Informative line

### Lc Oscillations

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.

# Total Energy in an LC Circuit

• Consider an LC circuit in which a capacitor is connected to an inductor as shown in figure.
• Suppose the initial charge on capacitor be Qo.
• 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

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

#### In the LC circuit with inductance L = 1 H and capacitance  $$C=1\,\mu F$$ , the instantaneous charge is Q = 2 mC and instantaneous current is $$I$$ = 1 A. Calculate the total energy of the system at that instant, after the switch is closed, as shown in figure.

A 2.5 J

B 3 J

C 4 J

D 2 J

×

Total energy (U) in an LC circuit is given as

$$U = \dfrac {Q^{2}}{2C} +\dfrac{1}{2}\;LI^{2}$$

where,

Q = Charge

L = Inductance

$$I$$ = Current

C = Capacitance

Given : Q = 2 mC, L = 1 H, $$C=1\,\mu F$$ , I = 1 A

$$U=\dfrac {(2 × 10 ^{-3})^{2}}{2 × 1 × 10^{-6}}\;+\; \dfrac {1}{2}× 1 ×1 ^{2}$$

$$U=2 +\dfrac {1}{2}$$

$$U=\dfrac {5}{2}=2.5\,J$$

### In the LC circuit with inductance L = 1 H and capacitance  $$C=1\,\mu F$$ , the instantaneous charge is Q = 2 mC and instantaneous current is $$I$$ = 1 A. Calculate the total energy of the system at that instant, after the switch is closed, as shown in figure.

A

2.5 J

.

B

3 J

C

4 J

D

2 J

Option A is Correct

# Derivative of Total Energy in an LC Circuit

• Consider an LC circuit in which a capacitor is connected to an inductor as shown in figure.
• Suppose the initial charge on capacitor be Qo.
• 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.

#### For an LC circuit, the inductance L = 2 H and the capacitance C = 0.5 F. When the derivative of total energy is done, which of the following result comes out?

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 $$\dfrac {d^{2}Q}{dt^{2}} -Q = 0$$

×

Total energy (U) in an LC circuit is given as

$$U = \dfrac {Q^{2}}{2C} +\dfrac{1}{2}\;LI^{2}$$

where,

Q = Charge

L = Inductance

$$\text I$$ = Current

C = Capacitance

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.

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

Given : L= 2 H , C = 0.5 F

$$\dfrac {d^{2}Q}{dt^{2}} = \dfrac{-1}{2×0.5}Q$$

$$\dfrac {d^{2}Q}{dt^{2}} = -Q$$

$$\dfrac {d^{2}Q}{dt^{2}} + Q =0$$

### For an LC circuit, the inductance L = 2 H and the capacitance C = 0.5 F. When the derivative of total energy is done, which of the following result comes out?

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

$$\dfrac {d^{2}Q}{dt^{2}} -Q = 0$$

Option A is Correct

#### In the figure shown, the battery has an emf  $$\Delta V=10\,V$$, the inductance L = 2 mH and the capacitance C = 2 mF. The switch has been set to position 'a' for a long time so that the capacitor is charged. The switch is then thrown to position 'b' removing the battery from the circuit and connecting the capacitor directly across the inductor. Find the frequency of oscillation of the circuit.

A $$\dfrac{100}{\pi} Hz$$

B $$\dfrac {60}{\pi} Hz$$

C $$\dfrac {50}{\pi} Hz$$

D $$\dfrac {250}{\pi}Hz$$

×

The frequency of oscillation of an LC circuit is given as,

$$f= \dfrac {\omega }{2 \pi}$$

$$f= \dfrac{1}{2\pi \sqrt{LC}}$$

where,

L = Inductance

C = Capacitance

Given : C = 2 mF, L = 2 mH

$$f = \dfrac {1}{2 \pi\sqrt{(2× 10 ^{-3})(2 × 10 ^{-3})}}$$

$$f = \dfrac {1}{2 \pi × 2 × 10 ^{-3}}$$

$$f = \dfrac {1000}{4 \pi}$$

$$f = \dfrac {250}{\pi} Hz$$

### In the figure shown, the battery has an emf  $$\Delta V=10\,V$$, the inductance L = 2 mH and the capacitance C = 2 mF. The switch has been set to position 'a' for a long time so that the capacitor is charged. The switch is then thrown to position 'b' removing the battery from the circuit and connecting the capacitor directly across the inductor. Find the frequency of oscillation of the circuit.

A

$$\dfrac{100}{\pi} Hz$$

.

B

$$\dfrac {60}{\pi} Hz$$

C

$$\dfrac {50}{\pi} Hz$$

D

$$\dfrac {250}{\pi}Hz$$

Option D is Correct

# Maximum Charge across Capacitor in an LC Circuit

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

$$Q_{max}=C\Delta V$$

where,

C = Capacitance

$$\Delta V$$ = Voltage

#### In the given figure, the battery has an emf  $$\Delta V=10\,V$$, the inductance L = 2 mH and the capacitance C = 2 mF. The switch has been set to position 'a' for a long time so that the capacitor is charged. The switch is then thrown to position 'b' removing the battery from the circuit and connecting the capacitor directly across the inductor. Find the maximum value of charge on the capacitor.

A 40 mC

B 20 mC

C 2 mC

D 4 mC

×

Maximum charge across capacitor in an LC circuit is given as

$$Q_{max}=C\Delta V$$

where,

C = Capacitance

$$\Delta V$$ = Voltage

Given : $$\Delta V=10\,V$$ ,  C = 2 mF

Qmax = 2 × 10-3 × 10

= 20 mC

### In the given figure, the battery has an emf  $$\Delta V=10\,V$$, the inductance L = 2 mH and the capacitance C = 2 mF. The switch has been set to position 'a' for a long time so that the capacitor is charged. The switch is then thrown to position 'b' removing the battery from the circuit and connecting the capacitor directly across the inductor. Find the maximum value of charge on the capacitor.

A

40 mC

.

B

20 mC

C

2 mC

D

4 mC

Option B is Correct

# Comparison of an LC Circuit With mass - spring system

## Spring - Mass System

• 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.

### Comparison

Case 1

•   At t = 0, the spring is stretched and released in the spring - mass system.

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

•  At $$t=\dfrac {T}{4}$$, when the spring is released (set free), at the back motion it gains v= vmax 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

•  At $$t=\dfrac {T}{2}$$, the spring is compressed to $$x_0$$ . At the same time the capacitor of an LC circuit gets fully charged ,i.e.,current $$I=0$$ .

Case 4

•  Again at $$t=\dfrac {3T}{4}$$,  the spring is free and moved with velocity $$\vec v_{max}$$ and the current in an LC circuit reaches to maximum.

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

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

Case 5

•  At $$t=T$$  ,

For LC circuit : $$I=0\,,Q=Q_{max}$$

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

#### If an LC circuit is compared with mass-spring system, then potential energy stored in stretched spring is related to which quantity in LC circuit?

A Potential energy in capacitor

B Kinetic energy of spring

C Magnetic energy stored in inductor

D Half of potential energy stored in capacitor

×

When the spring is stretched, the total energy of the spring - mass system gets stored in the form of potential energy of the string,

$$U = \dfrac{1}{2}kx^{2}$$

At the same time, the current in the LC circuit reduces to zero. Thus, the total energy of LC circuit gets stored in the form of potential energy of capacitor.

Hence, option (A) is correct.

### If an LC circuit is compared with mass-spring system, then potential energy stored in stretched spring is related to which quantity in LC circuit?

A

Potential energy in capacitor

.

B

Kinetic energy of spring

C

Magnetic energy stored in inductor

D

Half of potential energy stored in capacitor

Option A is Correct

# Calculation of Charge

• 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.
• 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.

#### In an LC circuit at any instant t, current  $$I=I_0$$  cos $$\omega \;t$$. Calculate the charge.

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

×

Since, the charge lags current by $$\dfrac {\pi}{2}$$ phase.

$$Q = \int Idt$$

So, charge Q = $$\dfrac {I_0}{\omega} cos \left(\omega t-\dfrac{\pi}{2}\right)$$

Hence, option  'A'  is correct.

### In an LC circuit at any instant t, current  $$I=I_0$$  cos $$\omega \;t$$. Calculate the charge.

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

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

Option A is Correct