1.

The current () through a 10 Ω resistor in series with an inductance is given by () = 3 + 4 sin (100 + 45°) + 4 sin (300 + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are

Answer» Power = I2R = 25 X 10 = 250 Watts.


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