In a series LCR circuit R-4 - XL-5 - and X C -8- the current

The correct option is A leads the voltage by tan ^ 1 (3/4) #160;φ =arctan X _ L X _ C R #160; With the given values, we get φ=arctan( 34) ...

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Byju's Answer

Standard XII

Physics

Phasor Diagrams in AC Circuits

In a series L...

Question

In a series LCR circuit R=4 $Ω$ , XL=5 $Ω$ and $X_{C} = 8 Ω$ , the current

leads the voltage by

{t a n}^{- 1} (3 / 4)

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leads the voltage by

{t a n}^{- 1} (5 / 8)

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leads the voltage by

{t a n}^{- 1} (3 / 5)

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leads the voltage by

{t a n}^{- 1} (5 / 9)

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Solution

The correct option is A leads the voltage by ${t a n}^{- 1} (3 / 4)$
$φ = arctan \frac{X_{L} - X_{C}}{R}$
With the given values, we get $φ = arctan (- \frac{3}{4})$

Thus, the current leads the voltage by ${tan}^{- 1} (\frac{3}{4})$ .

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In a series LCR circuit R=4Ω, XL=5Ω and XC=8Ω, the current

The correct option is A leads the voltage by tan−1(3/4)φ=arctanXL−XCRWith the given values, we get φ=arctan(−34)Thus, the current leads the voltage by tan−1(34).

In a series LCR circuit R=4 $Ω$ , XL=5 $Ω$ and $X_{C} = 8 Ω$ , the current

The correct option is A leads the voltage by ${t a n}^{- 1} (3 / 4)$
$φ = arctan \frac{X_{L} - X_{C}}{R}$
With the given values, we get $φ = arctan (- \frac{3}{4})$

Thus, the current leads the voltage by ${tan}^{- 1} (\frac{3}{4})$ .