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 (integer)
3
Solution
<p>$${X_C} = {1 \over {wc}} = {{\pi \sqrt x } \over {2\pi \times 50 \times 50}} \times {10^6}$$</p>
<p>$v_R^2 + v_C^2 = {(200)^2}$</p>
<p>$v_C^2 = {200^2} - {100^2}$</p>
<p>${v_C} = 100\sqrt 3 \,V$</p>
<p>${v_R} = 100\,V$</p>
<p>$P = {{{V^2}} \over R}$</p>
<p>$R = {{100 \times 100} \over {50}} = 200\,\Omega$</p>
<p>${i_{rm}} = {1 \over 2}\,A$</p>
<p>$${1 \over 2} \times {x_C} = 100\sqrt 3 \Rightarrow {10^{ - 6}} \times {{\sqrt x } \over {5000}} \times {1 \over 2} = 100\sqrt 3 $$</p>
<p>${{{{10}^{ - 6}}\sqrt x } \over {10000 \times 100}} = \sqrt 3$</p>
<p>$\sqrt x = \sqrt 3$</p>
<p>$x = 3$</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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