Consider the AC circuit in the Figure B3 below, with the following voltage sources: v = 20 cos(2000r - 36.87) V 22 = 10 cos(5000 + 16.26) V 100 F 1 mH m 100 Figure B3 Because the frequency of the AC sources is different, we need to use superposition technique to solve the circuit. Using superposition and phasor, find a) the impedance values for components and re-draw the circuit (when only one of the AC sources is one, and the other one is off. You need to draw two circuits). [8 marks] b) the steady-state expression for vo with the effect of both AC sources. [16 marks]