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| 1. |
In figure, the voltage V = 4∠0 V. The source voltage V is |
| Answer» IR = 2∠OA, I1 = 2∠-90° = -j 2A. Total current = 2 - j2. V2 = 2 (2 - j2) + 4∠0 = 8 - j4 volts. | |