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How much of the power is true
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The above simple formulas allow you to figure out, given the resistance, reactance, and VA power, how many watts are true or real power, and how many watts are imaginary or reactive power. This is important in radio-frequency (RF) equipment, because RF wattmeters will usually display VA power, and this reading is exaggerated when there is reactance in a circuit.
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314 Power and resonance in ac circuits
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A circuit has 50 of resistance and 30 of inductive reactance in series. A wattmeter shows 100 watts, representing the VA power. What is the true power First, calculate the power factor. You might use either the phase-angle method or the R/Z method. Suppose you use the phase-angle method, then, Phase angle arctan (X/R) arctan (30/50) 31 degrees The power factor is the cosine of the phase angle. Thus, PF Remember that PF watts. cos 31 0.86 86 percent
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PT/PVA. This means that the true power, PT, is equal to 86
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A circuit has a resistance of 1,000 in parallel with a capacitance of 1000 pF. The frequency is 100 kHz. If a wattmeter reads a VA power of 88.0 watts, what is the true power This problem requires several steps in calculation. First, note that the components are in parallel. This means that you have to find the conductance and the capacitive susceptance, and then combine these to get the admittance. Convert the frequency to megahertz: f 100 kHz 0.100 MHz. Convert capacitance to microfarads: C 1000 pF 0.001000 F. From the previous chapter, use the equation for capacitive susceptance: BC 6.28fC 6.28 0.100 0.001000 0.000628 siemens
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The conductance of the resistor, G, is found by taking the reciprocal of the resistance, R: G 1/R 1/1000 0.001000 siemens
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Although you don t need to know the actual complex admittance vector to solve this problem, note in passing that it is G + jB 0.001000 + j0.000628
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Now, use the formula for calculating the resistance and reactance of this circuit, in terms of the conductance and susceptance. First, find the resistance: R G/(G2 + B2) 0.001000/(0.0010002 + 0.0006282) 0.001000/0.000001394 717 Then, find the reactance: X B/(G2 + B2) 0.000628/0.000001394 451
Power transmission 315 Therefore, R 717 and X 451. Using the phase-angle method to solve this (the numbers are more manageable that way than they are with the R/Z method), calculate Phase angle arctan (X/R) arctan ( 451/717) arctan ( 0.629) 32.2 degrees Then the power factor is PF cos 32.2 0.846 84.6 percent PT/PVA. Therefore, the true power
The VA power, PVA, is given as 88.0 watts, and PF is found this way:
PT/PVA 0.846 PT/88.0 0.846 0.846 88.0 74.4 watts
This is a good example of a practical problem. Although there are several steps, each requiring careful calculation, none of the steps individually is very hard. It s just a matter of using the right equations in the right order, and plugging the numbers in. You do have to be somewhat careful in manipulating plus/minus signs, and also in placing decimal points.
Power transmission
One of the most multifaceted, and important, problems facing engineers is power: transmission. Generators produce large voltages and currents at a power plant, say from turbines driven by falling water. The problem: getting the electricity from the plant to the homes, businesses, and other facilities that need it. This process involves the use of long wire transmission lines. Also needed are transformers to change the voltages to higher or lower values. A radio transmitter produces a high-frequency alternating current. The problem: getting the power to be radiated by the antenna, located some distance from the transmitter. This involves the use of a radio-frequency transmission line. The most common type is coaxial cable. Two-wire line is also sometimes used. At ultra-high and microwave frequencies, another kind of transmission line, known as a waveguide, is often employed. The overriding concern in any power-transmission system is minimizing the loss. Power wastage occurs almost entirely as heat in the line conductors and dielectric, and in objects near the line. Some loss can also take the form of unwanted electromagnetic radiation from a transmission line. In an ideal transmission line, all of the power is VA power; that is, it is in the form of an alternating current in the conductors and an alternating voltage between them.
316 Power and resonance in ac circuits It is undesirable to have power in a transmission line exist in the form of true power. This translates either into heat loss in the line, radiation loss, or both. The place for true power dissipation is in the load, such as electrical appliances or radio antennas. Any true power in a transmission line represents power that can t be used by the load, because it doesn t show up there. The rest of this chapter deals mainly with radio transmitting systems.
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