![]() To calculate the power, we must convert the current into amperes 855 mA = 855/1000 = 0.855 A. Suppose the voltage is 117 V, and the current is 855 mA. If the voltage across the resistance is caused by two flashlight cells in series, giving 3 V, and if the current through the resistance (a light bulb, perhaps) is 0.1 A, then E = 3 and I = 0.1, and we can calculate the power P, in watts, as: If it were a motor, some of the power would exist in the form of mechanical work. This would be the state of affairs if the resistor were an incandescent light bulb, for example. Or it might exist in several forms, such as heat, light and infrared. ![]() Then the power in watts dissipated by the resistance, call it P, is the product E X I. Suppose we call the voltage E and the current I, in volts and amperes, respectively. There's also electricity flowing through the resistance, not quantified in the diagram, either. ![]() There is a certain voltage across the resistor, not specifically given in the diagram. Some power always goes to waste, and this waste is almost all in the form of heat. This is because no equipment is 100-percent efficient. But heat is always present, in addition to any other form of power in an electrical or electronic device. In fact, there are dozens of different ways that power can be dissipated. Power can be manifested in many other ways, such as in the form of mechanical motion, or radio waves, or visible light, or noise. ![]() The heat can be measured in watts, abbreviated W, and represents electrical power. Whenever current flows through a resistance, heat results. A watt, in electrical terms, is the rate at which electrical work is done when one ampere (A) of current flows through one volt (V). A watt is a unit of power, named after engineer James Watt, which measures the rate of energy transfer. ![]()
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