Question is ⇒ The current i(t) through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4 sin (100t + 45°) + 4 sin (300t + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are, Options are ⇒ (A) 41 A, 410 W respectively, (B) 35 A, 350 W respectively, (C) 5 A, 250 W respectively, (D) 11 A, 1210 W respectively, (E) , Leave your comments or Download question paper.

Question is ⇒ The current i(t) through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4 sin (100t + 45°) + 4 sin (300t + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are, Options are ⇒ (A) 41 A, 410 W respectively, (B) 35 A, 350 W respectively, (C) 5 A, 250 W respectively, (D) 11 A, 1210 W respectively, (E) , Leave your comments or Download question paper. Previous question Next question


Q1.	The current i(t) through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4 sin (100t + 45°) + 4 sin (300t + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are
A.	41 A, 410 W respectively [Wrong Answer]
B.	35 A, 350 W respectively [Wrong Answer]
C.	5 A, 250 W respectively [Correct Answer]
D.	11 A, 1210 W respectively [Wrong Answer]
View Answer 2 -2 Explanation:- Answer : C Discuss it below : !! OOPS Login [Click here] is required to post your answer/result Help other students, write article, leave your comments

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