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 :
- (a) (I₁ + I₂)/√2
- (b) √((I₁² - I₂²)/2)
- (c) √((I₁² + I₂²)/2)
- (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.
- 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
- 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.
- 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.