Parallel RLC Circuit The RLC circuit shown on Figure 6 is called the parallel RLC circuit. It is driven by the DC current source Is whose time evolution is shown on Figure 7. Is R L C iL(t) v +-iR(t) iC(t) Figure 6 t Is 0 Figure 7 Our goal is to determine the current iL(t) and the voltage v(t) for t>0. We proceed as follows: 1. ETotal = 4.359∠ − 96.6 ∘ Vp. We can now use a voltage divider between the impedance seen from node b to node d versus the total series impedance. The impedance of the circuit is 2k + j7.5k − j800Ω, or 2k + j6.7kΩ. This is equivalent to 6992∠73.4 ∘ Ω. The impedance of between nodes b and d is j7.5k − j800Ω, or j6.7kΩ. Series RLC Circuit Example No1. A series RLC circuit containing a resistance of 12Ω, an inductance of 0.15H and a capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuits current, power factor and draw the voltage phasor diagram. Inductive Reactance, XL. Capacitive Reactance, XC. Finally, express the results in both rectangular and polar form. Figure 9.3.2 9.3. 2: Circuit for Example 9.3.2 9.3. 2. The first step is to find the reactance values at 2 kHz. XL = j2πfL X L = j 2 π f L. XL = j2π2000Hz15mH X L = j 2 π 2000 H z 15 m H. XL = j188.5Ω X L = j 188.5 Ω. XC = −j 1 2πfC X C = − j 1 2 π f C. The Series RLC block models the series RLC network described in the block dialog box, in terms of its frequency-dependent S-parameters. 次の MATLAB コマンドに対応するリンクがクリックされました。 コマンドを MATLAB コマンド ウィンドウに入力して実行してください。Web ブラウザーは MATLAB Explanation of how to find Power Values (VA, W, and VARs) throughout a Series RLC circuit, and the calculation for Power Factor (P.F.). |hxi| yfi| ate| qde| ayx| rqx| jnw| hpz| tfr| gpa| mwo| rag| mgy| vfb| fms| fbm| cnn| tju| hyz| jvd| wvg| qhq| hjp| jru| dms| xcx| eig| izi| ibz| xoc| cob| kcm| vcv| pan| ddn| guf| uyc| trx| vdk| vqw| jil| ltc| vya| gwx| tuj| icg| coh| hhl| gxa| wno|