JEE Mains · Physics · STD 12 -7. Alternating current
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........
- A \(2\)
- B \(3\)
- C \(1\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(P =\frac{ V ^{2}}{ R } \Rightarrow R =\frac{ V ^{2}}{ P }\) \(R =\frac{100 \times 10^{2}}{50}= R =200 \Omega\) \(V _{ R }^{2}+ V _{ C }^{2}= V ^{2}\) \((100)^{2}+ V _{ C }^{2}=(200)^{2}\) \(i =\frac{100}{200}=\frac{1}{2} ;\) \(V = I \times X _{ C }\) \(V _{ C }=100 \sqrt{3}\)…
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