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MHT CET · Physics · Alternating Current

With an alternating voltage source frequency ' f ', inductor ' \(L\) ', capacitor ' \(C\) ' and resistance ' \(R\) ' are connected in series. The voltage leads the current by \(45^{\circ}\). The value of ' \(L\) ' is \(\left(\tan 45^{\circ}=1\right)\)

  1. A \(\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)\)
  2. B \(\left(\frac{1-2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)\)
  3. C \(\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1+2 \pi \mathrm{fCR}}\right)\)
  4. D \(\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1-2 \pi \mathrm{fCR}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)\)

Step-by-step Solution

Detailed explanation

The phase difference between the current and the voltage is given by \(\tan \phi=\frac{\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}}{\mathrm{R}}\)
\(\begin{array}{ll}
\therefore & \omega L-\frac{1}{\omega C}=R \quad \ldots\left(\because \tan \phi=\tan 45^{\circ}=1\right) \\
\therefore & \omega L=R+\frac{1}{\omega C} \\
\therefore & L=\frac{R}{\omega}+\frac{1}{\omega^2 C}=\frac{R \omega C+1}{\omega^2 C} \\
\therefore & L=\frac{1+2 \pi f C R}{4 \pi^2 f^2 C} \quad \ldots(\because \omega=2 \pi f)
\end{array}\)