电路分析基础题目求大神指导 追加100分 谢谢了

P叫有功功率，平均功率啥玩意儿看不懂
对于原电路进行等效变换后得：
j4电感和4Ω电阻串联后与-j4电容并联，
并联后再与2Ω，j2电容串联。
故总阻抗表示为：
Z = 2-j2+[-j4//(j4+4)]
= 24/17 - j54/17 （数字有点妖）
= √[(24/17)^2 + (54/17)^2] ∠[arctg (4/9)]
≈ 3.476 ∠ [arctg (4/9)] 身边没有函数计算器，自己算下arctg (4/9)是多少吧
I = U / Z = 6∠0 / 3.476 ∠ [arctg (4/9)] = 6/3.476∠-[arctg (4/9)] ≈ 1.726∠-[arctg (4/9)]
功率因素角φ = arctg (4/9)
故：
有功功率P = U I cos [arctg (4/9)] = 6 * 1.726 cos[arctg (4/9)]
无功功率Q = U I sin [arctg (4/9)] = 6 * 1.726 sin[arctg (4/9)]
实在功率S = U I = 6 * 1.726
功率因素λ = cos [arctg (4/9)]
自己按按计算器吧