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Alternating Current question

2021 · 17 Mar · Shift 1 · Q55
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Alternating Current question

2021 · 17 Mar · Shift 1 · Q55

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
An AC current is given by I = I1 sin ω\omegaω t + I2 cos ω\omegaω t. A hot wire ammeter will give a reading :
  1. A
    I1+I22{{{I_1} + {I_2}} \over {\sqrt 2 }}2​I1​+I2​​
  2. B
    I12−I222\sqrt {{{I_1^2 - I_2^2} \over 2}}2I12​−I22​​​
  3. C
    I12+I222\sqrt {{{I_1^2 + I_2^2} \over 2}}2I12​+I22​​​
  4. D
    I1+I222{{{I_1} + {I_2}} \over {2\sqrt 2 }}22​I1​+I2​​
View written solutionFree

Correct answer: C

  1. A hot wire ammeter measures the r.m.s. current because its reading is based on the heating effect of current.

  2. The given current is

I(t)=I1sin⁡ωt+I2cos⁡ωtI(t)=I_1\sin \omega t + I_2\cos \omega tI(t)=I1​sinωt+I2​cosωt

We need

Irms=⟨I2(t)⟩I_{\text{rms}}=\sqrt{\langle I^2(t)\rangle}Irms​=⟨I2(t)⟩​

where ⟨⋅⟩\langle \cdot \rangle⟨⋅⟩ denotes time average over one complete cycle.

  1. First square the current:
I2=(I1sin⁡ωt+I2cos⁡ωt)2I^2=(I_1\sin \omega t + I_2\cos \omega t)^2I2=(I1​sinωt+I2​cosωt)2 I2=I12sin⁡2ωt+I22cos⁡2ωt+2I1I2sin⁡ωtcos⁡ωtI^2=I_1^2\sin^2\omega t + I_2^2\cos^2\omega t + 2I_1I_2\sin\omega t\cos\omega tI2=I12​sin2ωt+I22​cos2ωt+2I1​I2​sinωtcosωt
  1. Now take average over one full cycle:
  • ⟨sin⁡2ωt⟩=12\langle \sin^2 \omega t\rangle=\frac12⟨sin2ωt⟩=21​
  • ⟨cos⁡2ωt⟩=12\langle \cos^2 \omega t\rangle=\frac12⟨cos2ωt⟩=21​
  • ⟨sin⁡ωtcos⁡ωt⟩=0\langle \sin \omega t\cos \omega t\rangle=0⟨sinωtcosωt⟩=0

Therefore,

⟨I2⟩=I12⋅12+I22⋅12+2I1I2⋅0\langle I^2\rangle=I_1^2\cdot \frac12 + I_2^2\cdot \frac12 + 2I_1I_2\cdot 0⟨I2⟩=I12​⋅21​+I22​⋅21​+2I1​I2​⋅0 ⟨I2⟩=I12+I222\langle I^2\rangle=\frac{I_1^2+I_2^2}{2}⟨I2⟩=2I12​+I22​​
  1. Hence,
Irms=I12+I222I_{\text{rms}}=\sqrt{\frac{I_1^2+I_2^2}{2}}Irms​=2I12​+I22​​​
  1. Comparing with the options:
  • A: Incorrect
  • B: Incorrect
  • C: Correct
  • D: Incorrect

So the hot wire ammeter reading is

I12+I222\boxed{\sqrt{\frac{I_1^2+I_2^2}{2}}}2I12​+I22​​​​
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