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barcode generator in vb.net Power Calculations in Software
Power Calculations Code39 Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Encoding USS Code 39 In None Using Barcode creator for Software Control to generate, create Code 3/9 image in Software applications. You can calculate the power P, in watts, in a dc circuit such as that shown in Fig. 47, by using the formula P = EI. This formula tells us that the power in watts is the product of the voltage in volts and the current in amperes. If you are not given the voltage directly, you can calculate it if you know the current and the resistance. Recall the Ohm s Law formula for obtaining voltage: E = IR. If you know I and R but you don t know E, you can get the power P this way: P = EI = (IR)I = I 2R Suppose you re given only the voltage and the resistance. Remember the Ohm s Law formula for obtaining current: I = E/R. Therefore: P = EI = E(E/R) = E 2/R Decoding Code 39 Full ASCII In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Code 39 Printer In C#.NET Using Barcode drawer for .NET framework Control to generate, create Code39 image in Visual Studio .NET applications. Problem 410 Suppose that the voltmeter in Fig. 47 reads 12 V and the ammeter shows 50 mA. What is the power dissipated by the potentiometer Use the formula P = EI. First, convert the current to amperes, getting I = 0.050 A. (Note that the last 0 counts as a significant digit.) Then multiply by 12 V, getting P = EI = 12 0.050 = 0.60 W. Code39 Creator In .NET Framework Using Barcode printer for ASP.NET Control to generate, create Code 39 image in ASP.NET applications. Printing Code39 In .NET Using Barcode encoder for .NET framework Control to generate, create Code 3 of 9 image in .NET framework applications. 62 DirectCurrent Circuit Basics
Create Code 3/9 In Visual Basic .NET Using Barcode creator for Visual Studio .NET Control to generate, create Code 39 Full ASCII image in .NET framework applications. Creating Code39 In None Using Barcode maker for Software Control to generate, create Code39 image in Software applications. Problem 411 If the resistance in the circuit of Fig. 47 is 999 and the voltage source delivers 3 V, what is the power dissipated by the potentiometer Use the formula P = E 2/R = 3 3/999 = 9/999 = 0.009 W = 9 mW. You are justified in going to only one significant figure here. Problem 412 Suppose the resistance in Fig. 47 is 47 k and the current is 680 mA. What is the power dissipated by the potentiometer Use the formula P = I 2R, after converting to ohms and amperes. Then P = 0.680 0.680 47,000 = 22,000 W = 22 kW. (This is an unrealistic state of affairs: an ordinary potentiometer, such as the type you would use as the volume control in a radio, dissipating 22 kW, several times more than a typical household!) Problem 413 How much voltage would be necessary to drive 680 mA through a resistance of 47 k , as is described in the previous problem Use Ohm s Law to find the voltage: E = IR = 0.680 47,000 = 32,000 V = 32 kV. That s the level of voltage you d expect to find on a major utility power line, or in a highpower tubetype radio broadcast transmitter. Creating UPC Symbol In None Using Barcode encoder for Software Control to generate, create UCC  12 image in Software applications. Create Barcode In None Using Barcode drawer for Software Control to generate, create bar code image in Software applications. Resistances in Series
Encode Bar Code In None Using Barcode generation for Software Control to generate, create bar code image in Software applications. EAN13 Generation In None Using Barcode generator for Software Control to generate, create UPC  13 image in Software applications. When you place resistances in series, their ohmic values add together to get the total resistance. This is easy to imagine, and it s easy to remember! ISBN Creation In None Using Barcode encoder for Software Control to generate, create ISBN  13 image in Software applications. Printing Bar Code In Java Using Barcode printer for Android Control to generate, create barcode image in Android applications. Problem 414 Suppose resistors with the following values are connected in series, as shown in Fig. 48: 112 , 470 , and 680 . What is the total resistance of the series combination Simply add up the values, getting a total of 112 + 470 + 680 = 1262 . You might round this off to 1260 . It depends on the tolerances of the resistors how precise their actual values are to the ones specified by the manufacturer. UPC  13 Creator In Java Using Barcode creator for Java Control to generate, create EAN13 Supplement 5 image in Java applications. Printing Barcode In None Using Barcode maker for Online Control to generate, create bar code image in Online applications. 48 Three resistors in series.
UCC128 Reader In Visual C# Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET framework applications. Code39 Generator In Visual C# Using Barcode drawer for .NET framework Control to generate, create Code39 image in VS .NET applications. Illustration for Problem 414. Resistance values are in ohms.
Encoding EAN 13 In VS .NET Using Barcode maker for .NET framework Control to generate, create EAN / UCC  13 image in VS .NET applications. 2D Barcode Creator In Visual Basic .NET Using Barcode creator for .NET framework Control to generate, create Matrix Barcode image in Visual Studio .NET applications. Resistances in Parallel 63
Resistances in Parallel
When resistances are placed in parallel, they behave differently than they do in series. One way to look at resistances in parallel is to consider them as conductances instead. In parallel, conductances add up directly, just as resistances add up in series. If you change all the ohmic values to siemens, you can add these figures up and convert the final answer back to ohms. The symbol for conductance is G. This figure, in siemens, is related to the resistance R, in ohms, by these formulas, which you learned in Chap. 2: G = 1/R R = 1/G Problem 415 Consider five resistors in parallel. Call them R1 through R5, and call the total resistance R as shown in Fig. 49. Let the resistance values be as follows: R1 = 100 , R2 = 200 , R3 = 300 , R 4 = 400 , and R5 = 500 . What is the total resistance, R, of this parallel combination 49 Five resistors of values R1 through R5, connected in parallel, produce a net resistance R. Illustration for Problems 415 and 416. Converting the resistances to conductance values, you get: G1 = 1/100 = 0.01 S, G2 = 1/200 = 0.005 S, G3 = 1/300 = 0.00333 S, G4 = 1/400 = 0.0025 S, and G5 = 1/500 = 0.002 S. Adding these gives G = 0.01 + 0.005 + 0.00333 + 0.0025 + 0.002 = 0.0228 S. The total resistance is therefore R = 1/G = 1/0.0228 = 43.8 . Problem 416 Suppose you have five resistors, called R1 through R5, connected in parallel as shown in Fig. 49. Suppose all the resistances, R1 through R5, are 4.70 k . What is the total resistance, R, of this combination When you have two or more resistors connected in parallel and their resistances are all the same, the total resistance is equal to the resistance of any one component divided by the number of components. In this example, convert the resistance of any single resistor to 4700 , and then divide this by 5. Thus, you can see that the total resistance is 4700/5 = 940 . In a situation like this, where you have a bunch of resistors connected together to operate as a single unit, the total resistance is sometimes called the net resistance. Take note, too, that R is not italicized when it means resistor, but R is italicized when it means resistance!

