JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 12)
A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 $$\mu$$F and R = 100 $$\Omega$$. The phase difference ($$\Phi $$) between the applied voltage and resultant current will be :
tan$$-$$1(0.17)
tan$$-$$1(9.46)
tan$$-$$1(0.30)
tan$$-$$1(13.33)
Explanation
$${X_L} = 3000 \times 10 \times {10^{ - 3}} = 30\,\Omega $$
$${X_C} = {1 \over {3000 \times 25}} \times {10^6} = {{40} \over 3}\,\Omega $$
So $${X_L} - {X_C} = 30 - {{40} \over 3} = {{50} \over 3}\,\Omega $$
$$\tan \theta = {{{X_L} - {X_C}} \over R} = {{50/3} \over {100}} = {1 \over 6}$$
So $$\theta = {\tan ^{ - 1}}(0.17)$$
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