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Starter. 1. T = 4.00 sec. 2. f = 1/T = .250 Hz. 3. 2.00 Volts. What is the period of the AC voltage? What is the frequency? What is the amplitude? What is the Peak-to-Peak voltage?. 4. 4.00 Volts. AC Waveform. V(t) = V max sin(2 p ft) . f = 1/T. AC Waveforms.
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Starter 1. T = 4.00 sec 2. f = 1/T = .250 Hz 3. 2.00 Volts What is the period of the AC voltage? What is the frequency? What is the amplitude? What is the Peak-to-Peak voltage? 4. 4.00 Volts
AC Waveform V(t) = Vmaxsin(2pft) f = 1/T
AC Waveforms The form for an AC voltage is given by: V(t) = Vmaxsin(2pft) where: f = frequency in Hz, and Vmax = the amplitude. Example: An AC voltage is given by : V(t) = 25V sin(500t) What’s the frequency? What’s the period? What is the Peak-to-Peak Voltage? 1. 500 = 2pf so f = 500/(2p) = 79.6Hz 2. T = 1/f = 1/79.6 = .0126 sec 3. Peak-to-Peak Voltage = 2 x Vmax = 50V
STARTER Find the frequency of this waveform. f = 1/T = 1/.4 = 2.5 Hz
Ohm’ Law For AC , Ohm’s Law becomes: Imax = Vmax/R Example: What’s the maximum current in this circuit? Imax = Vmax/R = 50/100 = .500Amps
For DC, the power in a resistor is I2R.But for AC, I2 looks like this: So we use the average value of I2 which turns out to be ½ of the maximum value of I2. Average of I2 = ½ of maximum value of I2
Average of I2 = ½ of maximum value of I2 Take the square root of each side: IRMS = Imax/ Also: VRMS = Vmax/ All the power formulas still work – just use RMS values. P = (IRMS)2R = (IRMS)( VRMS) = (VRMS)2/R
Example What’s the power sent to the resistor? VRMS = Vmax/ = 50/ P = (VRMS)2/R = (2500/2) / 100 = 12.5 Watts
Equation Summary AC circuits Imax = Vmax/R IRMS = Imax/VRMS = Vmax/ P = (IRMS)2R = (IRMS)( VRMS) = (VRMS)2/R V(t) = Vmaxsin(2pft) T = 1/f
Connection A diode bridge is used to convert the output of an AC generator to D.C. What application would this have in an automobile?
EXIT Draw an AC waveform and label the peak-to-peak voltage, the amplitude, and the period.