Solution: Disconnect ZL from the circuit and find a Thevenin equivalent circuit of one-port network at this time. Let the port nodes be a and b.

us(t)= 12√2 cos( 1000t)= 12√2 cos(- 1000t)= 12√2 sin[90-(- 1000t]。

At the same time: Is (phasor) = 4 ∠ 0 A, ω =1000 rad/s.xc =1(ω c) =1(1000× 250//kloc-).

There is only one loop in the circuit: US(t)-4ω- 12ω-IS(t), and the loop current is the current source current. Because the capacitor is also open, there is no current and no voltage, and the port voltage Uab (phasor) is the voltage across the 4 Ω resistor series voltage source.

Uoc (phasor) =Uab (phasor) =-Is (phasor) ×R+Us (phasor) =-4 ∠ 0× 4+12 ∠ 90 =-16+j1.

Then the voltage source is short-circuited, and the current source is open-circuited, thus obtaining: zeq =-jxc+r = 4-j4 = (ω).

According to the maximum power transmission theorem, when ZL=Zeq's * * * yoke is complex, ZL can obtain the maximum power, that is, when ZL=4+j4, the maximum power of ZL is: PLmax=Uoc? /(4×Re(ZL))=20？ /(4×4)=25 (width).