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| 1. |
The current (), though a 10 Ω resistor in series with an inductance, is given by () = 3 + 4 sin (10 + 45°) + 4 sin (300 + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are |
| Answer» Irms = = 6.4 Power P = I2R = 25 x 10 = 41 x 10 = 410 W. | |