Calculate the total energy of the system at that instant, after the switch is closed. Practice derivative of total energy in an LC circuit and find the maximum value of charge on the capacitor.

- Consider an LC circuit in which a capacitor is connected to an inductor as shown in figure.
- Suppose the initial charge on capacitor be Q
_{o}. - When the switch is closed, the capacitor starts discharging which leads to decrease in electric energy of capacitor.
- The current generated from capacitor's discharging, generates magnetic energy which gets stored in an inductor.
- Thus, energy is transferred from electric field of capacitor to magnetic field of the inductor.
- In the absence of resistor, the total energy in the circuit is transformed back and forth between them.
- Energy stored in capacitor at some instant of time,

\(U_C = \dfrac {Q^{2}}{2C}\)

Q is an instantaneous charge.

- Energy stored in inductor is given as

\(U_L = \dfrac {1}{2} LI^{2}\)

- Thus, total energy of the circuit

\(U_C=\dfrac {Q^{2}}{2C}+\dfrac {1}{2} LI^{2}\)

- Consider an LC circuit in which a capacitor is connected to an inductor as shown in figure.
- Suppose the initial charge on capacitor be Q
_{o}. - When the switch is closed, the capacitor starts discharging which leads to decrease in electric energy of capacitor.
- The current generated from capacitor's discharging, generates magnetic energy which gets stored in an inductor.
- Thus, energy is transferred from electric field of capacitor to magnetic field of the inductor.
- In the absence of resistor, the total energy in the circuit is transformed back and forth between them.
- Energy stored in capacitor at some instant of time ,

\(U_C = \dfrac {Q^{2}}{2C}\)

Q is an instantaneous charge.

- Energy stored in inductor is given as

\(U_L = \dfrac {1}{2} LI^{2}\)

- Thus, total energy of the circuit

\(U=\dfrac {Q^{2}}{2C}+\dfrac {1}{2} LI^{2}\)

- Since there is no resistance in the circuit so, no energy is radiated away from circuit and also by ignoring the electromagnetic radiation, the total energy of the system remains constant. (U = Constant)

\(\dfrac{dU}{dt} = 0\)

\(\dfrac {d}{dt} \left(\dfrac {Q^{2}}{2C}\;+\;\dfrac{1}{2} LI^{2}\right ) = 0\)

\(\dfrac {Q}{C}\;\dfrac {dQ}{dt} + LI \dfrac {dI}{dt} = 0\;\;\;\;\;\;\;—(1)\)

We know that, \(I = \dfrac {dQ}{dt}\;\;\;\;\;\;—(2)\)

\(\dfrac{dI}{dt}\;=\;\dfrac{d^{2}Q}{dt^{2}}\;\;\;\;\;\;\;\;—(3)\)

Put (2) and (3) in (1),

\(\dfrac {Q}{C} + \dfrac {Ld^{2}Q}{dt^{2}}=0\)

This is a differential equation of second order whose degree is 1.

\(\dfrac {d^{2}Q}{dt^{2}}\;=\;\dfrac{-1}{LC}Q\)

**Order → **The highest derivative in the differential equation is known as order.

**Degree → **The power of the highest derivative is known as degree.

A \(\dfrac {d^{2}Q}{dt^{2}} +Q = 0\)

B \(\dfrac {d^{2}Q}{dt^{2}} +\dfrac {1}{2}\;Q = 0\)

C \(\dfrac {d^{2}Q}{dt^{2}} +2Q = 0\)

D \(\dfrac {d^{2}Q}{dt^{2}} -Q = 0\)

- If the mass m is moving with speed v and the spring having spring constant k, is displaced from its equilibrium by \(x\), then total energy

\(U=K.E.+U_{PE}\)

For spring mass system \(U=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\)

For an LC circuit \(U=\dfrac{Q^2}{2C}+\dfrac{1}{2}LI^2\)

- Potential energy stored in stretched spring is similar as energy stored in capacitor.
- The kinetic energy \((KE=\dfrac {1}{2}mv^{2})\) of moving block is similar as the magnetic energy \(\left(\dfrac {1}{2}LI^{2}\right )\) stored in an inductor.

**Case 1**

Also, at t = 0, the current in an LC circuit is zero as capacitor is fully charged. So, all the energy is stored as electric potential energy in the capacitor.

- Thus, at t = 0, the electric potential energy stored in capacitor is same as potential energy of the spring - mass system.

**Case 2 **

_{0 }= v_{max}velocity by the time the capacitor is discharging and maximum current flows in the circuit.- Thus, at t = T/4, the kinetic energy of the spring - mass system will be same as magnetic energy \(\left(\dfrac {1}{2}LI^{2}_{max}\right)\) stored in the inductor of an LC circuit.

**Case 3 **

**Case 4**

For LC circuit \(I=I_{max} \,\,,\,\,Q=0\)

For spring mass system \(x=0\,\,,\,\,v=v_{max}\)

**Case 5**

For LC circuit : \(I=0\,,Q=Q_{max}\)

For spring mass system : \(x=x_0\,,\,v=0\)

A Potential energy in capacitor

B Kinetic energy of spring

C Magnetic energy stored in inductor

D Half of potential energy stored in capacitor

- Consider an LC circuit in which a capacitor is connected to an inductor, as shown in figure.
- Suppose the initial charge on capacitor be Q
_{o }. - When the switch is closed, the capacitor starts discharging which leads to decrease in electric energy of capacitor.
- The current generated from capacitor's discharging generates magnetic energy which gets stored in inductor.
- Thus, energy is transferred from electric field of capacitor to magnetic field of the inductor.
- In the absence of resistor, the total energy in the circuit is transformed back and forth between them.
- Energy stored in capacitor at some instant of time

\(U_C = \dfrac {Q^{2}}{2C}\)

Q is an instantaneous charge .

- Energy stored in inductor is given as

\(U_L = \dfrac {1}{2} LI^{2}\)

- Thus, total energy of the circuit

\(U_C=\dfrac {Q^{2}}{2C}+\dfrac {1}{2} LI^{2}\)

Since there is no resistance in the circuit, so no energy is radiated away from circuit and also by ignoring the electromagnetic radiation, the total energy of the system remains constant. (U = Constant)

\(\dfrac{dU}{dt} = 0\)

\(\dfrac {d}{dt} \left(\dfrac {Q^{2}}{2C}\;+\;\dfrac{1}{2} LI^{2}\right ) = 0\)

\(\dfrac {Q}{C}\;\dfrac {dQ}{dt} + LI \dfrac {dI}{dt} = 0\;\;\;\;\;\;\;—(1)\)

We know that, \(I = \dfrac {dQ}{dt}\;\;\;\;\;\;—(2)\)

\(\dfrac{dI}{dt}=\dfrac{d^{2}Q}{dt^{2}}\;\;\;\;\;\;\;\;—(3)\)

Put (2) and (3) in (1),

\(\dfrac {Q}{C} + \dfrac {Ld^{2}Q}{dt^{2}}=0\)

This is the differential equation of second order whose degree is 1.

\(\dfrac {d^{2}Q}{dt^{2}}=\dfrac {-1\,Q} {LC}\)

**Order → **The highest derivative in the differential equation is known as order.

**Degree → **The power of the highest derivative is known as degree.

\(\dfrac {d^{2}Q}{dt^{2}} = \dfrac {-1}{L} \dfrac {Q}{C}\)

- The general solution of this equation \(Q(t)=Q_0\;\;cos(\omega_0t+\phi) \).

where \(Q_0\) is the amplitude of charge and \(\phi\) is the phase.

- The angular frequency is given by, \(\omega_0=\dfrac{1}{\sqrt {LC}}\)?

The corresponding current,

\(I = \dfrac {dQ}{dt} = \dfrac{d}{dt}(Q_0\;cos\;(\omega_0t\;+\phi ))\)

\(I(t) = - \omega_0 Q_0 \;sin(\omega_0t + \phi)\)

\(I(t) = -I_0\; sin (\omega_0t+\phi)\)

Here \(\phi = 0\),

\(Q(t) = Q_0\; cos \;\omega_0t\)

**Conclusion:- **The charge lags the current by \(\dfrac{\pi}{2}\) phase.

A \(Q= \dfrac {I_0}{\omega }cos \left(\omega t -\dfrac {\pi}{2}\right)\)

B \(Q= \dfrac {I_0}{\omega }cos \left(\omega t -\dfrac {\pi}{4}\right)\)

C \(Q= \dfrac {I_0}{\omega }cos \left(\omega t -\dfrac {\pi}{3}\right)\)

D \(Q= \dfrac {I_0}{\omega }sin \left(\omega t -\dfrac {\pi}{2}\right)\)

A \(\dfrac{100}{\pi} Hz\)

B \(\dfrac {60}{\pi} Hz\)

C \(\dfrac {50}{\pi} Hz\)

D \(\dfrac {250}{\pi}Hz\)

Maximum charge across capacitor in an LC circuit is given as,

\(Q_{max}=C\Delta V\)

where,

C = Capacitance

\(\Delta V\) = Voltage