QJEEM17Ph9478: JEE Main 2017 Physics - Class 12 JEEM17Ph9478 Solution: A sinusoidal voltage of peak value 283 V and angular frequency…

It is given that e₀ = 283 V; $\omega$ = 320.

The inductor reactance is X_L = 320 $\times$ 25 $\times$ 10 ^{$-$ 3} = 8 $\Omega$

The capacitor reactance is

${X_C} = {1 \over {\omega C}} = {1 \over {320 \times 1000 \times {{10}^{ - 6}}}} = {{1000} \over {320}} = 3.1\,\Omega$

It is given that R = 5 $\Omega$ . Therefore, the total impedance is

$Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = \sqrt {50} = 7\,\Omega$

and the phase difference between the voltage across the source and the current is

$\tan \phi = {{{X_L} - {X_C}} \over R} = {{8 - 3.1} \over 5} \approx 1 \Rightarrow \theta = 45^\circ$