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This voltage is commonly employed with heavy appliances, such as the kitchen oven/stove (if they are electric), heating (if it is electric), and the laundry washer and dryer A 117-V rms outlet supplies just one phase, appearing between two of the three slots in the outlet The third opening in the outlet leads to an earth ground (Fig 18-9B) In Electronic Devices The smallest power transformers are found in electronic equipment such as television sets, ham radios, and home computers Most solid-state devices use low voltages, ranging from about 5 V up to perhaps 50 V This equipment needs step-down power transformers in its power supplies Solid-state equipment usually (but not always) consumes relatively little power, so the transformers are usually not very bulky The exception is high-powered AF or RF amplifiers, whose tran-.

Unlike cellular service, however, users are stationary Consequently, LMDS cells do not need to support roaming Antenna/transceiver units are generally placed on rooftops, as they need unobstructed line-of-sight to operate properly In fact, this is one of the disadvantages of LMDS (and a number of other wireless technologies): besides suffering from unexpected physical obstructions, the service suffers from rain fade caused by the absorption and scattering of the transmitted microwave signal by atmospheric moisture Even some forms of foliage will cause interference for LMDS, so the transmission and reception equipment must be mounted high enough to avoid such obstacles, hence the tendency to mount the equipment on rooftops Because of its high-bandwidth capability, many LMDS implementations interface directly with an ATM backbone to take advantage of both its bandwidth and its diverse QoS capability.

Problem 11-1

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Capacitances in parallel add like resistances in series. The total capacitance is the sum of the individual component values. If two or more capacitors are connected in parallel, and one of the capacitances is far larger than any of the others, the total capacitance can be taken as approximately the value of the biggest one.

Two capacitors, with values of C1 0.10 F and C2 (Fig. 11-3). What is the total capacitance 0.050 F, are connected in series

If ATM is indeed the transport fabric of choice, then the LMDS service becomes a broadband access alternative to a network capable of transporting a full range of services including voice, video, image, and data the full suite of multimedia applications..

Using the above formula, first find the reciprocals of the values. They are 1/C1 and 1/C2 20. Then 1/C 10 20 30, and C 1/30 0.033 F.

Problem 11-4 Suppose three capacitors are in parallel, having values of C1 = 0.100 F, C2 = 0.0100 F, and C3 = 0.001000 F, as shown in Fig. 11-4. What is the total capacitance Add them up: C = 0.100 + 0.0100 + 0.001000 = 0.111000. Because two of the values are given to only three significant figures, the final answer should be stated as C = 0.111 F.

Multichannel, Multipoint Distribution System (MMDS)

Problem 11-2

Illustration for Problem 11-4.

Two capacitors with values of 0.0010 F and 100 pF are connected in series. What is the total capacitance Convert to the same size units. A value of 100 pF represents 0. 000100 F. Then you can say that C1 0.0010 F and C2 0.00010 F. The reciprocals are 1/C1 1000 and

MMDS got its start as a wireless cable television solution. In 1963, a spectrum allocation known as the Instructional Television Fixed Service (ITFS) was carried out by the FCC as a way to distribute educational content to schools and universities. In the 1970s, the FCC established a two-channel metropolitan distribution service called the Multipoint

Problem 11-5 Suppose two capacitors are in parallel, one with a value of 100 F and one with a value of 100 pF. What is the net capacitance In this case, you can say right away that the net capacitance is 100 F for practical purposes. The 100-pF capacitor has a value that is only one-millionth of the capacitance of the 100- F component. The smaller capacitance contributes essentially nothing to the net capacitance of this combination.

Capacitors in parallel 203 1/C2 10,000. Therefore, 1/C 1,000 10,000 11,000, and C 0. 000091 F. This number is a little awkward, and you might rather say it s 91 pF. In the above problem, you could have chosen pF to work with, rather than F. In either case, there is some tricky decimal placement involved. It s important to doublecheck calculations when numbers get like this. Calculators will take care of the decimal placement problem, sometimes using exponent notation and sometimes not, but a calculator can only work with what you put into it! If you put a wrong number in, you will get a wrong answer, perhaps off by a factor of 10, 100, or even 1,000.

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