DHS 18 Alternating Current (Lecture Slides)

DHS Y6 Physics H2 [Alternating Currents] 02/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 1 ALTERNATING CURRENTS SECTION V ELECTRICITY AND MAGNETISM T opic 18 ALTERNATING CURRENTS • Characteristics of alternating currents • The transformer • Rectification with a diode 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? Content •Characteristics of alternating currents •The transformer •Rectification with a diode 1 2 3 4

DHS Y6 Physics H2 [Alternating Currents] 02/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 2 Learning Outcomes • understand and use • period (T) • frequency (f) • peak value (I0, V0) • root-mean-square value (Irms, Vrms) • deduce <P> = P0 / 2 for sinusoidal I • represent a.c. by • solve problems using for sinusoidal case 0 sin tI I ω rms 0 2I I Learning Outcomes (continue) • understand principle of operation of a simple iron - cored transformer • for an ideal transformer, recall and solve problems using • explain the use of a single diode for the half-wave rectification of an a.c. s p s p p sN N V V  I I Recall: A simple alternator θ cos cos sin B B BA BA t dN NBA tdt      Φ θ Φ ω ω ωε sinusoidal e.m.f. is induced in a coil rotating with constant speed in a uniform B field Recall: A d.c. generator the ring halves are attached to the loop and rotate with it; the commutator reverses the connections to the external circuit at regular positions where the e.m.f. reverses (arrangement of split rings) θ 5 6 7 8

DHS Y6 Physics H2 [Alternating Currents] 02/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 3 Difference between AC & DC Alternating current (A.C.) : an electric current which periodically reverses direction Direct current (D.C.) : an electric current which only flows in a single direction around a circuit Characteristics of A.C. period 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 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. peak value Maximum absolute value of the alternating current or voltage 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. Characteristics of A.C. T = 0.5 s f = 2 Hz ω = 2πf = 13 rad s−1 peak value = V0 peak to peak value = 2V0 Characteristics of A.C. T = 2 s f = 0.5 Hz positive peak value = V1 negative peak value = V4 9 10 11 12

DHS Y6 Physics H2 [Alternating Currents] 02/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 4 Voltage across a sinusoidal source V t 0 Equations for sinusoidal A.C. 0 sin tI I ω 0 sinV V t ω instantaneous value peak value 2 2T fπω= π mathematical symbols physical situation mathematical symbols physical situation 13 14 15 16

DHS Y6 Physics H2 [Alternating Currents] 02/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 5 mathematical symbols physical situation mathematical symbols physical situation mathematical symbols physical situation mathematical symbols physical situation 17 18 19 20

DHS Y6 Physics H2 [Alternating Currents] 02/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 6 mathematical symbols physical situation mathematical symbols physical situation mathematical symbols physical situation Self-assessment (2 min) 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 first reaches 4.8 A is t1 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 t2 = ………… .………… .. ms. 21 22 23 24

DHS Y6 Physics H2 [Alternating Currents] 04/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 1 Self-assessment (Answer) Self-assessment (working) 3 o o 1 0 2 2 4003 s 4004. 4.8 5.0sin 3 400 4.8sin 0.9603 5.0 400 73.74 or 106.263 400 0.409 1 7 or 0 . rad 9 s35 10 s 400sin 5.0sin .5 03 r 3 ad3 3.07 t t t t t T t t                                                I I or 4.43 s Note: If θ ≤ 180°, sin (θ) = sin (180° - θ) Quiz The graph shows the variation with time t of the current I in a wire. How would you represent the variation with time t of the net amount of charge Q which has flowed past a point in the wire? Q t0 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. 1 2 3 4

DHS Y6 Physics H2 [Alternating Currents] 04/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 2 0 sin t I ω Mean power in a resistive load for a sinusoidala.c. mean power in a resistive load is half the maximum power for a sinusoidal a.c.   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 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.   2 1 1sin 1 cos2 2 2t t   ω ω 0 sin t I ω Mean power in a resistive load for a sinusoidala.c. (derived in terms of V) mean power in a resistive load is half the maximum power for a sinusoidal a.c.       2 2 2 0 2 2 0 2 0 2 0 max mean value of sin sin 1 2 1 2 1 2 V P R V t R V t R V R V R P             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.   2 1 1sin 1 cos2 2 2t t   ω ω 0 sin t I ω Mean power in a resistive load for a sinusoidala.c. (derived in terms of V) mean power in a resistive load is half the maximum power for a sinusoidal a.c.       2 2 2 0 2 2 0 2 0 2 0 max mean value of sin sin 1 2 1 2 1 2 V P R V t R V t R V R V R P             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.   2 1 1sin 1 cos2 2 2t t   ω ω . . . . . . . . . Power in a resistive load for a sinusoidal a.c. P P I V 2 0 6.0sin 6. si 06.0 W, 3.0 2 sin W n3 . .0 2 0 P t P t P tV         ω I ω ω E.g. I is in phase with V: crests and troughs occur together Peaks are related as for a d.c. circuit: V0 = I0R 5 6 7 8

DHS Y6 Physics H2 [Alternating Currents] 04/04/2025 [For Internal Use Only] lim.boonsiong@dhs.edu.sg 3 Root-mean-square (r.m.s.) value 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. Root-mean-square (r.m.s.) value 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 Root-mean-square (r.m.s.) value Since <P> delivered by source is converted to internal energy in the resistor, can we define an equivalent constant current that would produce the same heating effect? How can we express this current in a simple, single value? Root-mean-square (r.m.s.) value Mean power dissipated in a resistor over one cycle:   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.) curre