〈〉 = = 2P=I Rdc

9749 H2 Physics Lecture Notes Nanyang Junior College 6 18.5.1 Root Mean Square Value (Graphical) As the area under a P - t graph would yield the energy dissipated in the given resistor, the area under the Pinstantaneous - t graph must be equal to the area under the Pmean - t graph such that the energy dissipated is the same. For a sinusoidal alternating current From the above graphs, it can be seen that for a sinusoidal alternating current, Hence for a sinusoidal alternating current, the r.m.s. value of the current I rms is related to the peak current Io by the expression 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐼𝐼𝑜𝑜 √2 I0 -I0 I t t P I02R t I02R P A B ½ I02 R T T T I = I0 sinωt I2 R= I02 R sin2ωt 〈𝑃𝑃〉 * Mathematically, it can be shown that Shaded Area A = Shaded Area B 2 2 ( 2 2 2 2 2 IP = )R I R II II o rms o rms o rms 〈〉 = = =

9749 H2 Physics Lecture Notes Nanyang Junior College 7 Similarly, General Knowledge The alternating electrical supply from wall sockets in Singapore is 240 V, 50 Hz. Is 240 V the peak value, or the r.m.s. value of the supply? Problem Solving Skill Set (PS3) 1. Consider one period of the alternating quantity. 2. Square the quantity. 3. Mean (average) the squared quantity. 4. Square root the mean-square quantity. Since r.m.s. values are the effective values of a.c., = rms rmsP IV or 2= rmsPIR or 2 rmsVP R= For a sinusoidal a.c., rms rms 00 00 0 P =I V IV= 22 IV= 2 P= 2      which is the same result as shown previously. Note: 1. The expressions 0 rms II= 2 and rms VV = 0 2 are only true for sinusoidal a.c. 2. The three equations for mean power i.e. rms rmsP =I V , 2 rmsP =I R and 2 rmsVP= R can be used for all types of a.c. 3. Current and voltage ratings in electrical appliances/supplies are expressed in r.m.s. values. 4. Power ratings in electrical appliances/supplies refer to the mean (average) power. rms VV = 0 2

9749 H2 Physics Lecture Notes Nanyang Junior College 8 Example 1 Determine the mean value and r.m.s. value for each of the alternating current shown (a) (b) Mean value: 〈𝑰𝑰〉 = ∫ 𝑰𝑰𝑰𝑰𝑰𝑰 𝑻𝑻 𝟎𝟎 𝑻𝑻 = 𝟒𝟒(𝟎𝟎. 𝟎𝟎𝟎𝟎) − 𝟒𝟒(𝟎𝟎. 𝟎𝟎𝟎𝟎) 𝟎𝟎. 𝟎𝟎𝟎𝟎 = 𝟎𝟎 𝐀𝐀 r.m.s. value: 〈𝑰𝑰𝟎𝟎〉 = ∫ 𝑰𝑰𝟎𝟎𝑰𝑰𝑰𝑰𝑻𝑻 𝟎𝟎 𝑻𝑻 = 𝟎𝟎𝟏𝟏(𝟎𝟎.𝟎𝟎𝟎𝟎) 𝟎𝟎.𝟎𝟎𝟎𝟎 = 𝟎𝟎𝟏𝟏 𝐀𝐀𝟎𝟎 ∴ 𝑰𝑰𝒓𝒓𝒓𝒓𝒓𝒓 = �〈𝑰𝑰𝟎𝟎〉 = 𝟒𝟒 𝐀𝐀 Mean value: 〈𝑰𝑰〉 = ∫ 𝑰𝑰𝑰𝑰𝑰𝑰 𝑻𝑻 𝟎𝟎 𝑻𝑻 = 𝟎𝟎(𝟎𝟎. 𝟎𝟎𝟎𝟎) − 𝟒𝟒(𝟎𝟎. 𝟎𝟎𝟎𝟎) 𝟎𝟎. 𝟎𝟎𝟎𝟎 = −𝟎𝟎 𝐀𝐀 r.m.s. value: 〈𝑰𝑰𝟎𝟎〉 = ∫ 𝑰𝑰𝟎𝟎𝑰𝑰𝑰𝑰𝑻𝑻 𝟎𝟎 𝑻𝑻 = 𝟒𝟒(𝟎𝟎.𝟎𝟎𝟎𝟎)+𝟎𝟎𝟏𝟏(𝟎𝟎.𝟎𝟎𝟎𝟎) 𝟎𝟎.𝟎𝟎𝟎𝟎 = 𝟎𝟎𝟎𝟎 𝐀𝐀𝟎𝟎 ∴ 𝑰𝑰𝒓𝒓𝒓𝒓𝒓𝒓 = �〈𝑰𝑰𝟎𝟎〉 = 𝟑𝟑. 𝟎𝟎𝟏𝟏 𝐀𝐀