MCQ
An $AC$ current is given by $I = I_0 + I_1$ $sin\, wt$ then its $rms$ value will be
- ✓$\sqrt {{I_0}^2 + 0.5\,{I_1}^2} $
- B$\sqrt {{I_0}^2 + 0.5\,{I_0}^2} $
- C$0$
- D$I_0/ \sqrt 2 $
✓
Answer
Correct option: A.
$\sqrt {{I_0}^2 + 0.5\,{I_1}^2} $
a
$I_{r m s}=\sqrt{\frac{1}{T} \int_{0}^{T}\left(I_{0}+I_{1} \sin \omega t\right)^{2} d t}$
$I_{r m s}=\sqrt{\frac{1}{T} \int_{0}^{T}\left(I_{0}+I_{1} \sin \omega t\right)^{2} d t}$
$\sqrt{\frac{1}{T} \int_{0}^{T} I_{0}^{2} d t+\int_{0}^{T} I_{1}^{2} \sin ^{2} \omega t d t+\int_{0}^{T} \frac{2 I_{0} I_{1}}{T} \sin 2 \omega t d t}$
$=\sqrt{I_{0}^{2}+\left(I_{1}^{2} / 2\right)}=\sqrt{I_{0}^{2}+0.5 I_{1}^{2}}$
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