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

Show that in the free oscillations of an $LC$ circuit,the sum of energies stored in the capacitor and the inductor is constant in time.

Obtain the differential equation for an $LC$ circuit.

The natural frequency of an $L-C$ circuit is $1,25,000 \text{ cycle/s}$. Then the capacitor $C$ is replaced by another capacitor with a dielectric medium of dielectric constant $K$. In this case,the frequency decreases by $25 \text{ kHz}$. The value of $K$ is

$A$ $4 \mu F$ capacitor is charged to $10 \ V$. The battery is then disconnected and a pure $10 \ mH$ coil is connected across the capacitor so that $LC$ oscillations are set up. The maximum current in the coil is (in $A$)

In the given circuit, the resonant frequency is (in $ Hz$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module