The charge of the capacitor in an LC oscillat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The total energy $U$ in an $LC$ circuit is constant and is given by $U = \frac{1}{2} \frac{Q_{0}^{2}}{C}$.When the energy stored in the inductor $

Vedclass · Accepted Answer

$\frac{Q_{0}}{\sqrt{2}}$

Vedclass · Answer

(D) The total energy $U$ in an $LC$ circuit is constant and is given by $U = \frac{1}{2} \frac{Q_{0}^{2}}{C}$.When the energy stored in the inductor $(U_L)$ and the capacitor $(U_C)$ are equal,each must be half of the total energy.So,$U_C = \frac{U}{2}$.Substituting the expressions for energy: $\frac{1}{2} \frac{q^2}{C} = \frac{1}{2} \left( \frac{1}{2} \frac{Q_{0}^{2}}{C} \right)$.Simplifying the equation: $\frac{q^2}{C} = \frac{Q_{0}^{2}}{2C}$.$q^2 = \frac{Q_{0}^{2}}{2}$.Taking the square root on both sides,we get $q = \frac{Q_{0}}{\sqrt{2}}$.

The charge of the capacitor in an $LC$ oscillatory circuit,when the energy associated with the inductor and capacitor are equal,is . . . . . . . ($Q_{0}$ is the initial charge on the capacitor.)

Similar Questions

In an oscillating $LC$ circuit, the maximum charge on the capacitor is $Q$. When the energy is stored equally between the electric and magnetic fields, the charge on the capacitor becomes

$A$ $1 \mu F$ capacitor is charged to $50 \ V$ and is then discharged through a $10 \ mH$ inductor of negligible resistance. The maximum current in the inductor is (in $A$)

The positive feedback is required by an amplifier to act as an oscillator. The feedback here means ...........

In an $LC$ circuit,the capacitor has maximum charge $q_0$. The value of ${\left| {\frac{{di}}{{dt}}} \right|_{\max }}$ is

Initially,the key was placed on $(1)$ until the capacitor got fully charged. Now,the key is placed on $(2)$ at $t = 0$. Find the minimum time when the energy in both the capacitor and the inductor will be the same.

Vedclass Test Series

Exam Paper Generator

Online Exam Module