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Current i(t), through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4sin (100t + 45°) + 4sin (300 t + 60°) A RMS value of the current and power dissipated in the circuit respectively are
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- √41 A , 410 W
- √35 A , 350 W
- 5 A , 250 W
- 11 A , 1210 W
- √41 A , 410 W
Correct Option: C
i = 3+ 4 sin (1000 + 45°) + 4 sin (300 + 60°) amp 
Power = i2 R(RMS)
= 25 × 10 = 250 W
