In an oscillation of L-C circuit,the maximum | vedclass

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

Question

(B) The total energy in an $L-C$ circuit is constant and is given by $U_{total} = \frac{Q^2}{2C}$,where $Q$ is the maximum charge.At any instant,the t

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The total energy in an $L-C$ circuit is constant and is given by $U_{total} = \frac{Q^2}{2C}$,where $Q$ is the maximum charge.At any instant,the total energy is the sum of electrostatic energy $(U_E)$ and magnetic energy $(U_B)$: $U_{total} = U_E + U_B = \frac{q^2}{2C} + \frac{1}{2}LI^2$.According to the problem,the energy is stored equally between the electric and magnetic fields,so $U_E = U_B$.Since $U_E + U_B = U_{total}$,we have $2U_E = U_{total}$.Substituting the expressions: $2 \left( \frac{q^2}{2C} \right) = \frac{Q^2}{2C}$.Simplifying this,we get $\frac{q^2}{C} = \frac{Q^2}{2C}$.Therefore,$q^2 = \frac{Q^2}{2}$,which gives $q = \frac{Q}{\sqrt{2}}$.

In an oscillation of $L-C$ 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 field is

Similar Questions

The circuit shown contains an inductor $L$,a capacitor $C$ with an initial charge $CE$,and a battery of $EMF$ $2E$. Find the maximum charge on the capacitor after the switch is closed at $t = 0$.

An oscillator circuit consists of an inductance of $0.5 \, mH$ and a capacitor of $20 \, \mu F$. The resonant frequency of the circuit is nearly (in $Hz$)

Obtain the differential equation for an $LC$ circuit.

In the $LC$ circuit shown below,the current is in the direction shown. At this time :-

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module