In a series LCR circuit, R=3Ω, XL=5Ω,XC=1Ω the applied voltage is given by V=4sin⁡(4t+π3)volt. The equation of current is

Tan⁡θ=XL−XCR=43⇒θ=53∘Z=R2+x2Z=5Ωi=i0sin⁡(ωt+θ−ϕ)i=V0Zsin⁡(ωt+60∘−53∘)

In a series LCR circuit, R=3Ω, XL=5Ω,XC=1Ω the applied voltage is given by V=4sin⁡(4t+π3)volt. The equation of current is

Tan⁡θ=XL−XCR=43⇒θ=53∘Z=R2+x2Z=5Ωi=i0sin⁡(ωt+θ−ϕ)i=V0Zsin⁡(ωt+60∘−53∘)

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In a series LCR circuit, $R = 3 Ω,$ $X_{L} = 5 Ω, X_{C} = 1 Ω$ the applied voltage is given by $V = 4 \sin (4 t + \frac{π}{3})$ volt. The equation of current is

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$i = 0.8 \sin (4 t + 7^{0})$

$i = 0.8 \sin (4 t + 97^{\circ})$

$i = 0.8 \sin (4 t + 113^{\circ})$

$i = 0.8 \sin (4 t - 23^{\circ})$

answer is D.

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$Tan θ = \frac{X_{L} - X_{C}}{R} = \frac{4}{3} \Rightarrow θ = 53^{\circ}$

$Z = \sqrt{R^{2} + x^{2}}$

$Z = 5 Ω$

$i = i_{0} \sin (ωt + θ - ϕ)$

$i = \frac{V_{0}}{Z} \sin (ωt + 60^{\circ} - 53^{\circ})$

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In a series LCR circuit, R=3Ω, XL=5Ω,XC=1Ω the applied voltage is given by V=4sin⁡(4t+π3)volt. The equation of current is

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In a series LCR circuit, $R = 3 Ω,$ $X_{L} = 5 Ω, X_{C} = 1 Ω$ the applied voltage is given by $V = 4 \sin (4 t + \frac{π}{3})$ volt. The equation of current is