JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 15)
In the given circuit the AC source has $$\omega$$ = 100 rad s-1. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?
_26th_August_Evening_Shift_en_15_1.png)
5.9 A
3.16 A
0.94 A
6 A
Explanation
$${Z_C} = \sqrt {{{\left( {{1 \over {\omega C}}} \right)}^2} + {R^2}} $$
$$ = \sqrt {{{\left( {{1 \over {100 \times 100 \times {{10}^{ - 6}}}}} \right)}^2} + {{100}^2}} $$
$${Z_C} = \sqrt {{{(100)}^2} + {{(100)}^2}} $$
$$ = 100\sqrt 2 $$
$${Z_L} = \sqrt {{{(\omega L)}^2} + {R^2}} $$
$$\sqrt {{{(100 \times 0.5)}^2} + {{50}^2}} $$
$$ = 50\sqrt 2 $$
$${i_C} = {{200} \over {{Z_C}}} = {{200} \over {100\sqrt 2 }} = \sqrt 2 $$
$${i_L} = {{200} \over {{Z_L}}} = {{200} \over {50\sqrt 2 }} = 2\sqrt 2 $$
$$\cos {\phi _1} = {{100} \over {10\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _1} = 45^\circ $$
$$\cos {\phi _2} = {{50} \over {50\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _2} = 45^\circ $$
_26th_August_Evening_Shift_en_15_2.png)
$$I = \sqrt {I_C^2 + I_L^2} $$
$$ = \sqrt {2 + 8} $$
$$ = \sqrt {10} $$
I = 3.16 A
$$ = \sqrt {{{\left( {{1 \over {100 \times 100 \times {{10}^{ - 6}}}}} \right)}^2} + {{100}^2}} $$
$${Z_C} = \sqrt {{{(100)}^2} + {{(100)}^2}} $$
$$ = 100\sqrt 2 $$
$${Z_L} = \sqrt {{{(\omega L)}^2} + {R^2}} $$
$$\sqrt {{{(100 \times 0.5)}^2} + {{50}^2}} $$
$$ = 50\sqrt 2 $$
$${i_C} = {{200} \over {{Z_C}}} = {{200} \over {100\sqrt 2 }} = \sqrt 2 $$
$${i_L} = {{200} \over {{Z_L}}} = {{200} \over {50\sqrt 2 }} = 2\sqrt 2 $$
$$\cos {\phi _1} = {{100} \over {10\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _1} = 45^\circ $$
$$\cos {\phi _2} = {{50} \over {50\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _2} = 45^\circ $$
$$I = \sqrt {I_C^2 + I_L^2} $$
$$ = \sqrt {2 + 8} $$
$$ = \sqrt {10} $$
I = 3.16 A
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