Question D.6: Find current i in the circuit in Fig. D.41.
Find current i in the circuit in Fig. D.41.
Recall that 20 sin 2t = 20 cos (2t − 90°) and that f = ω/2π = 2/2π = 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) |
| 3.18E-01 | 7.91E+00 | 4.35E+01 |
From the output file, we obtain I=7.906\underline{ / 43.49^{\circ}} 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.
