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

  • A
    $\left(\frac{1+2 \pi fCR}{4 \pi^2 f^2 C}\right)$
  • B
    $\left(\frac{1-2 \pi fCR}{4 \pi^2 f^2 C}\right)$
  • C
    $\left(\frac{4 \pi^2 f^2 C}{1+2 \pi fCR}\right)$
  • D
    $\left(\frac{4 \pi^2 f^2 C}{1-2 \pi fCR}\right)$

Explore More

Similar Questions

$A$ circuit when connected to an $A.C.$ source of $12 \; V$ gives a current of $0.2 \; A$. The same circuit when connected to a $D.C.$ source of $12 \; V$,gives a current of $0.4 \; A$. The circuit is

In a series $LCR$ circuit,operated with an $ac$ of angular frequency $\omega$,the total impedance is

$A$ resistor of $300 \Omega$ and an inductor of $\frac{1}{\pi} \text{ H}$ are connected in series with a $20 \text{ V}$,$200 \text{ Hz}$ $AC$ source. What is the phase difference between the current and the voltage?

In an $LC$ circuit,the inductance $L = 40 \; mH$ and capacitance $C = 100 \; \mu F$. If a voltage $V(t) = 10 \sin (314 t)$ is applied to the circuit,the current in the circuit is given by:

In an $LR$-circuit,the inductive reactance is equal to the resistance $R$ of the circuit. An e.m.f. $E = E_0 \cos(\omega t)$ is applied to the circuit. The power consumed in the circuit is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo