Alternating Current

JEE Main level, numerical answer, medium

32 of 190

Question 32 of 190

The alternating current is given by, i={42sin⁡(2πTt)+10}Ai=\left\{\sqrt{42}\sin \left(\dfrac{2\pi }{T}t\right)+10\right\}A. The R.M.S. value of this current is __________ AA.

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Answer: 11

The given function of current is, i={42sin⁡(2πTt)+10}Ai=\left\{\sqrt{42}\sin \left(\dfrac{2\pi }{T}t\right)+10\right\}A
It means, i2=42sin⁡2(2πtT)+100+2042sin⁡(2πtT){i}^{2}=42{\sin }^{2}\left(\dfrac{2\pi t}{T}\right)+100+20\sqrt{42}\sin \left(\dfrac{2\pi t}{T}\right)
⇒⟨i2⟩=422+100\Rightarrow \left\langle {i}^{2}\right\rangle =\dfrac{42}{2}+100
RMS value of the current is, irms=⟨i2⟩=121=11A{i}_{rms}=\sqrt{\left\langle {i}^{2}\right\rangle }=\sqrt{121}=11A

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