求救-問AC 的公式問題

[匿名]

2007-12-19 6:56 pm

問題係咁︰
A capacitor of 8uF takes a current of 1A when the AC voltage applied across it is 250.

a. The Frequency of the applied voltage
b. The resistance to be connected in series with the capacitor to reduce the currnet in the circuit to 0.5A at the same frequency
c. The phase angle of the resulting circuit

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想問以上3條數既公式應該係點?
a果題我諗極都唔明，無T，點樣可以求到個f出黎……
今晚考試，到家陣到未明，有無高人救下小弟……

更新1:

唔該哂一樓既朋友 A同C都明但係B就唔明 Z=500ohm R=3 Z:唔係(RxR)+(XcxXc)= 250ohm(即係開頭個Xc)，double 左佢即係等於將current改成0.5? R，問題無俾到R係幾多，請問呢度要用咩公式計的? 真係唔該哂你，今晚起碼唔使食零旦，應該都有幾分拎...

回答 (1)

魏王將張遼

2007-12-19 7:36 pm

✔ 最佳答案

(a) Applying the formula Vc = IXc where Xc is the reactance of the capacitor.
Then, Xc = 250 Ω
Also by formula Xc = 1/(2πfC) where f is the applied voltage frequency:
250 = 1/(2πf × 8 × 10-6)
2πf = 500
f = 79.6 Hz

(b) From the given condition, we should make the impedance value of the circuit to be double of that of Xc, then
Z = 500 Ω
√(R2 + Xc2) = 500
R = √3 × Xc
R = 433 Ω

(c) The phase angle is given by
tan-1 (Xc/R) = tan-1 (1/√3)
= 30°
with the voltage lagging the current.

2007-12-19 15:36:21 補充：
Z 係 impedance, 即係 resistance 和 reactance 合成的一個數值, 其 formula 如下:Z = √[R^2 (Xc - XL)^2]XL = Inductive reactance, Xc = Capacitative reactance, R = resistance

參考: My physics knowledge

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