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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

But if it measures 420 , it s outside the rated range, and is therefore a reject The second set, along with the first set, of numbers represents standard resistance values available in tolerances of plus or minus 5 percent ( 5%) A 470- , 5 percent resistor will have an actual value of 470 plus or minus 24 , or a range of 446 to 494 Some resistors are available in tolerances tighter than 5% These precision units are employed in circuits where a little error can make a big difference In most audio and radio-frequency oscillators and amplifiers, the 10% or 5% tolerance is good enough In many cases, even a 20% tolerance is satisfactory Power Rating All resistors are given a specification that determines how much power they can safely dissipate Typical values are 1 4 W, 1 2 W, and 1 W.

Problem 17-2

5

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A circuit contains a pure capacitive reactance of 40 , but no resistance. What is the power factor Here, the phase angle is 90 degrees (Fig. 17-6). A calculator will tell you that cos 90 0. Therefore, PF 0. This means that PT/PVA 0 0 percent. That is, none of the power is true power, and all of it is imaginary or reactive power.

Units also exist with ratings of 1 8 W or 2 W These dissipation ratings are for continuous duty, meaning they can dissipate this amount of power constantly and indefinitely You can figure out how much current a given resistor can handle by using the formula for power (P) in terms of current (I ) and resistance (R) That formula, you should recall, is P = I 2R Work this formula backward, plugging in the power rating in watts for P and the resistance in ohms for R, and solve for the current I in amperes Alternatively, you can find the square root of P/R The power rating for a given resistor can, in effect, be increased by using a network of 2 2, 3 3, 4 4, or more units in series-parallel.

bandwidth (336 Kbps) in the upstream direction Although this may seem odd, it makes sense given the requirements of most applications today that require modem access A Web session, for example, requires very little bandwidth in the upstream direction to request that a page be downloaded The page itself, however, may require significantly more, since it may be replete with text, graphics, Java applets, and even small video clips Since the majority of modem access today is for Internet surfing, asymmetric access is adequate for most users The limitations of 56K modems stem from a number of factors One of them is the fact that under current FCC regulations (specifically Part 68), line voltage supplied to a communications facility is limited such that the maximum achievable bandwidth is 53 Kbps in the downstream direction.

Problem 17-3

If you need a 47- , 45-W resistor, but all you have is a bunch of 47- , 1-W resistors, you can make a 7 7 network in series-parallel, and this will handle 49 W Resistor power dissipation ratings are specified with a margin for error A good engineer never tries to take advantage of this and use, say, a 1 4-W unit in a situation that needs to draw 027 W In fact, good engineers usually include their own safety margin Allowing 10 percent, a 1 4-W resistor should not be called upon to handle more than about 0225 W..

A circuit contains a resistance of 50 and an inductive reactance of 50 in series. What is the power factor The phase angle in this case is 45 degrees (Fig. 17-7).The resistance and reactance components are equal, and form two sides of a right triangle, with the complex impedance vector forming the hypotenuse. Find cos 45 0.707. This means that PT/PVA 0.707 70.7 percent.

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