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If Rn = 10 , then En = IRn = 0.1 10 = 1.0 V. If Rn = 20 , then En = IRn = 0.1 20 = 2.0 V. Let s check to be sure all of these voltages add up to the supply voltage. There are five resistors with 1.0 V across each, for a total of 5.0 V; there are also five resistors with 2.0 V across each, for a total of 10 V. So the sum of the voltages across the resistors is 5.0 + 10 = 15 V.

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In a circuit containing both resistance and reactance, the relationships among true power PT, apparent or VA power PVA, and imaginary or reactive power PX are PVA2 PT2 PX2 PT < PVA PX < PVA

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Problem 5-2 In the circuit of Fig 5-2, what will happen to the voltages across the resistances if one of the 20- resistances is replaced with a short circuit In this case the total resistance becomes R = (10 5) + (20 4) = 50 + 80 = 130 The current is therefore I = E/R = 15/130 = 012 A This is the current at any point in the circuit, rounded off to two significant figures The voltage En across any of the 10- resistances Rn is equal to IRn, which is 012 10 = 12 V The voltage En across any of the 20- resistances Rn is equal to IRn, which is 012 20 = 24 V Checking the total voltage, add (5 12) + (4 24) = 60 + 96 = 156 V.

310 Power and resonance in ac circuits Engineers often strive to eliminate, or at least minimize, the reactance in a circuit. This is particularly true for radio antenna systems, or when signals must be sent over long spans of cable. It is also important in the design of radio-frequency amplifiers. To a lesser extent, minimizing the reactance is important in audio work and in utility power transmission.

processor, or populating a spreadsheet, or manipulating a digital photograph, the file lives in RAM until the person saves, at which time it is written to Main memory, which is nonvolatile, as we ll see in a moment. Modern systems typically have a minimum of 128 Megabytes (MB) of RAM.

This rounds off to 16 V when we cut it down to two significant figures A Rounding-Off Bug Compare the result for total voltage in Problem 5-2 with the result for total voltage in Problem 5-1 What is going on here Where does the extra volt come from in the second calculation Certainly, shorting out one of the resistances cannot cause the battery voltage to change! This is an example of what can happen when you round off to a certain number of significant figures after calculating the value of some parameter X in a circuit, then change a different parameter Y in the circuit, and finally calculate the value of X again, rounding off to the same number of significant digits as you did the first time The discrepancy is the result of a roundingoff bug.

The ratio of the true power to the VA power, PT/PVA, is called the power factor in an ac circuit. If there is no reactance, the ideal case, then PT PVA, and the power factor (PF) is equal to 1. If the circuit contains all reactance and no resistance of any significance (that is, zero or infinite resistance), then PT 0, and PF 0. If you try to get a pure reactance to dissipate power, it s a little like throwing a foam-rubber ball into a gale-force wind. The ball will come right back in your face. A pure reactance cannot, and will not, dissipate power. When a load, or a circuit in which you want power to be dissipated, contains some resistance and some reactance, then PF will be between 0 and 1. That is, 0 < PF < 1. PF might be expressed as a percentage between 0 and 100, written PF%. Mathematically,

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