JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2021
- Medium
- MCQ
A \(100 \,\Omega\) resistance, a \(0.1\, \mu {F}\) capacitor and an inductor are connected in series across a \(250 \,{V}\) supply at variable frequeny. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is \(60\, {Hz}\). (In \({H}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(70.3\)
Step-by-step Solution
Detailed explanation
\(C=0.1\, \mu {F}=10^{-7}\, {F}\)
Resonant frequency \(=60\, {Hz}\)
at resonance
\(\omega_{0}=\frac{1}{\sqrt{L C}}\)
\(2 \pi f_{0}=\frac{1}{\sqrt{L C}} \Rightarrow \frac{1}{4 \pi^{2} f_{0}^{2} C}\)
By putting values \({L}=70.3\, {Hz}\)
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