✔ 最佳答案
Assume the coil is 50mH, not 50 ohm.
jwL = j (2pf)L = j [2 x 3.1416 x 50 x 50 x 10^(-3)] = 15.708j
1) Impedance, Z = 50 + 15.708j
2) In polar form is sqrt(50^2 + 15.708^2) arg ( arctan 15.708/50)
= 52.409 arg ( 17.44)
3) Voltage = 240 arg (0)
So current, I = V/Z = 240/52.409 arg(0 - 17.44) = 4.58 arg ( - 17.44)
4)
Voltage across the resistor = IR = (4.58)(50)arg(0 -17.44) = 229 arg ( - 17.44)
Impedance of coil in polar form = 15.708 arg ( 90)
So voltage across the coil = IZ = (4.58)(15.708) arg ( 90 - 17.44)
= 71.9 arg ( 72.56)
2012-12-24 08:01:13 補充:
Note : If impedance of the coil is really 50 ohm, then impedance of the circuit will simply be 50 + 50j = 70.71 arg ( 45). The frequency of 50Hz is then not necessary. Other calculation is the same.