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Question: Answered & Verified by Expert
With an alternating voltage source of frequency ' $\mathrm{f}$ ', inductor ' $\mathrm{L}$ ', capacitor ' $\mathrm{C}$ ' and resistance ' $R$ ' are connected in series. The voltage leads the current by $45^{\circ}$. The value of ' $L$ ' is $\left(\tan 45^{\circ}=1\right)$
PhysicsAlternating CurrentMHT CETMHT CET 2023 (10 May Shift 2)
Options:
  • A $\left(\frac{1+2 \pi f \mathrm{CR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$
  • B $\left(\frac{1-2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$
  • C $\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1+2 \pi \mathrm{fCR}}\right)$
  • D $\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1-2 \pi \mathrm{fCR}}\right)$
Solution:
2213 Upvotes Verified Answer
The correct answer is: $\left(\frac{1+2 \pi f \mathrm{CR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$
$\begin{aligned} & \quad \tan \phi=\frac{X_L-X_C}{R} \\ & \Rightarrow \tan 45^{\circ}=\left[\frac{2 \pi \mathrm{fL}-\frac{1}{2 \pi \mathrm{fC}}}{R}\right] \quad \ldots .\left(\because \phi=45^{\circ}\right) \\ & \quad R=\frac{(2 \pi \mathrm{f})^2 \mathrm{LC}-1}{2 \pi \mathrm{fC}} \\ & \therefore \quad L=\frac{2 \pi \mathrm{fCR}+1}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\end{aligned}$

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