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Rgr electrical engineering solution. Solving problems in electrical engineering (TOE). Custom Electrical Solution |
To convert values to actual ones you need to: Dot above I means it is complex. In order not to be confused with current, in electrical engineering a complex unit is denoted by the letter “j”. For a given voltage we have: When solving problems, they usually operate with effective values. New elements are introduced in alternating current:
Their resistances (reactances) are found as: For example, we have a circuit, it is connected to a voltage of 200 V, having a frequency of 100 Hz. We need to find the current. Element parameters are set: To find the current, you need to divide the voltage by the resistance (from Ohm's law). The main task here is to find resistance. The complex resistance is found as: We divide the voltage by the resistance and get the current. All these actions are conveniently carried out in MathCad. The complex unit is put "1i" or "1j". If this is not possible, then:
Also, let's say a few words about power. Power is the product of current and voltage for DC circuits. For alternating current circuits, another parameter is introduced - the phase shift angle (or rather its cosine) between voltage and current. Suppose we have found the current and voltage (in complex form) for the previous circuit. Power can also be found using another formula: In this formula is the conjugate current complex. Conjugate means that its imaginary part (the one with j) changes its sign to the opposite (minus/plus). These were the formulas for active (useful) power. In AC circuits, there is also reactive power (generated by capacitors, consumed by coils). ![]() Im– the imaginary part of a complex number (the one with j). Knowing the reactive and active power, you can calculate the total power of the circuit: To simplify the calculation of direct and alternating current circuits containing a large number of branches, use one of the simplified methods of circuit analysis. Let's take a closer look at the loop current method. Loop current method (MCT)This method is suitable for solving circuits containing more nodes than independent circuits (for example, the circuit from the section on direct current). The principle of the solution is as follows: ![]() This method, like others (for example, the method of nodal potentials, equivalent generator, superposition) is suitable for both direct and alternating current circuits. When calculating alternating current circuits, the resistances of the elements are reduced to a complex form of notation. The system of equations is also solved in complex form. LiteratureCustom Electrical SolutionAnd remember that our solvers are always ready to help you with TOE. . Assignment for calculation and graphic work. For a three-phase circuit in Figure 1, containing non-sinusoidal periodic (T=1/f=1/50=0.02s), emf. e A (t), e B (t), e C (t) of equal amplitude E m, differing from each other only by a time shift by t f =2π/3ω=T/3, it is necessary to obtain:
1. Initial data. fundamental harmonic frequency f=50 Hz. Shape of e.m.f. – rectangular. Load connection diagram: ^
2. Fourier series expansion.
Figure 2. – Specified non-sinusoidal E.M.F. Figure 3. – harmonics that make up the voltage eA(t) Figure 4 shows the value Instantaneous values of linear voltages: Let's find the effective value of linear voltages: ^
3. Calculation of resistances:
Let us determine the complex amplitudes of the harmonics of the current phase “ab”: Let us determine the complex amplitudes of the harmonics of the current phase “bс”: Let us determine the complex amplitudes of the harmonics of the current phase “ca”: Instantaneous values of phase currents:
Effective values of phase currents: Let us determine the complex amplitudes of the harmonics of the current line “a”: Let us determine the complex amplitudes of the harmonics of the line “b” current: Let us determine the complex amplitudes of the harmonics of the line “c” current: Instantaneous values of line currents: Figure 6. – linear currents ^5. Power: Reactive power of phase “ab”: Power factor of phase “ab”: Active power of phase “bc”: Reactive power of phase “bc”: Power factor of phase “bc”: Active power of phase “ca”: Reactive power of phase “ca”: Phase power factor “ca”: Total active power of a three-phase system: Total reactive power of a three-phase system: Full power: Total apparent powers by phase: Apparent power: Apparent total power is greater than actual power. Generalized power factor Phase “b” resistance: Phase “c” resistance: Determination of complex voltage amplitudes between neutral points: Effective neutral offset value: ^
7. Decomposition into symmetrical components:
Figure 9. – Circuit with a break in phase “a”
Current decomposition: Fifth harmonic: Using Ohm's law, we find the total complex resistances of direct, negative and zero sequence: Figure 10. – First voltage harmonic Figure 11. – Fifth harmonic voltage Fig12. First harmonic of currents Fig13. Fifth harmonic currents Conclusion: In the course of this work, I came to the conclusion that when performing complex calculations such as those presented above, almost absolute accuracy and care are needed, since one small error or inaccuracy entails a series of incorrect results, which has a detrimental effect on ultimately work. Bibliography Bessonov L.A. . Textbook - M.: Gardariki 2000, 638 p. Theoretical foundations of electrical engineering. T.I. Fundamentals of the theory of linear circuits. Ed. P.A. Ionkina. - M.: Higher School, 1976, 544 p. |
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