With an alternating voltage source of frequency ' $\mathrm{f}$ ', inductor ' $\mathrm{L}$ ', capacitor '…

With an alternating voltage source of frequency ' $\mathrm{f}$ ', inductor ' $\mathrm{L}$ ', capacitor ' $\mathrm{C}$ ' and resistance ' $\mathrm{R}$ ' are connected in series. The voltage leads the currently by $45^{\circ}$. The value of ' $L$ ' is ( $\left.\tan 45^{\circ}=1\right)$
  1. $\left(\frac{4 \pi^2 f^2 C}{1+2 \pi f C R}\right)$
  2. $\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$
  3. $\left(\frac{1-2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$
  4. $\left(\frac{4 \pi^2 f^2 C}{1-2 \pi f C R}\right)$

Solution

The correct option is (B). Concept: Consider the following Phasor diagram for a series LCR circuit: If the voltage leads the current by an angle $\theta=45^{\circ}$ then, $\begin{aligned} & \tan (\theta)=\frac{\left(X_L-X_C\right)}{R}=\tan \left(45^{\circ}\right) \\ & X_L-X_C=R\end{aligned}$ Introducing, inductive reactance $X_L=2 \pi \mathrm{fL}$, capacitive reactance $X_C=\frac{1}{2 \pi f C}$ and resistor $R$ $2 \pi f L-\frac{1}{2 \pi f C}=R$ On re-writing, $L=\left(\frac{1}{2 \pi f C}+R\right)\left(\frac{1}{2 \pi f}\right)$ or $\left(\frac{1+2 \pi f C R}{4 \pi^2 f^2 C}\right)$

Asked in: MHT CET 2022 (05 Aug Shift 1)

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