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Alternating Current and Voltage Electricity & Electronics 5: Alternating Current and Voltage AIM This unit looks at several aspects of alternating current and voltage including measurement of frequency and something called root-mean-square values. Some of the mathematics is quite tricky, but you are not required to reproduce it in the final exam. However, you may need to read through it several times – good luck! OBJECTIVES On completing this unit you should be able to: • describe how to measure frequency using an oscilloscope. • state the relationship between peak and rms values for a sinusoidally varying voltage and current. • carry out calculations involving peak and rms values of voltage and current. • state the relationship between current and frequency in a resistive circuit. Strathaven Academy -1- Electricity and Electronics Alternating Current and Voltage Alternating Current and Voltage Peak and r.m.s. values The graph of a typical alternating voltage is shown below. The maximum voltage is called the peak value. From the graph it is obvious that the peak value would not be a very accurate measure of the voltage available from an alternating supply. In practice the value quoted is the root mean square (r.m.s.) voltage. The r.m.s. value of an alternating voltage or current is defined as being equal to the value of the direct voltage or current which gives rise to the same heating effect (same power output). Consider the following two circuits which contain identical lamps. A A The variable resistors are altered until the lamps are of equal brightness. As a result the direct current has the same value as the effective alternating current (i.e. the lamps have the same power output). Both voltages are measured using an oscilloscope giving the voltage equation below. Also, since V=IR applies to the r.m.s. valves and to the peak values a similar equation for currents can be deduced. V r.m .s .= 1V p e a k 2 a n d 1 Ir.m .s .= Ip e a k 2 Note: a moving coil a.c. meter is calibrated to give r.m.s. values. Strathaven Academy -2- Electricity and Electronics Alternating Current and Voltage ACTIVITY 8 Title: Alternating Current – Peak and r.m.s. values Aim: To establish a relationship between peak and equivalent direct (r.m.s.) values of voltage. Apparatus: Lab pack, 6 V battery, oscilloscope, variable resistor (0-22 Ω), 2 × 2.5 V lamps, connecting leads. Circuit 1 Circuit2 to oscilloscope to oscilloscope variable a.c. supply Instructions • Set up Circuit 1. • Switch the time-base on the oscilloscope OFF. • Adjust the supply and the oscilloscope to give a measured peak alternating voltage of 1 V on the oscilloscope • Leave Circuit 1 switched on. • Set up Circuit 2. • Adjust the variable resistor until the lamp is the same brightness as the lamp in Circuit 1. • Use the oscilloscope to measure the direct voltage across this lamp. • Repeat the measurements for peak voltages of 2 V, 3 V, 4 V and 5 V. • Plot a graph of direct voltage against peak voltage. • Determine the gradient of the graph. • State the relationship between Vd.c. and Vpeak using the value obtained from the gradient of the graph. Strathaven Academy -3- Electricity and Electronics Alternating Current and Voltage Graphical method to derive relationship between peak and rms values of alternating current THIS IS NOT EXAMINABLE - IT IS HERE FOR INFORMATION ONLY! The power produced by a current I in a resistor R is given by I2 R. A graph of I2 against t for an alternating current is shown below. A similar method can be used for voltage. I2 The average value of I2 is Peak 2 An identical heating effect (power output) for a d.c. supply = I2r.m.s R [since I (d.c.) = Ir.m.s.] I2Peak R Average power output for a.c. = 2 I2 r.m.s R = I2 Peak R 2 Strathaven Academy hence I2r.m.s. = -4- I2 Peak 2 giving Ir.m.s. = I Peak Electricity and Electronics Alternating Current and Voltage Frequency of a.c. To describe the domestic supply voltage fully, we would have to include the frequency (i.e. 230 V 50 Hz). An oscilloscope can be used to find the frequency of an a.c. supply as shown below. Time base = 0.005 s cm-1 Wavelength = 4 cm Time to produce one wave = 4 × 0.005 = 0.02 s Frequency = y in p u t = 0 .0 0 5 ti m e b a se (s c m-1 ) 1 time to produce one wave 1 0.02 = 50 Hz Mains supply The mains supply is usually quoted as 230 V a.c. This is of course 230 V r.m.s. The peak voltage rises to approximately 325 V. Insulation must be provided to withstand this peak voltage. Example A transformer is labelled with a primary of 230 Vr.m.s. and secondary of 12 Vr.m.s. What is the peak voltage which would occur in the secondary? V peak = √ 2 × V r.m.s. V peak = 1.41 × 12 V peak = 17.0 V Frequency response of resistor The following circuit is used to investigate the relationship between current and frequency in a resistive circuit. A C u rren t (A ) S ig n a l gen erato r (co n s tan t e .m .f.) 0 freq u en cy (H z ) The results show that the current flowing through a resistor is independent of the frequency of the supply. Strathaven Academy -5- Electricity and Electronics Alternating Current and Voltage Alternating Current and Voltage 1. (a) (b) What is the peak voltage of the 230 V mains supply? The frequency of the mains supply is 50 Hz. How many times does the voltage fall to zero in one second? 2. The circuit below is used to compare the a.c. and d.c. supplies when the lamp is at the same brightness with each supply. The variable resistor is used to adjust the brightness of the lamp. A B (a) (b) (c) Explain how the brightness of the lamp is changed using the variable resistor. What additional apparatus would you use to ensure the brightness of the lamp was the same for each supply? In the oscilloscope traces shown below diagram 1 shows the voltage across the lamp when the switch is in position B and diagram 2 shows the voltage when the switch is in position A. 3 .6 c m 1 0 .2 c m D ia g r a m 1 D ia g r a m 2 Y Gain set to 1 V cm-1 From the oscilloscope traces, how is the root mean square voltage numerically related to the peak voltage. (d) Redraw diagrams 1 and 2 to show what would happen to the traces if the time base was switched on. 3. The root mean square voltage produced by a low voltage power supply is 10 V r.m.s. (a) Calculate the peak voltage of the supply. (b) If the supply was connected to an oscilloscope, Y-gain set to 5 V cm-1 with the time base switch off, describe what you would see on the screen. Strathaven Academy -6- Electricity and Electronics Alternating Current and Voltage 4. (a) (b) 5. A transformer has a peak output voltage of 12 V. What is the r.m.s. output voltage? A vertical line 6 cm long appears on an oscilloscope screen when the Y gain is set to 20 V cm-1. Calculate: (i) the peak voltage of the input (ii) the r.m.s. voltage of the input. The following trace appears on an oscilloscope screen when the time base is set at 2.5 ms cm-1. 6 ± 0 .1 c m (a) (b) 6. What is the frequency of the input including the uncertainty to the nearest Hz? Sketch what you would see on the screen if the time base was changed to (i) 5 ms cm-1 (ii) 1.25 ms cm-1 An a.c. input of frequency 20 Hz is connected to an oscilloscope with time base set at 0.01 s cm-1. What would be the wavelength of the waves appearing on the screen? Strathaven Academy -7- Electricity and Electronics