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barcode printing using vb.net Resistance calculations in Software
Resistance calculations QR Code ISO/IEC18004 Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QR Code Generator In None Using Barcode creator for Software Control to generate, create QR Code JIS X 0510 image in Software applications. Ohms Law can be used to find a resistance between two points in a dc circuit, when the voltage and the current are known. For the following problems, imagine that both the voltmeter and ammeter scales in Fig. 47 are visible, but that the potentiometer is uncalibrated. Denso QR Bar Code Reader In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. QR Code JIS X 0510 Generator In Visual C#.NET Using Barcode generator for VS .NET Control to generate, create QR Code JIS X 0510 image in .NET framework applications. 72 Basic dc circuits
Denso QR Bar Code Generator In VS .NET Using Barcode encoder for ASP.NET Control to generate, create QR Code image in ASP.NET applications. QR Code 2d Barcode Maker In .NET Using Barcode generation for .NET framework Control to generate, create QRCode image in Visual Studio .NET applications. Problem 47 Make Denso QR Bar Code In Visual Basic .NET Using Barcode creator for VS .NET Control to generate, create QR Code JIS X 0510 image in VS .NET applications. Generating UPC Code In None Using Barcode creator for Software Control to generate, create GS1  12 image in Software applications. If the voltmeter reads 24 V and the ammeter shows 3.0 A, what is the value of the potentiometer Use the formula R E/I and plug in the values directly, because they are expressed in volts and amperes: R 24/3.0 8. 0 . Note that you can specify this value to two significant figures, the eight and the zero, rather than saying simply 8 . This is because you are given both the voltage and the current to two significant figures. If the ammeter reading had been given as 3 A (meaning some value between 21/2 A and 31/2 A), you would only be entitled to express the answer as 8 (somewhere between 71/2 and 81/2 ). A zero can be a significant figure, just as well as the digits 1 through 9. Paint Code128 In None Using Barcode generation for Software Control to generate, create Code 128 Code Set A image in Software applications. Paint ECC200 In None Using Barcode maker for Software Control to generate, create DataMatrix image in Software applications. Problem 48 EAN 13 Printer In None Using Barcode creator for Software Control to generate, create GTIN  13 image in Software applications. Generating Bar Code In None Using Barcode creator for Software Control to generate, create barcode image in Software applications. What is the value of the resistance if the current is 18 mA and the voltage is 229 mV First, convert these values to amperes and volts. This gives I 0.018 A and E 0.229 V. Then plug into the equation R E/I 0.229/0.018 13 . You re justified in giving your answer to two significant figures, because the current is only given to that many digits. Code11 Creator In None Using Barcode generation for Software Control to generate, create Code 11 image in Software applications. Print Code 128 Code Set B In None Using Barcode creator for Word Control to generate, create Code 128A image in Office Word applications. Problem 49 Read Universal Product Code Version A In VS .NET Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications. UCC128 Generator In C#.NET Using Barcode printer for .NET Control to generate, create UCC128 image in .NET applications. Suppose the ammeter reads 52 uA and the voltmeter indicates 2.33 kV. What is the resistance Convert to amperes and volts, getting I 0.000052 A and E 2330 V. Then plug into the formula: R 2330/0.000052 45,000,000 45 M . Creating EAN13 In ObjectiveC Using Barcode creation for iPad Control to generate, create EAN 13 image in iPad applications. Decoding Barcode In Visual Basic .NET Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET applications. Power calculations
Printing Code 39 In .NET Framework Using Barcode maker for Reporting Service Control to generate, create Code39 image in Reporting Service applications. Print Bar Code In C#.NET Using Barcode drawer for VS .NET Control to generate, create barcode image in .NET framework applications. You can calculate the power, in watts, in a dc circuit such as that shown in Fig. 47, by the formula P EI or the product of the voltage in volts and the current in amperes. You might not be given the voltage directly, but can calculate it if you know the current and the resistance. Remember the Ohm s Law formula for obtaining voltage: E IR. If you know I and R, but don t know E, you can get the power P by means of the formula P (IR)I I 2R. That is, you take the current in amperes, multiply this figure by itself, and then multiply the result by the resistance in ohms. You can also get the power if you aren t given the current directly. Suppose you re given only the voltage and the resistance. Remember the Ohm s Law formula for obtaining current: I E /R. Therefore, P E(E/R) E 2/R. Take the voltage, multiply it by itself, and divide by the resistance. Stated all together, these power formulas are: P EI I 2R E 2/R Now you are ready to do some problems in power calculations. Refer once again to Fig. 47. Resistances in series 73
Problem 410 Suppose that the voltmeter 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 zero counts as a significant digit.) Then P EI 12 0.050 0.60 W. You might say that this is 600 mW, although that is to three significant figures. It s not easy to specify the number 600 to two significant digits without using a means of writing numbers called scientific notation. That subject is beyond the scope of this discussion, so for now, you might want to say 600 milliwatts, accurate to two significant figures. (You can probably get away with 600 milliwatts and nobody will call you on the number of significant digits.) Problem 411 If the resistance in the circuit of Fig. 47 is 999 and the voltage source delivers 3 V, what is the dissipated power 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 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 a ridiculous state of affairs. An ordinary potentiometer, such as the one you would get at an electronics store, dissipating 22 kW, several times more than a typical household. The voltage must be phenomenal. It s not too hard to figure out that such a voltage would burn out the potentiometer so fast that it would be ruined before the little Pow! could even begin to register. Problem 413 Just from curiosity, what is the voltage that would cause so much current to be driven through such a large resistance Use Ohm s Law to find the current: E IR 0.680 47, 000 32,000 V 32 kV. That s the sort of voltage you d expect to find only in certain industrial/commercial applications. The resistance capable of drawing 680 mA from such a voltage would surely not be a potentiometer, but perhaps something like an amplifier tube in a radio broadcast transmitter.

