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qr code vb.net source Fig. 86 in .NET framework
Fig. 86 USS Code 128 Decoder In .NET Framework Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in VS .NET applications. Code 128A Creator In Visual Studio .NET Using Barcode generator for Visual Studio .NET Control to generate, create USS Code 128 image in VS .NET applications. CHAPTER 5 Sinusoidal Waves. rms Value
Code 128B Decoder In .NET Framework Using Barcode recognizer for .NET framework Control to read, scan read, scan image in .NET applications. Bar Code Creator In .NET Framework Using Barcode maker for .NET framework Control to generate, create barcode image in Visual Studio .NET applications. information alone does not permit us to write the equations of the two waves, even if their peak values are given. This is because the location of the origin of the axes, relative to the waves, is not shown in the gure. To illustrate this, suppose the peak value of A is 10 and the peak value of B is 7, and suppose it is given that curve A passes through the origin (in the manner of the sine wave in Fig. 85). With this information the equations of the two waves can now be written thus (in radians): for A: for B: y 10 sin !t y 7 sin !t =4 92 93 Recognize Barcode In VS .NET Using Barcode decoder for .NET Control to read, scan read, scan image in .NET applications. Printing Code 128A In Visual C# Using Barcode drawer for .NET Control to generate, create USS Code 128 image in Visual Studio .NET applications. Equation (93) is the mathematical way of showing that sinusoid B lags sinusoid A by =4 radians (458). Another point to be emphasized is that the curves A and B in Fig. 86 represent two sinusoidal functions having the SAME FREQUENCY. If two sinusoidal functions do not have the same frequency, then no xed phase relationship exists between the two functions, and the term phase shift would have little meaning. This is illustrated in Fig. 87, in which A and B denote curves of two sinusoids having unequal frequencies. Encode Code 128 Code Set A In Visual Studio .NET Using Barcode maker for ASP.NET Control to generate, create Code 128A image in ASP.NET applications. Code128 Creation In Visual Basic .NET Using Barcode generation for Visual Studio .NET Control to generate, create Code 128B image in .NET applications. Fig. 87
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Bar Code Reader In Java Using Barcode Control SDK for BIRT reports Control to generate, create, read, scan barcode image in Eclipse BIRT applications. Barcode Generator In None Using Barcode generation for Microsoft Excel Control to generate, create bar code image in Excel applications. In section 2.3 we showed that in a dc (directcurrent) circuit, the power P in watts, expended in a resistance of R ohms, is given by any of the formulas Printing EAN / UCC  13 In None Using Barcode printer for Software Control to generate, create European Article Number 13 image in Software applications. UCC  12 Generator In ObjectiveC Using Barcode drawer for iPad Control to generate, create GTIN  128 image in iPad applications. where V is dc voltage and I is dc current.
CHAPTER 5 Sinusoidal Waves. rms Value
Now let s consider the problem of how to calculate power in an ALTERNATING CURRENT (ac) circuit. We at once see that there is a problem here, because the power in an ac circuit is not constant but changes from instant to instant throughout the cycle, because the voltage and current are both continually changing during the cycle. This di culty is resolved by de ning that by power in an ac circuit we will always mean the AVERAGE POWER in the circuit. This de nition leads to what is called the e ective or rms value of an ac voltage or current. The development proceeds as follows. To begin, let us make the following slight change in notation. In eqs. (87) and (88) we used V and I to denote the peak values of sine waves of voltage and current. Let us now, for convenience later on, change that notation and, hereafter, always denote peak values by Vp and Ip , instead of by plain V and I. Upon making this change in notation, eqs. (87) and (88) become v Vp sin !t i Ip sin !t 94 95 In the above equations, v and i denote instantaneous voltage and current at any time t seconds after we start positive time at t 0. It then follows, from the basic considerations used to derive eq. (15) in Chap. 2, that instantaneous power, p, is equal to instantaneous voltage times instantaneous current, that is p vi or, in terms of a load resistance of R ohms, eq. (96) can be written in the forms p v2 =R and p i2 R: 96 But, as already mentioned, we are not interested in instantaneous power; instead, we are interested in nding the AVERAGE POWER obtained over ONE COMPLETE CYCLE of the sine waves of eqs. (94) and (95). (The average power over any one complete cycle is the same for all cycles, and is the average power as long as the waves continue to exist.)* To continue, let us now substitute, into eq. (96), the values of v and i from eqs. (94) and (95). Doing this, we have that the instantaneous power p in an ac circuit is equal to p Vp Ip sin !t 2 Vp Ip sin2 !t 97 in which we ll assume that the peak values, Vp and Ip , will remain constant in any given problem. From eq. (87), the angle !t is in radians. We recall that the function sin !t goes through one complete cycle in the period from !t 0 to !t 2 (Fig. 85). Our problem, therefore, now is to nd the AVERAGE VALUE of eq. (97) when the voltage and current waves, v and i (eqs. (94) and (95)), go through one complete cycle from !t 0 to !t 2. One way this can be done is to make use of the trigonometric identity 1 sin2 1 cos 2 2 98 { * Power does not accumulate in a circuit, because power is a measure of the RATE at which work is being done. This could be compared to an automobile moving at a constant speed; if the speed is measured for a period of, say, 1 minute, the same speed is measured for all intervals of 1 minute, and is the speed for the entire trip. { See note 6 in Appendix.

