The alternating current is given by i = √(42) sin (2π/T t) + 10 A The r.m.s. value of of this current is................. A. Numerical answer type. Work it out, then open the answer.

Answer: 11. i² = 42sin ²ω t + 20√(42)sin ω t + 100 Over a cycle sin ²ω t = 1/2 and sin ω t = 0. irms² = 21 + 0 + 100 = 121 irms = 11 A

The alternating current is given by i = √(42) sin (2π/T t) + 10 A The… | JEE Main 2021 (27 Aug, Shift 1)

Q37JEE Main 2021 · 27 AUG · SHIFT 1Difficult

AC Basics: Peak, RMS & Average Values

The alternating current is given by $i = \left\{\sqrt{42} \sin\left(\frac{2\pi}{T} t\right) + 10\right\} A$

The r.m.s. value of of this current is ................. A.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 11

$i^2 = 42\sin^2\omega t + 20\sqrt{42}\sin\omega t + 100$
Over a cycle $\langle\sin^2\omega t\rangle = \frac{1}{2}$ and $\langle\sin\omega t\rangle = 0$ .
$i_{rms}^2 = 21 + 0 + 100 = 121$
$i_{rms} = 11$ A