EJC Physics H218 Alternating Current - 1. Notes (2024) Full

Commemorative silver coin. In 2018, the Serbian Mint began a coin series to honor Nikola Tesla. It features a portrait of Tesla, and his induction generator on the other side. The generator generates Alternating Current and revolutionized industrial mechanics 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.) values 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 of the form 0 s tinxx = (d) distinguish between r.m.s. and peak values and recall and solve problems using the relationship rms 0 / 2=II for the sinusoidal case (e) show an understanding of the principle of operation of a simple iron-core transformer and recall and solve problems using / / / S P S P P SN N V V== II for an ideal transformer (f) explain the use of a single diode for the half-wave rectification of an alternating current. An alternating current (a.c.) is an electric current which has a flow direction that reverses periodically with time . At the microscopic level, an a.c. can be considered as having the charge carriers oscillate about fixed points. Compare that with a d.c. where the direction of flow of current is maintained in the same direction. A typical AC/DC adaptor. As seen in topic H215, this particular “handphone charger” outputs 5V D.C. at a maximum of 1 A. It accepts an input alternating voltage of 180 to 240 V rms, at a frequency of 50Hz or 60 Hz, taking up an current of 0.15 A.

Quantity Symbol Description period T time taken for a.c. to complete one cycle frequency f number of complete cycles of the a.c. per unit time angular frequency  (for a sinusoidal a.c.), product of 2π and the frequency of the a.c., 2 f= peak value (amplitude) 0I maximum value of the a.c. in either direction within a 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 average value of an a.c. over a given time interval root-mean- square (r.m.s.) value rmsI value of a steady direct current that will dissipate thermal energy at the same average rate as the a.c. in a given resistor A common representation of an a.c. is 0 si n t=II : I The mean (or average) current value is zero for an a.c. where 0 si n t=II . 0 for s n i0 t==I I I Within a complete cycle, there is a positive value of current and a corresponding negative value, so mean is zero. But a resistor heats up when an a.c. flows through it. The mean current value of an a.c. does not effectively characterise an a.c. t I 0 T peak value peak-to-peak value t I 0 T

Example 1 The mean value of an alternating current is zero. Explain why heating occurs when there is an alternating current in a resistor. Heating effect is due to power dissipated in resistor: For constant resistance R, observe above we [Squared] squared the current and [Mean] found the average. Comparing to an equivalent d.c. for the same heating effect: 22 0 dc 2 20 dc 0 rms square-root for next li 1 2 2 ne 2 II I I III == = = = avg dc P R R [Root] Method 1 • power dissipated is directly proportional to square of current, • square of current is always positive Method 2 • a.c. changes direction every half cycle • but heating effect is independent of current direction I t t P I02R t P A B V T T T We represent the sine a.c. as . Except at , some current is always flowing in the resistor. Energy output for a constant power output is . If power output is changing, energy output is area under P-t graph I02R I02R When working with power P, the squaring of current makes all values positive. average power Area A is equal to Area B.

rmsI To process r.m.s, we work backwards i.e. • [S] square the function • [M] find the average value • [R] square-root averaged value Example 2 Find the (i) mean value and (ii) r.m.s. value for each of the a.c. below (a) (b) Note: Determine carefully one period of a.c before performing r.m.s. calculations. I /A t /s 4 - 4 0.01 0.02 0.03 0.04 (i) mean value: (ii) r.m.s. value: Graph of ([S]quared): [M]ean value is = 16 A2 [R]oot t /s 16 0.01 0.02 0.03 0.04 I /A t /s 2 - 4 0.01 0.02 0.03 0.04 (i) mean value: (ii) r.m.s. value: Graph of ([S]quared): [M]ean value is [R]oot t /s 16 0.01 0.02 0.03 0.04 4 r.m.s. value of an alternating current is the value of a steady direct current that will dissipate thermal energy at the same average rate as the a.c. in a given resistor for sinusoidal a.c. 0 rms 0 rms mean max 2 2 1 2 VV PP = = = II (includes both sin and cos)

Most methods of circuit analysis applied in d.c. can be used in a.c. circuits as well. Consider: D.C. Quantity A.C. (sinusoidal) Instantaneous, peak and mean p.d. are the same constant value dcV instantaneous voltage across resistor ac 0 sin V tV = peak voltage across resistor 0V r.m.s. voltage across resistor 0 rms 2 VV = dc dc R V=I Instantaneous, peak and mean current are the same constant value instantaneous current flowing through resistor ( ) ( ) ac ac 0 0 0 1 sin sin sin VV t t R R V R t   ==   =  = I I peak current flowing through resistor 0 0 V R=I r.m.s. current flowing through resistor 2 rms ac 0 2 = = II I dc dc dc 2 dc 2 dc PV R V R = = = I I Instantaneous, peak and mean power are the same constant value instantaneous power dissipated in resistor ( ) ( ) 2 ac ac 2 0 22 0 sin sin R Rt P Rt   = = = I I I peak power dissipated in resistor 2 00P R= I mean power dissipated in resistor ( ) rms rms 00 00 0 22 1 2 1 2 PV V V P        =  = = = I I I d.c. source of emf E resistor of resistance R 0.02 a.c. source of emf resistor of resistance R 0.02 ~

Example 3 A tourist from the U.S.A. brought along an electric water kettle designed to operate with 110 V to Singapore and plugs it into a local 240 V outlet. The heater breaks down due to overheating. (a) If the heater typically draws 500 W, find the resistance of the heating coil. (b) Determine the power consumption when operated using a 240 V outlet. (c) Calculate the r.m.s. current when operated using the 240 V outlet. Note: Household electrical outlets are a.c. and are typically labelled with their r.m.s. value i.e. (i) the a.c. supply is rated at 240 Vrms (iI) the peak p.d. is about 340 V. ( ( )0 2 2 240 V 339== rmsVV ) Example 4 A steady current of 2.0 A dissipates a certain power in a variable resistor. The variable resistor is now connected to a sinusoidal alternating current. (a) If power dissipation remains the same, state and explain the r.m.s. value of the a.c.. (b) The resistance has to be halved in order to maintain the same power dissipation. Find the new rmsI . (a) 2.0 A. The r.m.s. value of an a.c. is the value of a steady direct current that will dissipate thermal energy at the same average rate as the a.c. in a given resistor. (b) power dissipated in resistor is 2P R= I for same power dissipation ( ) ac dc 2 rms 2 dc 22 rms dc rms dc 1 1 2 2 2 0 2 8 A 2 2 I I II II =  = = = =   = P R R . P . Note: For exams, compute final answers according to the least no. of sf. Do not leave them in surds. (a) (b) Assume R remains constant as heating coil heats up (Note that this is almost