An AC current is given by I = I1 sin ω t + I2 cos ω t. A hot wire ammeter will give a reading: (a) (I₁ + I₂)/(√(2)) (b) √((I₁² - I₂²)/(2)) (c) √((I₁² + I₂²)/(2)) (d) (I₁ + I₂)/(2√(2))

Answer: (c). A hot wire ammeter reads the rms value: Irms² = I². I² = I₁²sin ²ω t + I₂²cos ²ω t + 2I₁I₂sin ω tcos ω t Over a cycle sin ² = cos ² = 1/2 and sin ω tcos ω t = 0. Irms = √((I₁² + I₂²)/(2)) So the answer is (c).

An AC current is given by I = I1 sin ω t + I2 cos ω t. A hot wire… | JEE Main 2021 (17 Mar, Shift 1)

Q35JEE Main 2021 · 17 MAR · SHIFT 1Medium

AC Basics: Peak, RMS & Average Values

An AC current is given by I = I1 sin $\omega$ t + I2 cos $\omega$ t. A hot wire ammeter will give a reading:

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Answer: (c)

A hot wire ammeter reads the rms value: $I_{rms}^2 = \langle I^2 \rangle$ .
$I^2 = I_1^2\sin^2\omega t + I_2^2\cos^2\omega t + 2I_1I_2\sin\omega t\cos\omega t$
Over a cycle $\langle\sin^2\rangle = \langle\cos^2\rangle = \frac{1}{2}$ and $\langle\sin\omega t\cos\omega t\rangle = 0$ .
$I_{rms} = \sqrt{\frac{I_1^2 + I_2^2}{2}}$
So the answer is (c).

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