Higher Physics Monitoring And Measuring Alternating Current-Books Download

Higher Physics Monitoring and measuring alternating current
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3 1 Electrons and Energy,Monitoring and measuring alternating current. Alternating current, Previously you learned that an alternating current a c periodically changes direction. The electrons flow in one direction before changing to flow in the opposite way This. contrasts with direct current d c which flows in one direction only. Alternating current is produced by rotating an electromagnet in a coil of wire This. means that the induced voltage which is constantly changing in the form of a sine. function pushes the current one way and then the other. An oscilloscope shows a trace on a screen of how a voltage varies with time. The changing voltage is plotted on the y axis against time on the x axis. Examples of each type of current when displayed on an oscilloscope screen are shown. a c waveform d c waveform,Measuring frequency and peak voltage. The values for voltage and time on an oscilloscope can be read from the number of. divisions on the screen multiplied by the scale shown on one of two dials on the control. panel at the side of the screen, The voltage axis scale called the y gain is usually given as volts div or volts cm and the. time axis scale known as the time base is shown as multiples of seconds div or. seconds cm, The frequency of an a c signal is calculated from its period This is the time for one.
complete cycle of the current as it moves in one direction then the other so we. measure the horizontal distance on the screen between crests. The peak voltage is defined as the distance from the maximum positive value to the. centre line or from the maximum negative value to the centre line This can be. obtained from an oscilloscope by counting the number of volts from the maximum value. to the minimum value and dividing by two,Worked example. Calculate i the frequency and ii the peak voltage of the waveform shown on the CRO. screen below Each box on the CRO screen has a side of length 1 cm. Timebase 5 ms,Volts div 2 V cm,i Frequency ii Peak Voltage. The distance between crests is 4 cm The distance from bottom to top is 8 cm. The time base is set at 5 ms cm 1 The volts div is set at 2 V cm 1. Period T 4 5 ms 4 0 005 0 02 s V peak 8 2,f 50 Hz V peak 8 V. Timebase switched off, If the time base is switched off the a c signal will not spread along the x axis However. the voltage variation will continue to oscillate up and down meaning a straight vertical. line will be displayed on the screen, The height of this line from bottom to top can be processed in exactl y the same way as.
above ie halved then converted into voltage using the volts div. Alternating current calculating an average, Electricity is a method of transferring energy so alternating current a c is capable of. transferring energy in the same way as direct current d c. The instantaneous amount of power being supplied can be calculated from the voltage. and or current at a particular time using the equation P I x V. However in a c these values are changing,Peak voltage. constantly so it is necessary to calculate an Voltage. average However for a sine wave there is,an equal number of peaks above the centre. line positive and below the line negative,Therefore the average value of the voltage Time. during any complete cycle is zero,It is necessary to use a different average.
Alternating current peak and rms, The amount of energy transmitted by a c will depend on some average value of the. voltage The average value of an a c voltage is called the root mean square rms. voltage V rms, The rms voltage is defined as the value of direct voltage that produces the same power. eg heating or lighting as the alternating voltage, The rms value is what is quoted on a power supply so that a fair comparison be tween. a c and d c can be made eg a 6 V battery d c will produce the same brightness of. light bulb as a 6 V rms a c supply, The rms current has a similar definition to the rms voltage. Consider the following two circuits which contain identical lamps. d c voltage a c voltage, The variable resistors are altered until the lamps are of equal brightnes s As a result the.
d c has the same value as the effective a c ie the lamps have the same power output. By measuring the voltages shown on the two oscilloscopes it is possible to determine. the relationship between the rms voltage and the peak voltag e Also since V IR. applies to the rms values and to the peak values a similar equation for current can be. V rms 1 V peak and Irms 1 Ipeak, Example A transformer is labelled with a primary coil of 230 Vrms and a secondary coil of. 12 Vrms What is the peak voltage which would occur in the secondary. V peak 2 V rms,V peak 1 41 12,V peak 17 0 V,Graphical method to derive the relationship. between peak and rms values of alternating current. The power produced by a current I in a resistor of resistance R is given by P I 2 R. A graph of I 2 against t for an alternating current is shown below A similar method can. be used for voltage,I peak I peak, The average value of I 2 is and therefore the average power supplied is P R. An identical heating effect power output for a d c supply is P I rms 2 R. since I rms is defined as the value of d c current that will supply equivalent power. Setting both of these equal to each other gives,Important notes. 1 Readings on meters that measure a c are rms values not peak values. e g a multimeter switched to a c mode will display rms values. 2 For power calculations involving a c always use rms values. P I rmsVrms I rms, 3 The mains supply is usually quoted as 230 V a c This is of course 230 V rms.
The peak voltage rises to approximately 325 V This voltage is dangerous. therefore electrical insulation must be provided to withstand this peak voltage. Current voltage power and resistance,Basic definitions. In the circuit below when S is closed the free electrons in the conductor experience a. force that causes them to move Electrons will tend to drift away from the negatively. charged end of the battery to the positively charged end of the battery. Current is a flow of electrons and is given by the flow of charge coulombs per second. The energy required to drive the electron current around this circuit is provided by a. chemical reaction in the battery The electrical energy that is supplied by the source is. transformed into other forms of energy in the components that ma ke up the circuit. Voltage is defined as the energy transferred per unit charge. Where E W is the work done ie energy transferred therefore. 1 volt V 1 joule per coulomb J C 1, When energy is being transferred from an external source to the circuit the voltage is. referred to as an electromotive force emf, When energy is transformed into another form of energy by a component in the circuit. the voltage is referred to as a potential difference pd. In the battery,Energy is supplied to the circuit, In the component Chemical energy electrical energy. Energy is provided by the circuit An emf voltage can be measured. Electrical energy light heat energy,A pd voltage can be measured.
Sources of emf, Emfs can be generated in a great variety of ways See the table below for examples. Chemical energy drives the current,Chemical cell,eg battery. Heat energy drives the current,Thermocouple,eg temperature sensor in an oven. Mechanical vibrations drive the current,Piezo electric generator. eg acoustic guitar pickup,Light energy drives the current.
Solar cell,eg solar panels on a house,Changes in magnetic field drive the current. Electromagnetic generator,eg power stations, In any circuit providing the resistance of a component remains constant if the. potential difference V across the component increases the current I through the. component will increase in direct proportion,At a fixed temperature for a given conductor. ie V I constant,The constant of proportion is defined to be. the resistance,ie R or V IR,This is Ohm s law, A component that has a constant resistance when the current through it is increased is.
said to be ohmic, Some components do not have a constant resistance their resistance change s as the pd. across the component is altered for example a light bulb a transistor or a diode These. are said to be non ohmic, A graph of V versus I for such components will not be a straight line. Circuit rules, You must be able to apply the circuit rules that you learned last year to series and parallel. circuits both alone and in combination The circuit rules are summarized in the table below. Series Parallel,Current IS I1 I 2 I3 I S I1 I 2 I3. Voltage VS V1 V2 V3 VS V1 V2 V3,Resistance RT R1 R2 R3.
RT R1 R 2 R 3, where V S supply voltage I S current from the supply and R T total resistance. 1 Find the readings on the meters in the following circuit. Step 1 Calculate the total resistance in the circuit. RT R1 R2 15 9 24,Step 2 Calculate current,I V R 12 24 0 5 A. Step 3 Calculate the voltage across the 9 resistor. V IR 0 5 x 9 4 5 V, 2 Find the readings on the meters in the following circuit. 12 V Solution, Step 1 Calculate the total resistance in the circuit. 1 RT 1 R1 1 R2 1 3 1 5 8 15,Step 2 Calculate current.
I V R 12 1 9 6 3 A,Step 3 Voltage across each branch is equal to the. supply voltage so V 12 V,Electric power, The power is defined as the electrical energy transferred in one second. work done QV Q,power V IV,Therefore P I x V,From Ohm s law V IxR and I. Substituting P I IR P V,Therefore P I2R P, The expression I 2 R gives the energy transferred in one second due to resistive heating. Apart from obvious uses in electric fires cookers toasters etc consideration has to be. given to heating effects in resistors transistors and integrated circuits and care taken. not to exceed the maximum ratings for such components. The expression V 2 R is particularly useful when the voltage of a power supply is fixed. and you are considering changes in resistance eg different po wer of heating elements. Example An electric heater has two heating elements allowing three settings low. medium and high Show by calculation which element s would be switched. on to provide each power setting,For 30 P 1800 W low power.
For 15 P 3500 W medium power,Combined resistance R T 10. RT 15 30 10,For both P 5300 W high power,Addition of power. By conservation of energy the total power used in both series and parallel circuits is. the sum of the power used in each component, Example Calculate the total power dissipated in each of the following circuits. a Step 1 Calculate the total current in circuit,I V S R T 9 18 0 5 A. Step 2 Calculate power in each resistor,P 1 I 2 R 0 5 2 x 12 3 W.
P 2 I 2 R 0 5 2 x 6 1 5 W,Step 3 Calculate total power. P T P 1 P 2 3 1 5 4 5 W,b Step 1 Calculate power in each resistor. P 1 V 2 R 12 2 24 6 W,P 2 V 2 R 12 2 6 24 W,Step 2 Calculate total power. P T P 1 P 2 6 24 30 W,Identical resistors in parallel. Two resistors of the same value R wired in parallel R. R P half the value of one of them R,Examples 2 1000 in parallel R T 500.
2 250 in parallel R T 125,Voltage and current by proportion. The voltage across an individual resistor in a series circuit is proportional to the. resistance, The current through one of two resistors in parallel is inversely proportional to the. resistance ie proportional to the other resistance. Consider the following examples,i Current in parallel ii Voltage in series. 3A will split in the ratio R S 8 4 12 24,ie For 8 V1 12 4 V. 4 2 4 2 3 3 24,For 2 I 1 3 2 A For 4 V2 12 2 V,For 4 I 2 3 1 A For 12 V3 12 6 V.
I 1 2 A I 2 1 A,The smallest resistor takes,the largest current. Worked example 1,In the circuit shown opposite calculate. a the current in each resistor,b the pd across each resistor. c the power dissipated in each resistor 6 4,Solution 10. a R S R 1 R 2 4 6 10,Rp 10 10 5,R T 5 5 10,IT 4 A current through 5.
I 1 meets a resistance of 10 and so does I 2,so I 1 I 2 2 A. b For the 5 resistor V IR 4 5 20 V,for the 4 resistor V IR 2 4 8 V. for the 6 resistor V IR 2 6 12 V,for the 10 resistor V IR 2 10 20 V. c For 5 P IV 4 20 80 W,for 4 P I 2 R 2 2 4 16 W,for 6 P 24 W. for 10 P IV 2 20 40 W,Worked example 2,A 10 W model car motor operates on a.
12 V supply In order to adjust the speed 12 V,of the car it is connected in series with a. controller rheostat,The controller is adjusted to reduce the. motor s power to 4 W M, Assuming that the resistance of the motor does not change calculate. a the resistance of the controller at this setting. b the power wasted in the controller, a We cannot find the resistance of the controller directly We need t o find the. resistance of the motor and the total resistance of the circuit when the motor is. operating at 10 W, At 10 W the full 12 V acts across the motor The voltage across the rheostat is 0 V.
i e the rheostat is set to zero Ohms,For the motor at 10 W P 10 W V 12 V. V2 V 2 122,For the motor at 4 W P I 2 R so I 0 527 A. For the whole circuit RT 22 8,Resistance of controller 22 8 14 4 8 4.

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