JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2021
- Hard
- Numerical
In an \(LCR\) series circuit, an inductor \(30 \,{mH}\) and a resistor \(1 \, \Omega\) are connected to an \(AC\) source of angular frequency \(300 \, {rad} / {s}\). The value of capacitance for which, the current leads the voltage by \(45^{\circ}\) is \(\frac{1}{x} \times 10^{-3} \, {F}\). Then the value of \(x\) is ..... .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(\tan \phi=\frac{x_{C}-x_{L}}{R}\)
\(\tan 45=\frac{x_{C}-x_{L}}{R}\)
\(X_{C}-X_{L}=R\)
\(\frac{1}{\omega C}-\omega L=R\)
\(\frac{1}{\omega C}-300 \times 0.03=1\)
\(\frac{1}{\omega C}=10\)
\(C=\frac{1}{10 \omega}=\frac{1}{10 \times 300}\)
\(C=\frac{1}{3} \times 10^{-3}\)
\(X=3\)
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