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A wall wart power supply can be used to power the receiver/display unit or a 9- to 12-V rechargeable battery can be used for portable operation if desired A small hole is drilled on the top of the case to allow the piezo buzzer to soundoff through the case The buzzer is epoxied to the underside of the enclosure A coaxial power jack is mounted on the rear of the receiver chassis, along with an antenna jack to accept an external antenna A quarter-wave vertical whip or half-wave dipole can be used for the antenna on the receiver display unit Be sure to use the same type and polarity of antennas between the transmitter and receiver units to ensure maximum range For example, if the transmitter utilizes a vertical antenna, the receiver section should also use a vertical antenna.

EXHIBIT 4-2

Suppose we have a series circuit with a supply of 150 V and three resistors: R1 330 , R2 680 , and R3 910 . What is the power dissipated by R2 You must find the current in the circuit. To do this, calculate the total resistance first. Because the resistors are in series, the total is R 330 680 910 1920 . Then the current is I 150/1920 0. 07813 A 78.1 mA. The power in R2 is P2 I 2R2 0.07813 0.07813 680 4.151 W. Round this off to two significant digits, because that s all we have in the data, to obtain 4.2 W. The total power dissipated in a series circuit is equal to the sum of the wattages dissipated in each resistor. In this way, the distribution of power in a series circuit is like the distribution of the voltage.

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If you elect to use a half-wave dipole for the transmitter, then a half-wave dipole should be used at the receiver unit as well..

Problem 5-8

Calculate the total power in the circuit of Problem 5-7 by two different methods. The first method is to figure out the power dissipated by each of the three resistors separately, and then add the figures up. The power P2 is already known. Let s bring it back to the four significant digits while we calculate: P2 4.151 W. Recall that the current in the circuit is I 0.07813 A. Then P1 0.07813 0. 07813 330 2.014 W, and P3 0.07813 0.07813 910 5.555 W. Adding these gives P 2.014 4.151 5.555 11.720 W. Round this off to 12 W. The second method is to find the series resistance of all three resistors. This is R 1920 , as found in Problem 5-7. Then P I 2R 0.07813 0.07813 1920 11.72 W, again rounded to 12 W. You might recognize this as an electrical analog of the distributive law you learned in junior-high-school algebra.

measures as The Five Chinese Brothers (taken from the Chinese folk tale of the same name). The cash flow measures and the initial estimates for Microsoft are: the revenue growth rate 11 percent, the net operating profit margin 42 percent, the net fixed capital investment rate (equal to the investment rate minus the depreciation rate see below), the incremental working capital investment rate 9.8 percent, and the company s income tax rate 32 percent. Briefly, these may be described as:

13 15

When resistances are wired in parallel, they each consume power according to the same formula, P I 2R. But the current is not the same in each resistance. An easier method to find the power Pn, dissipated by resistor Rn, is by using the formula Pn E 2/Rn where E is the voltage of the supply. Recall that this voltage is the same across every resistor in a parallel circuit.

Pin # 1 2 3 4 5 6 7

Problem 5-9

The Five Chinese Brothers: Revenue Growth Rate Net Operating Profit Margin Tax Rate Working Capital Investment Fixed Capital Investment Annual Growth in Revenue Operating Income/Revenue Taxes/Pretax Income Change in Working Capital/Revenue Net Capital Investment/Revenue

A circuit contains three resistances R1 22 , R2 47 , and R3 age E 3.0 V. Find the power dissipated by each resistor. 68 across a volt-

Pin # 8, 9 10 11 12 13 14 15

Kirchhoff s first law 89 First find E 2, because you ll be needing that number often: E 2 3.0 3.0 9.0. Then P1 9.0/22 0.4091 W, P2 9.0/47 0.1915 W, P3 9.0/68 0.1324 W. These can be rounded off to P1 0.41 W, P2 0.19 W, and P3 0.13 W. But remember the values to four places for the next problem. In a parallel circuit, the total power consumed is equal to the sum of the wattages dissipated by the individual resistances. In this respect, the parallel circuit acts like the series circuit. Power cannot come from nowhere, nor can it vanish. It must all be accounted for.

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