MHT CET · Physics · Alternating Current
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 ( \(\left.\tan 45^{\circ}=1\right)\)
- A \(\left(\frac{4 \pi^2 f^2 C}{1+2 \pi f C R}\right)\)
- B \(\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)\)
- C \(\left(\frac{1-2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)\)
- D \(\left(\frac{4 \pi^2 f^2 C}{1-2 \pi f C R}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)\)
Step-by-step Solution
Detailed explanation
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 \(\left(\frac{1+2 \pi f C R}{4 \pi^2 f^2 C}\right)\)
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 \(\left(\frac{1+2 \pi f C R}{4 \pi^2 f^2 C}\right)\)
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