JEE Mains · Physics · STD 12 -7. Alternating current
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 ..... .
- A \(3\)
- B \(5\)
- C \(7\)
- D \(9\)
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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