With an alternating voltage source frequency ' f ', inductor ' $L$ ', capacitor ' $C$ ' and resistance ' $R$…

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. $\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$
  2. $\left(\frac{1-2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$
  3. $\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1+2 \pi \mathrm{fCR}}\right)$
  4. $\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1-2 \pi \mathrm{fCR}}\right)$

Solution

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

Asked in: MHT CET 2024 (11 May Shift 1)

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