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1/( j10) = (1/ j )(1 10) = (1/ j)0.1 = j 0.1 When you want to find an imaginary value of susceptance in terms of an imaginary value of reactance, first take the reciprocal of the real-number part of the expression, and then multiply the result by 1.

328 Transformers and impedance matching inductances of the primary and secondary depend on the frequency of operation, and also on the resistive part of the impedance in the circuit. As the frequency increases, the needed inductance decreases. At high-resistive impedances, more inductance is generally needed than at low-resistive impedances

Problem 15-5 Suppose you have a capacitor of 100 pF at a frequency of 3.00 MHz. What is BC First, find XC by the formula for capacitive reactance:

6

The turns ratio in a transformer is the ratio of the number of turns in the primary, Tpri, to the number of turns in the secondary, Tsec. This ratio is written Tpri:Tsec or Tpri/Tsec. In a transformer with excellent primary-to-secondary coupling, the following relationship always holds: Epri/Esec = Tpri/Tsec That is, the primary-to-secondary voltage ratio is always equal to the primary-to-secondary turns ratio (Fig. 18-2).

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XC = 1/(6.28f C ) Note that 100 pF = 0.000100 F. Therefore: XC = 1/(6.28 3.00 0.000100) = 1/0.001884 = 531 The imaginary value of XC is equal to j531. The susceptance, BC, is equal to 1/X C. Thus, BC = 1/( j531) = j0.00188, rounded to three significant figures. The general formula for capacitive susceptance in siemens, in terms of frequency in hertz and capacitance in farads, is: BC = 6.28f C This formula also works for frequencies in megahertz and capacitances in microfarads.

enhanced services billing, SS7 signal interconnection, and so on. Until these issues are clearly resolved, ATM-based, feature-rich telephony will not become a mainstream service, but will instead be limited to simple voice, and there is a difference. FRBS refers to the capability of ATM to interwork with Frame Relay. Conceptually, the service implies that an interface standard enables an ATM switch to exchange data with a Frame Relay switch, thus allowing for interoperability between frame and cell-based services. Many manufacturers are taking a slightly different tack, however; they build switches with soft, chewy cell technology at the core and surround the core with hard, crunchy interface cards to suit the needs of the customer. For example, an ATM switch might have ATM cards on one side to interface with other ATM devices on the network, but it might have Frame Relay cards on the other side to enable it to communicate with other Frame Relay switches, as shown in Figure 6-11. Thus, a single piece of hardware can logically serve as both a cell and Frame Relay switch. This design is becoming more and more common, because it helps to avoid a future rich with forklift upgrades. LANE enables an ATM network to move traffic transparently between two similar LANs, but it also enables ATM to transparently slip into the LAN arena. For example, two Ethernet LANs could communicate across the fabric of an ATM network, as could two Token Ring LANs. In effect, LANE enables ATM to provide a bridging function between similar LAN environments. In LANE implementations the ATM net-

Problem 18-1

Problem 15-6 Suppose an inductor has L = 163 H at a frequency of 887 kHz. What is BL Note that 887 kHz = 0.887 MHz. You can calculate XL from the formula for inductive reactance:

A transformer has a primary-to-secondary turns ratio of exactly 9:1. The voltage at the primary is 117 V. What is the voltage at the secondary This is a step-down transformer. Simply plug in the numbers in the above equation and solve for Esec:

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XL = 6.28fL = 6.28 0.887 163 = 908 The imaginary value of XL is equal to j908. The susceptance, BL = is equal to 1/XL. It follows that BL = 1/j 908 = j0.00110. The general formula for inductive susceptance in siemens, in terms of frequency in hertz and inductance in henrys, is: BL = 1/(6.28f L) This formula also works for frequencies in kilohertz and inductances in millihenrys, and for frequencies in megahertz and inductances in microhenrys.

Problem 18-2

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