Start Free Mock LoginAlready registered? Login here

Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACLC oscillationMedium2 minQB

PhysicsMediumsingle choice

In $L C$ circuit the inductance $L = 40 mH$ and capacitance $C = 100 μF .$ If a voltage $V (t) = 10 s in (314 t)$ is applied to the circuit, the current in the circuit is given as:

Options:

Answer:

Solution:

$z = R^{2} + (X_{C} - X_{L})^{2}$

$R = 0$

$Z = X_{C} - X_{L}$

$= \frac{1}{ω C} - ω L$

$= \frac{1}{314 \times 100 \times 1 0 ^{- 6}} - 314 \times 40 \times 1 0^{- 3}$

$= 31.84 - 12.56$

$= 19.28Ω$

$i = \frac{V _{0}}{Z} sin (314 t + \frac{π}{2})$

$∴ i = \frac{V _{0}}{Z} cos (314 t) \Rightarrow i = \frac{10}{19.28} cos (314 t)$

$\Rightarrow i = 0.52 cos (314 t)$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:LC oscillation

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...