Question D.6: Find current i in the circuit in Fig. D.41....

Recall that 20 \sin 2t = 20 \cos (2t  –  90°)   and that ƒ = ω / 2π = 2 / 2 π = 0.31831 .  The schematic is shown in Fig. D.42. The attributes of V1 are
set  as ACMAG = 20  ,  ACPHASE = – 90  ; while the attributes of IAC are set as   AC = 5 . The current-controlled current source is connected in such a way as to conform with the original circuit in Fig. D.41; its gain is set equal to 2. The attributes of the pseudocomponent IPRINT are set as AC = yes  , MAG = yes  , PHASE = ok  , REAL = , and IMAG = .  Since this is a single-frequency ac analysis, we select Analysis/ Setup/AC Sweep and enter Total Pts = 1  , Start Freq = 0.31831  , and  Final Freq = 0.31831 . We save the circuit and select Analysis/Simulate for simulation. The output file includes

FREQ                 IM(V_PRINT3)                IP(V_PRINT3)
3.183E-01         7.906E+00                        4.349E+01

From the output file, we obtain I = 7.906 \underline{/ 43.49°} A    or  i(t) = 7.906 \cos (2t +  43.49°) A    This example is for a single-frequency ac analysis; Example D.7 is for AC sweep over a range of frequencies.