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Dialog boxes are great for asking the user what to do, as you ve seen so far in this chapter But often, you ll need to get a piece of information from the user for example, their name or the quantity of widgets their company wants to order To add a text-entry field to a dialog, add the optional default answer parameter to the display dialog command If you want to provide a default answer in the text-entry field, add the appropriate string after the default answer parameter; if you want to leave the text-entry field blank, just place "" (two pairs of double quotation marks indicating an empty string) after the parameter instead For example, the following statement produces the dialog box shown in Figure 8-11:

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{NULL, MinInt ... 9}, {10 ... 19}, {20 ... 29}, {30 ... MaxInt}

display dialog "Please enter your employee code:" default answer "CZ 1234" with title "Employee Code"

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Let s add a new partition with a boundary value 40. We do that by ALTERing the partition function and using the SPLIT function:

ALTER PARTITION FUNCTION PFNExample1_Right()

When you add the default answer parameter, the dialog reply record includes a text returned property containing the text in the text-entry field as well as the button returned property containing the button clicked For example, the dialog box shown in Figure 8-11 returns a dialog reply record like this:

CHAPTER 31 SPLIT RANGE (40);

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Problem 16-20 Suppose a series RX circuit (shown by the generic block diagram of Fig 16-11) has a resistance of R = 500 and a capacitive reactance of X = 500 Suppose 100-V rms ac is applied to this circuit What is the current First, calculate Z 2 = R 2 + X 2 = 5002 + ( 500)2 = 2500 + 2500 = 5000 Then Z is the square root of 5000, or 707 Therefore, I = E/Z = 100/707 = 141 A rms Problem 16-21 What are the rms ac voltages across the resistance and the reactance, respectively, in the circuit described in Problem 16-20 The Ohm s Law formulas for dc will work here Because the current is I = 141 A rms, the voltage drop across the resistance is equal to ER = IR = 141 500 = 705 V rms The voltage drop across the reactance is the product of the current and the reactance: EX = IX = 141 ( 500) = 705 V rms This is an rms ac voltage of equal magnitude to that across the resistance But the phase is different Note that voltages across the resistance and the reactance a capacitive reactance in this case, because it s negative don t add up to 100 V rms, which is placed across the whole circuit This is because, in an RX ac circuit, there is always a difference in phase between the voltage across the resistance and the voltage across the reactance The voltages across the components always add up to the applied voltage vectorially, but not always arithmetically Problem 16-22 Suppose a series RX circuit (Fig 16-11) has R = 100 and X = 400 The applied voltage is 100V rms ac What is the current Calculate Z 2 = R 2 + X 2 = 100 + 1600 = 1700 This means that Z is the square root of 1700, or 412 Therefore, I = E/Z = 100/412 = 243 A rms Problem 16-23 What are the rms ac voltages across the resistance and the reactance, respectively, in the circuit described in Problem 16-22 Knowing the current, calculate ER = IR = 243 100 = 243 V rms Also, EX = IX = 243 400 = 972 V rms If you add ER + EX arithmetically, you get 243 + 972 = 1215 V as the total.

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{text returned:"CZ 9288", button returned:"OK"}

What the ALTER PARTITION FUNCTION command with the SPLIT option has done is add a new boundary value of 40, and therefore with four boundary values we now have five partitions:

{NULL, MinInt ... 9}, {10 ... 19}, {20 ... 29}, {30 ... 39}, {40 ... MaxInt}

containing resistance and reactance Illustration for Problems 16-20 through 16-23

Figure 8-11 You can add a text-entry field to any dialog box This dialog box includes a default value to show the user a sample response

You can verify this by running the following queries:

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