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