JEE Main 2021 · 17 March · morning · Single correct · +4 / −1

An AC current is given by I = I₁ sinωt + I₂ cosωt. A hot wire ammeter will give a reading :

  1. (a) (I₁ + I₂)/√2
  2. (b) √((I₁² - I₂²)/2)
  3. (c) √((I₁² + I₂²)/2)
  4. (d) (I₁ + I₂)/2√2
Answer and solution

Answer (c) √((I₁² + I₂²)/2)

Solution

Physics sir · 3 steps

A hot wire ammeter reads the rms value, so square the current, average it over a cycle, then take the root.

  1. What it measures. A hot wire ammeter works by heating, and heat depends on I². So it shows the rms current, Iᵣₘₛ = √(⟨ I² ⟩).

    I² = I₁²sin² ω t + I₂²cos² ω t + 2I₁I₂sin ω t cos ω t

  2. Average each term. Over one full cycle, ⟨ sin² ω t ⟩ = ⟨ cos² ω t ⟩ = 1/2. The cross term is I₁I₂sin 2ω t, which averages to zero. So ⟨ I² ⟩ = (I₁² + I₂²)/2.
  3. So. Iᵣₘₛ = √((I₁² + I₂²)/2), which is option (c).

The idea to keep. Sine and cosine of the same ω are 90° apart, so they combine like perpendicular vectors: the peak is √(I₁² + I₂²), and rms is that peak divided by √2.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Alternating Current

Asked about 140 times in JEE Main since 2020. This question is from 2021.