Special transformers and applications

Chapter 9.7 Special transformers and applications

Lessons In Electric Circuits Volume II – AC Book
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Lessons In Electric Circuits Volume II – AC Book

  • 9.7.SPECIAL TRANSFORMERS AND APPLICATIONS249“condition” AC power as an alternative to ferroresonant devices, but none can compete with thistransformer in terms of sheer simplicity.• REVIEW:• Voltage regulation is the measure of how well a power transformer can maintain constantsecondary voltage given a constant primary voltage and wide variance in load current. Thelower the percentage (closer to zero), the more stable the secondary voltage and the better theregulation it will provide.• A ferroresonant transformer is a special transformer designed to regulate voltage at a stablelevel despite wide variation in input voltage.9.7Special transformers and applications9.7.1Impedance matchingBecause transformers can step voltage and current to different levels, and because power is trans-ferred equivalently between primary and secondary windings, they can be used to “convert” theimpedance of a load to a different level. That last phrase deserves some explanation, so let’s inves-tigate what it means.The purpose of a load (usually) is to do something productive with the power it dissipates. Inthe case of a resistive heating element, the practical purpose for the power dissipated is to heatsomething up. Loads are engineered to safely dissipate a certain maximum amount of power, buttwo loads of equal power rating are not necessarily identical. Consider these two 1000 watt resistiveheating elements: (Figure 258,9.40)250 VI = 4 A62.5 ΩRloadPload = 1000 W125 VI = 8 A15.625 ΩRloadPload = 1000 WFigure 9.40: Heating elements dissipate 1000 watts, at different voltage and current ratings.Both heaters dissipate exactly 1000 watts of power, but they do so at different voltage and currentlevels (either 250 volts and 4 amps, or 125 volts and 8 amps). Using Ohm’s Law to determine thenecessary resistance of these heating elements (R=E/I), we arrive at figures of 62.5 Ω and 15.625Ω, respectively. If these are AC loads, we might refer to their opposition to current in terms ofimpedance rather than plain resistance, although in this case that’s all they’re composed of (noreactance). The 250 volt heater would be said to be a higher impedance load than the 125 voltheater.If we desired to operate the 250 volt heater element directly on a 125 volt power system, wewould end up being disappointed. With 62.5 Ω of impedance (resistance), the current would onlybe 2 amps (I=E/R; 125/62.5), and the power dissipation would only be 250 watts (P=IE; 125 x 2),or one-quarter of its rated power. The impedance of the heater and the voltage of our source wouldbe mismatched, and we couldn’t obtain the full rated power dissipation from the heater.