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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2022
  • Hard
  • Numerical

To light, a \(50\,W , 100\,V\) lamp is connected, in series with a capacitor of capacitance \(\frac{50}{\pi \sqrt{ x }} \mu F\), with \(200 V , 50 Hz AC\) source. The value of \(x\) will be........

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Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

\(P =\frac{ V ^{2}}{ R } \Rightarrow R =\frac{ V ^{2}}{ P }\)

\(R =\frac{100 \times 10^{2}}{50}= R =200 \Omega\)

\(V _{ R }^{2}+ V _{ C }^{2}= V ^{2}\)

\((100)^{2}+ V _{ C }^{2}=(200)^{2}\)

\(i =\frac{100}{200}=\frac{1}{2} ;\)

\(V = I \times X _{ C }\)

\(V _{ C }=100 \sqrt{3}\)

\(X C _{ C }=200 \sqrt{3}\)

\(200 \sqrt{3}=\frac{1}{\omega C }\)

\(C =\frac{ V ^{2}=40000}{20 \times 50 \times 20 \sqrt{3}}=\frac{ V _{ C }^{2}=30000}{\sqrt{ x }}\)

\(\sqrt{ x }=50 \times 10^{-6} \times 100 \times 200 \sqrt{3}\)

\(X =3\)

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