P=40W
S=220*0.41=90.2VA
所以cosφ=P/S=40/90.2=0.443 φ=63.7°、
功率因数是0.443
Z1=220∠0°/0.41∠-63.7°=536.6∠63.7°=237.8+j481.1
Z2=（237.8+j481.1）//(-jXc)
=(481.1Xc-j237.8Xc)/237.8+j(481.1-Xc)
=[Xc/237.8²+(481.1-Xc)²]*[237.8Xc+j(481.1Xc-288006.05)]
∴（237.8Xc）²/[（237.8Xc）²+(481.1Xc-288006.05)²]=0.9²
所以0.218Xc²-277Xc+82944=0
Xc=787.5 或 Xc=483.2
因为要求并联后,电流落后于电源电压,即电路要呈感性,
即Z2的虚部为正,故Xc=787.5
所以C=1/2πfXc=4.04μF