In an oscillating LC circuit,the maximum char | vedclass

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

Question

(D) In an oscillating $LC$ circuit,the total energy $U$ is constant and is given by the maximum energy stored in the capacitor: $U = \frac{1}{2} \frac

Vedclass · Accepted Answer

$\frac{Q}{\sqrt{2}}$

Vedclass · Answer

(D) In an oscillating $LC$ circuit,the total energy $U$ is constant and is given by the maximum energy stored in the capacitor: $U = \frac{1}{2} \frac{Q^2}{C}$.When the energy is stored equally between the electric field (capacitor) and the magnetic field (inductor),the energy in the capacitor $U_E$ is half of the total energy $U$.$U_E = \frac{1}{2} U$Substituting the expressions for energy: $\frac{1}{2} \frac{q^2}{C} = \frac{1}{2} \left( \frac{1}{2} \frac{Q^2}{C} \right)$Simplifying the equation: $\frac{q^2}{C} = \frac{1}{2} \frac{Q^2}{C}$$q^2 = \frac{Q^2}{2}$$q = \frac{Q}{\sqrt{2}}$Therefore,the charge on the capacitor when the energy is shared equally is $\frac{Q}{\sqrt{2}}$.

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

Similar Questions

$A$ capacitor of capacity $20 \mu F$ is charged to a potential of $35 \text{ V}$. The battery is then disconnected. $A$ pure inductor coil of $200 \text{ mH}$ is connected across the capacitor so that $LC$ oscillations are set up. The maximum current in the coil is: (in $\text{ A}$)

$A$ fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self-inductance $L$ at $t = 0$. The time at which the energy is stored equally between the electric and the magnetic fields is:

For which two reasons is the discussion of $LC$ oscillations not realistic?

In the given circuit,when $S_1$ is closed,the capacitor gets fully charged. Now $S_1$ is opened and $S_2$ is closed. Then

The frequency of an $L-C$ circuit is $f_{1}$. If a resistance $R$ is also added to it,the frequency becomes $f_{2}$. The ratio $\frac{f_{2}}{f_{1}}$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module