JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 13)
A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 $$\Omega $$, L=25 mH and C=1000 $$\mu $$F. The total impedance, and phase difference between the voltage across the source and the current will respectively be :
10 $$\Omega $$ and tan$$-$$1 $$\left( {{5 \over 3}} \right)$$
$$7\,\Omega $$ and 45o
$$10\,\Omega $$ and tan$$-$$1$$\left( {{8 \over 3}} \right)$$
$$7\,\Omega $$ and tan$$-$$1$$\left( {{5 \over 3}} \right)$$
Explanation
It is given that e0 = 283 V; $$\omega$$ = 320.
The inductor reactance is XL = 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
$$\tan \phi = {{{X_L} - {X_C}} \over R} = {{8 - 3.1} \over 5} \approx 1 \Rightarrow \theta = 45^\circ $$
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