Alternating Current

JEE Main level, single correct, medium

107 of 190

Question 107 of 190

A 0.07H0.07H inductor and a 12Ω12\Omega resistor are connected in series to a 220V,50Hz220V,50Hz AC source. The approximate current in the circuit and the phase angle between current and source voltage are respectively.
[Take π\pi as 227]\left.\dfrac{22}{7}\right]

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Answer: (A) 8.8A8.8A and tan⁡−1(116){\tan }^{-1}\left(\dfrac{11}{6}\right)

ϕ=tan⁡−1(XLR)\phi ={\tan }^{-1}\left(\dfrac{{X}_{L}}{R}\right)
XL=ωL{X}_{L}=\omega L
XL=2×227×50×0.07=22Ω{X}_{L}=2\times \dfrac{22}{7}\times 50\times 0.07=22\Omega
ϕ=tan⁡−1(2212)R=12Ω\phi ={\tan }^{-1}\left(\dfrac{22}{12}\right)R=12\Omega
ϕ=tan⁡−1(116)\phi ={\tan }^{-1}\left(\dfrac{11}{6}\right)
Z=XL2+R2=25.059Z=\sqrt{{X}_{L}^{2}+{R}^{2}}=25.059
I=VZ=22025.059=8.77AI=\dfrac{V}{Z}=\dfrac{220}{25.059}=8.77A

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