Phase lag in a simple circuit - Part 2

For the parallel connection between the $4.7 \Omega$ resistor and the $2.5$ mH inductor, the admittance is given by

$Y = \dfrac{1}{4.7} - j \dfrac{1}{ 120 \pi \times 2.5 \times 10^{-3} }$

Since the current is $I = Y V$ , then the phase lag is $\phi = \tan^{-1} ( \dfrac{4.7}{ 120 \pi \times 2.5 \times 10^{-3} } ) = 78.67^{\circ}$