數學題 about AC circuit

2008-01-12 1:48 am
1. Capacitor of 8uF take a current of 1A when AC voltage applied across it is 250.

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



2. A coil a resistance of 6ohm & inductance of 0.03H connected across 50V, 60Hz supply

a. current
b. phase angle between the current and the applied voltage
c. apparent power
d. active power
更新1:

1a~~~~ Xc =250/1 =250ohm Xc =250 ohm, Xc 是咩黎???     because: Xc=1/2(pi)fc (pi) 是什麼??? fc 是否頻率? so: 250=1/2(pi) x f x 8e-6 why 8e-6 ??? 6係邊到黎?? 4e-3=50.3e-6 x f f=79.522 Hz

更新2:

1b~~~~~ Because: Z=V/I=500ohm R=(200^2-250^2)

更新3:

2b~~~ @=tan-1 XL/R @=tan-1(2pi x 60 x 0.03/6) @=60 度 4c~~ S= VI S 係代表咩????? =195.5VA 4d~~~ P=VI Cos@ =91.782 W

回答 (4)

2008-01-16 10:42 pm
✔ 最佳答案
因爲你既 V = 250, I = 1 要搵Ω,所以 Xc =250/1 =250Ω

Xc =250Ω, Xc 是咩黎???fc 是否頻率?
= 電容器接上交流電時既阻坑,f = 頻率Hz,c = 電容 1000000μF,
Capacitive reactance is inversely proportional to the signal frequency and the capacitance .

because: Xc=1/2(pi)fc, (pi) 是什麼??? 是 л ,半圓位既意思

方程式如下




so: 250=1/2(pi) x f x 8e-6 why 8e-6 ??? 6係邊到黎??
4e-3=50.3e-6 x f
f=79.522 Hz
因爲你要搵頻率,變成 f = 1/( 2 x л x Xc x F ) = 1/(2x3.14*****x250x8) = 79.***Hz

2008-01-16 20:06:51 補充:
Xc = 電容器接上交流電時既阻坑

2008-01-16 20:11:01 補充:
Xc = 係電容器接上交流電時既阻坑(Ω) ,f = 頻率(Hz),c = 電容 1000000μF(F)。

2008-01-16 20:15:14 補充:
小學係無呢D咁深既數學既,比心機讀多D書啦。

2008-01-16 20:20:13 補充:
S= VI S 係代表咩????? S = Apparent power = VA 咯P = Active power = W 咯

2008-01-16 20:31:53 補充:
because: Xc=1/2(pi)fc, (pi) 是什麼??? 是 л ,半圓位既意思方程式如下Xc = 1 / ( 2лFC)唔知點解,上面顯示唔出請參考 :http://en.wikipedia.org/wiki/ReactanceCapacitive reactance : Xc = 1 / ( 2лFC)

2008-01-17 22:12:58 補充:
呢題打錯字因爲你要搵頻率,變成 f = 1/( 2 x л x Xc x F ) = 1/(2x3.14*****x250x8) = 79.***Hz( 是但1個 F 字應該改爲 C )應該係 C = 1/( 2лXcF ) , C = 1/( 2лFXc ),Xc = 1/( 2лFC ),f = 1/( 2лXcC )因爲你要搵頻率,變成 f = 1/( 2 * л * Xc * C ) = 1/(2x3.14*****x250x8) = 79.***Hz
2008-01-27 1:57 am
好好!
2008-01-16 3:57 am
1a~~~~
Xc =250/1 =250ohm
because: Xc=1/2(pi)fc
so: 250=1/2(pi) x f x 8e-6
4e-3=50.3e-6 x f
f=79.522 Hz

1b~~~~~
Because: Z=V/I=500ohm
R=(200^2-250^2) <------再開方
R=433.012 ohm

1c~~~
@=tan-1(1/433.012 x 250)
@=30 度

2a~~~
I= 50/ {6^2+(2pi x 60 x 0.03)^2} <------{大括}開方
I=3.91 A

2b~~~
@=tan-1 XL/R
@=tan-1(2pi x 60 x 0.03/6)
@=60 度

4c~~
S= VI
=195.5VA

4d~~~
P=VI Cos@
=91.782 W
參考: .........
2008-01-12 6:12 am
1a~~~~
Xc =250/1 =250ohm
because: Xc=1/2(pi)fc
so: 250=1/2(pi) x f x 8e-6
4e-3=50.3e-6 x f
f=79.522 Hz

1b~~~~~
Because: Z=V/I=500ohm
R=(200^2-250^2) <------再開方
R=433.012 ohm

1c~~~
@=tan-1(1/433.012 x 250)
@=30 度

2a~~~
I= 50/ {6^2+(2pi x 60 x 0.03)^2} <------{大括}開方
I=3.91 A

2b~~~
@=tan-1 XL/R
@=tan-1(2pi x 60 x 0.03/6)
@=60 度

4c~~
S= VI
=195.5VA

4d~~~
P=VI Cos@
=91.782 W

呢份工課我都唔記左得交tin~


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