Question

有关电路分析实验的题目，求具体过程??步骤清晰，下图为题

Answer 1

既然 Ir = Ic，那么在数值上，容抗的值等于电阻值。有：
R = 1/(ωC)
z总 = R//zC + zL
= R * zC/(R + zC) + zL
= R * (-j)/(ωC)/[R - j/(ωC)] + jωL
= R * (-jR)/(R - jR) + jωL
= R * (-j)/(1-j) + jωL
= R * (-j) * (1+j)/[(1-j)(1+j)] + jωL
= R * [(-j) - j²]/(1-j²) + jωL
= R * (1 - j)/2 + jωL
= R/2 - j * (R/2) + jωL
= R/2 + j(ωL - R/2)
又因为 U 与 I 同相位，所以：
ωL - R/2 = 0
即 ωL = R/2
z总 = R/2
Ur = U * (R//zC)/z总

= U * [R *(-jR)/(R - jR)]/(R/2)
= U * (-j)R/(1-j)/(R/2)
= 2U * (-j)/(1-j)
= 2U * [(-j)(1+j)]/[(1-j)(1+j)]
= 2U * (-j - j²)/(1-j²)
= 2U * (1-j)/(1+1)
= U * (1-j)
= √2 * U * [√2/2 + j * (-√2/2)]
= √2 * U * [cos(-45°) + j * sin(-45°)]
= 200√2 ∠(-45°) (V)
Ir = Ur/R
= 200√2/R
所以，R = 200√2/Ir = 20√2 Ω

Answer 2

老师上课讲过 我记得是20欧😂