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In L-R series circuit, source voltage is 10V and voltage across R is 8V. Phase difference $(ϕ)$ between applied voltage and current is :

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$ϕ = \tan^{- 1} (\frac{4}{1})$

$ϕ = \tan^{- 1} (\frac{1}{4})$

$ϕ = \tan^{- 1} (\frac{3}{4})$

$ϕ = \tan^{- 1} (\frac{4}{3})$

answer is A.

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Detailed Solution

$\tan ϕ = \frac{V_{L}}{V_{R}}$ or $\tan ϕ = \frac{V_{L}}{V_{R}}$
$\Rightarrow ϕ = \tan^{- 1} (\frac{3}{4})$

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