JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 20)
To light, a $$50 \mathrm{~W}, 100 \mathrm{~V}$$ lamp is connected, in series with a capacitor of capacitance $$\frac{50}{\pi \sqrt{x}} \mu F$$, with $$200 \mathrm{~V}, 50 \mathrm{~Hz} \,\mathrm{AC}$$ source. The value of $$x$$ will be ___________.
Answer
3
Explanation
$${X_C} = {1 \over {wc}} = {{\pi \sqrt x } \over {2\pi \times 50 \times 50}} \times {10^6}$$
$$v_R^2 + v_C^2 = {(200)^2}$$
$$v_C^2 = {200^2} - {100^2}$$
$${v_C} = 100\sqrt 3 \,V$$
$${v_R} = 100\,V$$
$$P = {{{V^2}} \over R}$$
$$R = {{100 \times 100} \over {50}} = 200\,\Omega $$
$${i_{rm}} = {1 \over 2}\,A$$
$${1 \over 2} \times {x_C} = 100\sqrt 3 \Rightarrow {10^{ - 6}} \times {{\sqrt x } \over {5000}} \times {1 \over 2} = 100\sqrt 3 $$
$${{{{10}^{ - 6}}\sqrt x } \over {10000 \times 100}} = \sqrt 3 $$
$$\sqrt x = \sqrt 3 $$
$$x = 3$$
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