qr code generator vb.net Inductance and Capacitance in .NET

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Inductance and Capacitance
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In Fig. 125 it s evident that the total charge q delivered by the battery is the sum of the charges delivered to the individual capacitors, which is (making use of eq. (184)) equal to q V C1 C2 Cn The same charge in Fig. 126 is q VCT Comparison of this equation with the preceding equation shows that CT C1 C2 Cn 191
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Equation (191) is thus the formula for calculating the equivalent capacitance of a PARALLEL connection of n individual capacitors. Problem 113 Given four capacitors, of capacitances 0.09 mF, 0.17 mF, 0.12 mF, and 0.55 mF, (a) Find the equivalent capacitance if the four are connected in series. (b) Find the equivalent capacitance if the four are connected in parallel. Problem 114 A dc voltage of 450 volts is applied to three series-connected capacitors having capacitances of 0.15 mF, 0.06 mF, and 0.48 mF. Is it theoretically su cient, in this case, to specify that all three capacitors have a voltage rating of at least 300 volts Problem 115 In problem 114, nd the total amount of energy stored in the three capacitors.
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Reactance and Impedance. Algebra of ac Networks
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In this chapter we begin the study of the algebra of networks containing INDUCTANCE AND CAPACITANCE for SINUSOIDAL applied voltages. (The reasons why sinusoidal waves are so important are noted in the introduction to section 5.4.)
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Inductive Reactance. Impedance
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We begin with the basic case of inductance in series with resistance, to which a sine wave of peak voltage Vp is applied, as in Fig. 127.
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Fig. 127
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In Fig. 127, L is inductance in henrys, and R is resistance in ohms. Also in the gure, v and i denote instantaneous values of voltage and current, with the voltage and current arrows having their usual meaning.*
* Before continuing, it will be wise to rst review all of section 5.6.
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CHAPTER 8 Reactance and Impedance
In Fig. 127 the independent variable is time, t, in which t 0 at the instant the sine wave of voltage is rst applied to the circuit. Analysis shows, and experiment veri es, that the resulting current consists of the SUM OF TWO TERMS, one term being the TRANSIENT component of the current and the other term being the STEADY-STATE component of the current. As a matter of fact, analysis shows and experiment veri es that the current in Fig. 127 has the mathematical form i I0  Rt=L Ip sin !t  | {z } | {z } transient steady-state component
component
192
In the equation, note, rst, that the transient component is a NEGATIVE EXPONENTIAL FUNCTION OF TIME which rapidly decreases in value as time increases (see Fig. 18-A in note 13 in the Appendix). Thus the transient component vanishes very quickly, leaving only the permanent, sinusoidal, steady-state component. As a matter of fact, in much practical work the e ect of the transient component can be completely ignored; that is, only the PERMANENT SINUSOIDAL STEADY-STATE RESPONSE is of interest in most practical work. Let us therefore examine, in more detail, the steady-state component in eq. (192).* First, taking the applied voltage, Vp sin !t, as the reference wave, notice that the steady-state current wave is also a sine wave, but one that LAGS the reference voltage wave by  radians (!t is in radians). (At this point it should be noted that the current will ALWAYS lag the voltage in an inductive circuit; this fact will be made evident in the discussion which follows.) To begin, let us suppose, in eq. (192), that the transient component of current has died out, so that only the nal STEADY-STATE SINUSOIDAL CURRENT remains. It is this sinusoidal steady-state condition, for Fig. 127, that we now wish to study in detail. To begin, note that there are two voltage drops in Fig. 127, one across R, the other across L. These two voltage drops are depicted graphically, with respect to the current wave, in Fig. 128, where Ip is the peak value of the sinusoidal current.
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