Exam

JEE Main

Year

2020

Date

9 January

Shift

Evening Shift

Subject

Physics

Chapter

Alternating Current

JEE Main 2020 · 9 January · Evening Shift · Physics · Alternating Current

In LC circuit the inductance L = 40mH and capacitance C = 100μ F ....

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

Step-by-step solution

$z = \sqrt{{R}^{2} + {\left({X}_{C} - {X}_{L}\right)}^{2}}$
$R = 0$
$Z = {X}_{C} - {X}_{L}$
$= \dfrac{1}{\omegaC} - \omegaL$
$= \dfrac{1}{314 \times 100 \times {10}^{- 6}} - 314 \times 40 \times {10}^{- 3}$
$= 31.84 - 12.56$
$= 19.28Ω$

$i = \dfrac{{V}_{0}}{Z}\sin \left(314t + \dfrac{\pi}{2}\right)$
$∴ i = \dfrac{{V}_{0}}{Z}\cos \left(314t\right) ⇒ i = \dfrac{10}{19.28}\cos \left(314t\right)$
⇒ i = 0.52\cos \left(314t\right)