Question 92 of 190

To light, a $50W,100V$ lamp is connected, in series with a capacitor of capacitance $\dfrac{50}{\pi \sqrt{x}}\mu F$, with $200V,50HzAC$ source. The value of $x$ will be _____ .

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Answer: 3

Electrical power is given by $P=\dfrac{{V}^{2}}{R}\Rightarrow R=\dfrac{{V}^{2}}{P}$

Resistance of lamp is $R=\dfrac{100\times 100}{50}\Rightarrow R=200\Omega$Let ${V}_{C}$ and ${V}_{R}$ be the voltage across capacitor and resistor.
Here, ${V}_{R}^{2}+{V}_{C}^{2}={V}^{2}$
Current through lamp is $i=\dfrac{100}{200}=\dfrac{1}{2}A$
$\Rightarrow {\left(100\right)}^{2}+{V}_{C}^{2}={\left(200\right)}^{2}$
$\Rightarrow {V}_{C}^{2}=30000\Rightarrow {V}_{C}=100\sqrt{3}V$
We know that, $V=i\times {X}_{C}$
Then, ${X}_{C}=200\sqrt{3}\Omega =\dfrac{1}{\omega C}$
So, capacitance of capacitor is $C=\dfrac{1}{20\times 50\times 20\sqrt{3}}=\dfrac{50\times {10}^{-6}}{\sqrt{x}}$
$\sqrt{x}=50\times {10}^{-6}\times 100\times 200\sqrt{3}$
Thus, value of $x=3$.