Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 11 (Q.No. 12)
12.
The current i(t), though a 10 Ω resistor in series with an inductance, is given by i(t) = 3 + 4 sin (10t + 45°) + 4 sin (300t + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are
Answer: Option
Explanation:
Irms = 
= 6.4
Power P = I2R = 25 x 10 = 41 x 10 = 410 W.
Discussion:
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