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AC power的frequency及等等

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A capacitor of 8μF takes a current of 1A when the AC voltage applied across it is 250. Calculate:a. The frequency of the applied voltageb. The resistance to be connected in series with the capacitor to reduce the current in the circuit to 0.5A at the same frequencyc. The phase angle of the resulting... 顯示更多 A capacitor of 8μF takes a current of 1A when the AC voltage applied across it is 250. Calculate: a. The frequency of the applied voltage b. The resistance to be connected in series with the capacitor to reduce the current in the circuit to 0.5A at the same frequency c. The phase angle of the resulting circuit. 請問呢幾題要用咩公式計? a果條我知1/T=frequency，但係又唔見有T，所以唔知點計。 其他都唔多識… 更新: (b) From the given, we have to make the impedance value of the RC combination being double of that of the reactance of the capacitor, i.e. 500 Ω. So, Z = √(R2 + Xc2) 500 = √(R2 + 2502) R = 250√3 = 433 Ω 我想問250√3的√3係點計出來的?

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(a) First of all, the reactance across the capacitor is 250/1 = 250 Ω By formula Xc = 1/ωC where ω is the angular frequency, we have: 250 = 1/(2πfC) 2πf × 8 × 10-6 = 1/250 f = 79.6 Hz (b) From the given, we have to make the impedance value of the RC combination being double of that of the reactance of the capacitor, i.e. 500 Ω. So, Z = √(R2 + Xc2) 500 = √(R2 + 2502) R = 250√3 = 433 Ω (c) In the circuit, current is the common quantity and having the same phase as VR (voltage across the resistor) and leading the VC (voltage across the capacitor) by 90°. Therefore, the resulting angle of voltage is: tan-1 (1/√3) = 30° since voltage is proportional to resistance/reactance value. Finally, we have the current leading the voltage by 30°. 2008-01-21 20:04:22 補充： 500 = √(R^2 250^2)(2 x 250)^2 = R^2 250^24 x 250^2 = R^2 250^2R^2 = 3 x 250^2R = 250√3

其他解答:CF546184287637C5
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