LC Oscillations Questions in English

(N/A) The equation of motion for a forced damped mechanical oscillator is given by:
$m \frac{d^{2} x}{d t^{2}} + b \frac{d x}{d t} + k x = F_{0} \cos \omega_{d} t$
The equation for a driven $LCR$ circuit is given by:
$L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{q}{C} = V_{m} \sin \omega t$
By comparing these two differential equations, we can establish an analogy between mechanical and electrical systems as follows:

Mechanical System (Forced Oscillations)	Electrical System (Driven $LCR$ Circuit)
Displacement $x$	Charge $q$
Mass $m$	Inductance $L$
Damping constant $b$	Resistance $R$
Spring constant $k$	Inverse capacitance $1/C$
Driving force $F_{0} \cos \omega_{d} t$	Driving voltage $V_{m} \sin \omega t$
Natural frequency $\omega_{0} = \sqrt{k/m}$	Natural frequency $\omega_{0} = 1/\sqrt{LC}$

(B) An $LC$ circuit,also called a resonant circuit,tank circuit,or tuned circuit,is an electrical circuit consisting of an inductor,represented by the letter $L$,and a capacitor,represented by the letter $C$,connected together.
These circuits are used either for generating signals at a particular frequency or for picking out a signal at a particular frequency from a more complex signal.
In an ideal $LC$ circuit,the energy oscillates between the electric field of the capacitor and the magnetic field of the inductor at the resonant frequency $f = \frac{1}{2\pi\sqrt{LC}}$.

(A) An $LC$ oscillation is the flow of electric current that oscillates back and forth between a capacitor and an inductor in an $LC$ circuit.
When a charged capacitor is connected to an inductor,the energy stored in the electric field of the capacitor is transferred to the magnetic field of the inductor as current flows.
This process continues as the energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.
In an ideal $LC$ circuit with no resistance,these oscillations continue indefinitely with a constant frequency given by $f = \frac{1}{2\pi\sqrt{LC}}$.

(N/A) In an $LC$ circuit consisting of an inductor $L$ and a capacitor $C$,the sum of the potential drops across the inductor and the capacitor must be zero according to Kirchhoff's voltage law.
Let $q$ be the charge on the capacitor at any time $t$.
The potential difference across the capacitor is $V_C = \frac{q}{C}$.
The potential difference across the inductor is $V_L = L \frac{di}{dt}$.
Since $i = \frac{dq}{dt}$,the current is the rate of change of charge.
Thus,$V_L = L \frac{d^2q}{dt^2}$.
Applying Kirchhoff's loop rule: $V_L + V_C = 0$.
Substituting the expressions: $L \frac{d^2q}{dt^2} + \frac{q}{C} = 0$.
This is the differential equation for the $LC$ circuit,which can also be written as $\frac{d^2q}{dt^2} + \frac{1}{LC} q = 0$.

(N/A) In an $LC$ circuit consisting of an inductor $L$ and a capacitor $C$,the energy oscillates between the magnetic field of the inductor and the electric field of the capacitor.
The angular frequency of oscillation $\omega$ is given by the formula:
$\omega = \frac{1}{\sqrt{LC}}$
The frequency of oscillation $f$ is related to the angular frequency by $f = \frac{\omega}{2\pi}$.
Substituting the value of $\omega$,we get:
$f = \frac{1}{2\pi\sqrt{LC}}$
Here,$f$ is the frequency in Hertz $(Hz)$,$L$ is the inductance in Henry $(H)$,and $C$ is the capacitance in Farad $(F)$.

(C) In forced oscillations of a mechanical system,the equation of motion is $m(d^2x/dt^2) + b(dx/dt) + kx = F(t)$.
In an $LCR$ series circuit,the equation for charge $q$ is $L(d^2q/dt^2) + R(dq/dt) + (1/C)q = V(t)$.
Comparing these two equations:
$1$. Mass $m$ corresponds to Inductance $L$.
$2$. Damping constant $b$ corresponds to Resistance $R$.
$3$. Spring constant $k$ corresponds to the reciprocal of capacitance $1/C$.
$4$. Displacement $x$ corresponds to charge $q$.
Therefore,the physical quantity corresponding to the spring constant $k$ is $1/C$.

(A) In a harmonically oscillating spring-block system,the total energy is the sum of potential energy $\left(\frac{1}{2} kx^{2}\right)$ and kinetic energy $\left(\frac{1}{2} mv^{2}\right)$.
In an $LC$ circuit,the total energy is the sum of electrostatic energy stored in the capacitor $\left(\frac{1}{2} \frac{q^{2}}{C}\right)$ and magnetic energy stored in the inductor $\left(\frac{1}{2} LI^{2}\right)$.
Comparing these,the electrostatic energy $\left(\frac{1}{2} \frac{q^{2}}{C}\right)$ is analogous to the potential energy of the spring $\left(\frac{1}{2} kx^{2}\right)$,as both depend on the displacement/charge.
The magnetic energy $\left(\frac{1}{2} LI^{2}\right)$ is analogous to the kinetic energy of the block $\left(\frac{1}{2} mv^{2}\right)$,as both depend on the velocity/current.

(N/A) Applying Kirchhoff's voltage law to the $LCR$ circuit:
$L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = V_m \sin \omega t$
$(b)$ At $t = t_0$,the current is $i(t_0) = \frac{dq}{dt}|_{t=t_0}$ and the charge on the capacitor is $q(t_0)$.
The energy stored in the inductor is $U_L = \frac{1}{2} L [i(t_0)]^2$.
The energy stored in the capacitor is $U_C = \frac{1}{2} \frac{[q(t_0)]^2}{C}$.
$(c)$ After $t = t_0$,the circuit becomes an $LC$ circuit with $R=0$. The total energy $U = U_L + U_C$ is conserved. The charge $q(t)$ will oscillate sinusoidally with an angular frequency $\omega_0 = \frac{1}{\sqrt{LC}}$.

(B) The frequency of an ideal $L-C$ circuit is given by $f_{1} = \frac{1}{2\pi\sqrt{LC}}$.
When a resistance $R$ is added in series,the circuit becomes a damped $L-C-R$ circuit. The damped angular frequency is given by $\omega_{d} = \sqrt{\omega_{0}^{2} - \beta^{2}}$,where $\omega_{0} = \frac{1}{\sqrt{LC}}$ and $\beta = \frac{R}{2L}$.
Thus,the damped frequency $f_{2}$ is given by:
$f_{2} = \frac{1}{2\pi} \sqrt{\left(\frac{1}{\sqrt{LC}}\right)^{2} - \left(\frac{R}{2L}\right)^{2}}$
Taking the ratio $\frac{f_{2}}{f_{1}}$:
$\frac{f_{2}}{f_{1}} = \frac{\frac{1}{2\pi} \sqrt{\frac{1}{LC} - \frac{R^{2}}{4L^{2}}}}{\frac{1}{2\pi\sqrt{LC}}}$
$= \sqrt{\frac{\frac{1}{LC} - \frac{R^{2}}{4L^{2}}}{\frac{1}{LC}}}$
$= \sqrt{1 - \frac{R^{2}C}{4L}}$

(C) The resonant frequency of an $LC$ oscillator circuit is given by the formula:
$v_{0} = \frac{1}{2 \pi \sqrt{LC}}$
Given values are:
Inductance $L = 0.5 \, mH = 0.5 \times 10^{-3} \, H = 5 \times 10^{-4} \, H$
Capacitance $C = 20 \, \mu F = 20 \times 10^{-6} \, F$
Substituting these values into the formula:
$v_{0} = \frac{1}{2 \times 3.14 \times \sqrt{5 \times 10^{-4} \times 20 \times 10^{-6}}}$
$v_{0} = \frac{1}{6.28 \times \sqrt{100 \times 10^{-10}}}$
$v_{0} = \frac{1}{6.28 \times \sqrt{10^{-8}}}$
$v_{0} = \frac{1}{6.28 \times 10^{-4}}$
$v_{0} = \frac{10^{4}}{6.28} \approx 1592 \, Hz$

(B) In an $L-C$ oscillation circuit,the total time period $T$ for one complete cycle is given by:
$T = 2\pi \sqrt{LC}$
Given $L = 25 \, mH = 25 \times 10^{-3} \, H$ and $C = 10 \, \mu F = 10 \times 10^{-6} \, F$.
Substituting the values:
$T = 2\pi \sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}$
$T = 2\pi \sqrt{250 \times 10^{-9}} = 2\pi \sqrt{25 \times 10^{-8}}$
$T = 2\pi \times 5 \times 10^{-4} = 10\pi \times 10^{-4} = \pi \times 10^{-3} \, s = \pi \, ms$.
The energy in the capacitor is maximum at $t = 0$. The current in the circuit becomes maximum when the energy is fully transferred from the capacitor to the inductor,which occurs at $t = \frac{T}{4}$.
Therefore,the time required is:
$t = \frac{T}{4} = \frac{\pi \, ms}{4} = \frac{\pi}{4} \, ms$.

(B) An oscillator is a circuit that produces a continuous periodic waveform without any external input signal.
For an amplifier to act as an oscillator,it requires positive feedback.
Positive feedback means that a portion of the output signal (voltage or power) is fed back to the input in phase with the original input signal.
This feedback reinforces the input,allowing the circuit to sustain oscillations independently.

(A) The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.
Substituting the given values,$U = \frac{1}{2} \times (500 \times 10^{-6} \, F) \times (100 \, V)^2 = \frac{1}{2} \times 500 \times 10^{-6} \times 10^4 = 2.5 \, J$.
In an $LC$ circuit,the total energy is conserved. The maximum current $I_{max}$ occurs when the entire energy of the capacitor is transferred to the inductor.
Thus,$\frac{1}{2} LI_{max}^2 = U_{total}$.
$\frac{1}{2} \times (50 \times 10^{-3} \, H) \times I_{max}^2 = 2.5 \, J$.
$I_{max}^2 = \frac{2.5 \times 2}{50 \times 10^{-3}} = \frac{5}{0.05} = 100$.
$I_{max} = \sqrt{100} = 10 \, A$.

(A) At $t=0$,the capacitor is uncharged and the flux through the inductor is $\phi$.
Using the relation $\phi = L I$,the initial current in the circuit at $t=0$ is $I_{0} = \phi / L$.
The circuit forms an $LC$ oscillator. The differential equation for the charge $q$ on the capacitor is $L \frac{d^{2} q}{d t^{2}} + \frac{q}{C} = 0$,which simplifies to $\frac{d^{2} q}{d t^{2}} + \omega_{0}^{2} q = 0$,where $\omega_{0} = 1 / \sqrt{LC}$.
The general solution for the charge is $q(t) = A \sin(\omega_{0} t + \delta)$.
Since $q(0) = 0$,we have $\delta = 0$,so $q(t) = A \sin(\omega_{0} t)$.
The current is $I(t) = \frac{dq}{dt} = A \omega_{0} \cos(\omega_{0} t)$.
At $t=0$,$I(0) = A \omega_{0} = I_{0} = \phi / L$.
Therefore,$A = \phi / (L \omega_{0})$.
Substituting $A$ back,the current is $I(t) = (\phi / L) \cos(\omega_{0} t)$.
Thus,option $A$ is correct.

(A) When the switch is in position $A$ for a long time,the inductor acts as a short circuit (steady state). The current flowing through the inductor is $I_0 = \frac{E}{R}$.
At $t=0$,the switch is moved to position $B$. The circuit now consists of an inductor $L$ and a capacitor $C$ in series,forming an $LC$ oscillator circuit. The initial current in the inductor is $I_0 = \frac{E}{R}$ and the initial charge on the capacitor is $q_0 = 0$.
The energy stored in the inductor is converted into the energy stored in the capacitor. According to the conservation of energy:
$\frac{1}{2} L I_0^2 = \frac{1}{2} \frac{q_{\max}^2}{C}$
Substituting $I_0 = \frac{E}{R}$:
$L \left(\frac{E}{R}\right)^2 = \frac{q_{\max}^2}{C}$
$q_{\max}^2 = L C \frac{E^2}{R^2}$
$q_{\max} = \sqrt{L C} \frac{E}{R}$

(D) The total energy $U$ in an $LC$ circuit is constant and is given by the maximum energy stored in the capacitor: $U = \frac{Q^2}{2C}$.
When the charge on the capacitor is $Q' = \frac{Q}{2}$,the energy stored in the capacitor $U_C$ is given by:
$U_C = \frac{(Q')^2}{2C} = \frac{(Q/2)^2}{2C} = \frac{Q^2/4}{2C} = \frac{Q^2}{8C}$.
Since the total energy $U$ is the sum of the energy in the capacitor $U_C$ and the energy in the inductor $U_L$,we have $U = U_C + U_L$.
Therefore,the energy stored in the inductor $U_L$ is:
$U_L = U - U_C = \frac{Q^2}{2C} - \frac{Q^2}{8C}$.
Taking the common denominator:
$U_L = \frac{4Q^2 - Q^2}{8C} = \frac{3Q^2}{8C}$.
Since $U = \frac{Q^2}{2C}$,we can write $\frac{Q^2}{8C} = \frac{1}{4} U$.
Thus,$U_L = \frac{3}{4} U$.

(A) The total energy in an $L-C$ circuit is constant and is given by $U_{total} = \frac{Q^2}{2C}$.
At any instant,the total energy is the sum of electric energy $(U_E)$ and magnetic energy $(U_B)$: $U_{total} = U_E + U_B$.
Given that at a certain instant,$U_E = U_B$,we can write $U_{total} = U_E + U_E = 2U_E$.
Substituting the expressions for energy,we have $\frac{Q^2}{2C} = 2 \left( \frac{(Q')^2}{2C} \right)$,where $Q'$ is the charge on the capacitor at that instant.
Simplifying the equation: $Q^2 = 2(Q')^2$.
Therefore,$(Q')^2 = \frac{Q^2}{2}$,which gives $Q' = \frac{Q}{\sqrt{2}}$.

(A) The total energy in the $LC$ circuit is constant and is equal to the initial energy stored in the capacitor: $E_{\text{total}} = \frac{Q_0^2}{2C}$.
Let $E_c$ be the energy in the capacitor and $E_i$ be the energy in the inductor at any time $t$.
We are given that $E_c = 3 E_i$.
Since $E_{\text{total}} = E_c + E_i$,we can write $E_{\text{total}} = 3 E_i + E_i = 4 E_i$.
Substituting the total energy: $4 E_i = \frac{Q_0^2}{2C}$.
Therefore,$E_i = \frac{Q_0^2}{8C}$.
Since the energy in the inductor is given by $E_i = \frac{1}{2} L i^2$,we have $\frac{1}{2} L i^2 = \frac{Q_0^2}{8C}$.
Solving for $i$: $i^2 = \frac{Q_0^2}{4LC}$.
Taking the square root: $i = \frac{Q_0}{2 \sqrt{LC}}$.

(A) The resonant frequency of an $LC$ oscillator circuit is given by $\omega_0 = \frac{1}{\sqrt{LC}}$.
Given that the new inductance $L' = 2L$ and the new capacitance $C' = 8C$.
The new resonant frequency $\omega$ is given by $\omega = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(2L)(8C)}} = \frac{1}{\sqrt{16LC}}$.
Simplifying this,we get $\omega = \frac{1}{4\sqrt{LC}}$.
Since $\omega_0 = \frac{1}{\sqrt{LC}}$,we can substitute this into the expression for $\omega$ to get $\omega = \frac{\omega_0}{4}$.
Comparing this with $\omega = x\omega_0$,we find that $x = \frac{1}{4}$.

(B) In an oscillating $LC$ circuit,the total energy is conserved,oscillating between the electric field of the capacitor and the magnetic field of the inductor.
The maximum energy stored in the capacitor is equal to the maximum energy stored in the inductor:
$\frac{1}{2} L I_{\max}^2 = \frac{1}{2} \frac{Q_{\max}^2}{C}$
Rearranging for the maximum current $I_{\max}$:
$I_{\max} = Q_{\max} \sqrt{\frac{1}{LC}}$
Given values:
$L = 75 \times 10^{-3} \, H$
$C = 1.2 \times 10^{-6} \, F$
$Q_{\max} = 2.7 \times 10^{-6} \, C$
Calculating the angular frequency $\omega = \frac{1}{\sqrt{LC}}$:
$\omega = \frac{1}{\sqrt{75 \times 10^{-3} \times 1.2 \times 10^{-6}}} = \frac{1}{\sqrt{90 \times 10^{-9}}} = \frac{1}{\sqrt{9 \times 10^{-8}}} = \frac{1}{3 \times 10^{-4}} = \frac{10^4}{3} \, rad/s$
Now,$I_{\max} = Q_{\max} \times \omega$:
$I_{\max} = (2.7 \times 10^{-6}) \times \frac{10^4}{3} = 0.9 \times 10^{-2} \, A = 9 \times 10^{-3} \, A = 9 \, mA$.

(B) By the principle of conservation of energy in an $LC$ circuit,the maximum electrostatic energy stored in the capacitor is equal to the maximum magnetic energy stored in the inductor.
$\frac{1}{2} CV^2 = \frac{1}{2} LI_{\max}^2$
Rearranging the formula to solve for the maximum current $I_{\max}$:
$I_{\max} = V \sqrt{\frac{C}{L}}$
Given values:
$C = 100 \ \mu F = 100 \times 10^{-6} \ F = 10^{-4} \ F$
$L = 6.4 \ mH = 6.4 \times 10^{-3} \ H$
$V = 12 \ V$
Substituting the values:
$I_{\max} = 12 \times \sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}$
$I_{\max} = 12 \times \sqrt{\frac{10^{-4}}{6.4 \times 10^{-3}}} = 12 \times \sqrt{\frac{10^{-1}}{6.4}} = 12 \times \sqrt{\frac{0.1}{6.4}} = 12 \times \sqrt{\frac{1}{64}}$
$I_{\max} = 12 \times \frac{1}{8} = 1.5 \ A$

(A) The induced electromotive force $(emf)$ in the circuit due to the changing magnetic field is given by Faraday's law: $\varepsilon = -\frac{d\Phi}{dt} = -A \frac{dB}{dt}$.
Given $A = 1 \ m^2$ and $\frac{dB}{dt} = \beta = 0.04 \ T \ s^{-1}$,the magnitude of the induced $emf$ is $\varepsilon = 1 \times 0.04 = 0.04 \ V$.
This $emf$ acts as a voltage source in the $LC$ circuit. The current in an $LC$ circuit oscillates as $I(t) = I_0 \sin(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.
The maximum current $I_0$ in an $LC$ circuit driven by a constant $emf$ $\varepsilon$ is given by $I_0 = \varepsilon \sqrt{\frac{C}{L}}$.
Substituting the values: $I_0 = 0.04 \times \sqrt{\frac{10^{-3}}{0.1}} = 0.04 \times \sqrt{10^{-2}} = 0.04 \times 0.1 = 0.004 \ A$.
Converting to milliamperes: $I_0 = 0.004 \times 1000 \ mA = 4 \ mA$.

(C) In an $LC$ circuit,the charge on the capacitor varies as $q(t) = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.
At $t=0$,the energy is maximum,meaning the charge is maximum $(q = q_0)$.
The current in the circuit is given by $i(t) = -\frac{dq}{dt} = q_0 \omega \sin(\omega t)$.
The current is maximum when $\sin(\omega t) = 1$,which occurs at $\omega t = \frac{\pi}{2}$,or $t = \frac{\pi}{2\omega}$.
Given $L = 16 \text{ mH} = 16 \times 10^{-3} \text{ H}$ and $C = 10 \mu\text{F} = 10 \times 10^{-6} \text{ F} = 10^{-5} \text{ F}$.
Calculate $\omega = \frac{1}{\sqrt{16 \times 10^{-3} \times 10^{-5}}} = \frac{1}{\sqrt{16 \times 10^{-8}}} = \frac{1}{4 \times 10^{-4}} = 0.25 \times 10^4 = 2500 \text{ rad/s}$.
Now,$t = \frac{\pi}{2 \times 2500} = \frac{\pi}{5000} = \pi \times 2 \times 10^{-4} \text{ s} = 2 \pi \times 10^{-4} \text{ s}$.

(A) The resonant angular frequency of an $LC$ circuit is given by the formula: $\omega = \frac{1}{\sqrt{LC}}$.
Let the initial inductance be $L$ and the initial capacitance be $C$. The initial angular frequency is $\omega = \frac{1}{\sqrt{LC}}$.
According to the problem,the new inductance $L' = 2L$ and the new capacitance $C' = 4C$.
The new angular frequency $\omega'$ is given by: $\omega' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(2L)(4C)}} = \frac{1}{\sqrt{8LC}}$.
We can simplify this as: $\omega' = \frac{1}{\sqrt{8} \sqrt{LC}} = \frac{1}{2\sqrt{2} \sqrt{LC}}$.
Since $\omega = \frac{1}{\sqrt{LC}}$,we substitute this into the expression for $\omega'$:
$\omega' = \frac{\omega}{2\sqrt{2}}$.
Therefore,the correct option is $A$.

(B) The resonant angular frequency of an $LC$ circuit is given by the formula $\omega = \frac{1}{\sqrt{LC}}$.
Let the initial inductance be $L$ and capacitance be $C$. Then,$\omega = \frac{1}{\sqrt{LC}}$.
According to the problem,the new inductance $L' = 4L$ and the new capacitance $C' = 8C$.
The new resonant angular frequency $\omega'$ is given by $\omega' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(4L)(8C)}} = \frac{1}{\sqrt{32LC}}$.
Simplifying this,we get $\omega' = \frac{1}{\sqrt{16 \times 2 \times LC}} = \frac{1}{4 \sqrt{2} \sqrt{LC}}$.
Since $\omega = \frac{1}{\sqrt{LC}}$,we can substitute this into the equation to get $\omega' = \frac{\omega}{4 \sqrt{2}}$.

(C) When $S_1$ is closed,the capacitor $C$ is charged to a potential difference $V$. The energy stored in the capacitor is $U_E = \frac{1}{2}CV^2$.
When $S_1$ is opened and $S_2$ is closed,the capacitor $C$ and inductor $L$ form an $LC$ oscillating circuit.
The energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.
The maximum current $I_{max}$ in the circuit occurs when all the electric energy is converted into magnetic energy.
By the law of conservation of energy:
$\frac{1}{2}CV^2 = \frac{1}{2}LI_{max}^2$
$I_{max}^2 = \frac{C}{L}V^2$
$I_{max} = V\sqrt{\frac{C}{L}}$
Thus,the instantaneous current in the circuit can reach this maximum value.

(C) The maximum energy stored in the capacitor is given by $E_{max} = \frac{Q^2}{2C}$.
When the energy is stored equally between the electric and magnetic fields, the energy in the capacitor is half of the maximum energy, i.e., $E_{cap} = \frac{1}{2} E_{max}$.
Let $Q'$ be the charge on the capacitor at this instant. Then, $E_{cap} = \frac{Q'^2}{2C}$.
Equating the two expressions for $E_{cap}$:
$\frac{Q'^2}{2C} = \frac{1}{2} \left( \frac{Q^2}{2C} \right)$
$Q'^2 = \frac{Q^2}{2}$
$Q' = \frac{Q}{\sqrt{2}}$.

(C) The total energy in an $LC$ circuit is conserved.
Initially,the energy stored in the capacitor is $U_i = \frac{1}{2}CV_1^2$.
When the potential difference across the capacitor becomes $V_2$,the energy stored in the capacitor is $U_c = \frac{1}{2}CV_2^2$.
The energy stored in the inductor at this instant is $U_L = \frac{1}{2}LI^2$.
By the law of conservation of energy: $U_i = U_c + U_L$.
$\frac{1}{2}CV_1^2 = \frac{1}{2}CV_2^2 + \frac{1}{2}LI^2$.
$LI^2 = C(V_1^2 - V_2^2)$.
$I^2 = \frac{C}{L}(V_1^2 - V_2^2)$.
$I = \sqrt{\frac{C}{L}(V_1^2 - V_2^2)}$.
Thus,the correct option is $C$.

(D) Given: Capacitance $C = 20 \mu F = 20 \times 10^{-6} \text{ F}$, Voltage $V = 35 \text{ V}$, Inductance $L = 200 \text{ mH} = 200 \times 10^{-3} \text{ H}$.
In an $LC$ circuit, the energy stored in the capacitor is converted into the energy stored in the inductor.
The maximum energy in the capacitor is $U_E = \frac{1}{2} C V^2$.
The maximum energy in the inductor is $U_B = \frac{1}{2} L I_{\text{max}}^2$.
By conservation of energy, $U_E = U_B$, so $\frac{1}{2} C V^2 = \frac{1}{2} L I_{\text{max}}^2$.
Solving for $I_{\text{max}}$: $I_{\text{max}} = V \sqrt{\frac{C}{L}}$.
Substituting the values: $I_{\text{max}} = 35 \times \sqrt{\frac{20 \times 10^{-6}}{200 \times 10^{-3}}} = 35 \times \sqrt{\frac{1}{10000}} = 35 \times \frac{1}{100} = 0.35 \text{ A}$.

(C) In an $LC$ circuit, the total energy is conserved. The initial energy stored in the capacitor is $U_E = \frac{1}{2} C V^2$.
When the current in the coil is maximum $(I_{max})$, the energy stored in the inductor is $U_B = \frac{1}{2} L I_{max}^2$.
Since there is no resistance, the total energy remains constant, so the initial electrical energy is converted into maximum magnetic energy:
$\frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2$
$I_{max}^2 = \frac{C V^2}{L}$
$I_{max} = V \sqrt{\frac{C}{L}}$
Given: $C = 1 \, \mu F = 1 \times 10^{-6} \, F$, $V = 50 \, V$, $L = 10 \, mH = 10 \times 10^{-3} \, H = 10^{-2} \, H$.
$I_{max} = 50 \times \sqrt{\frac{1 \times 10^{-6}}{10^{-2}}}$
$I_{max} = 50 \times \sqrt{10^{-4}}$
$I_{max} = 50 \times 10^{-2} \, A = 0.5 \, A$.

(C) The energy stored in the capacitor is $U_E = \frac{1}{2}CV^2$.
When the capacitor is connected to the inductor,the energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.
At the moment of maximum current $i_0$,all the energy is stored in the inductor as magnetic energy $U_B = \frac{1}{2}Li_0^2$.
By the law of conservation of energy: $\frac{1}{2}CV^2 = \frac{1}{2}Li_0^2$.
Solving for $i_0$: $i_0^2 = \frac{CV^2}{L} \implies i_0 = V \sqrt{\frac{C}{L}}$.

(A) The frequency of an $LC$ oscillator is given by the formula:
$F = \frac{1}{2 \pi \sqrt{LC}}$
From this relation,we can see that the frequency $F$ is inversely proportional to the square root of the capacitance $C$ (i.e.,$F \propto \frac{1}{\sqrt{C}}$).
Let $F_1 = F$ and $C_1 = 0.1 \ \mu F$.
Let $F_2$ be the new frequency and $C_2 = 0.2 \ \mu F$.
Using the ratio method:
$\frac{F_2}{F_1} = \sqrt{\frac{C_1}{C_2}}$
$\frac{F_2}{F} = \sqrt{\frac{0.1}{0.2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Therefore,$F_2 = \frac{F}{\sqrt{2}} \ Hz$.

(A) The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.
Given $C = 4 \mu F = 4 \times 10^{-6} \ F$ and $V = 10 \ V$,the energy is $U = \frac{1}{2} \times 4 \times 10^{-6} \times (10)^2 = 2 \times 10^{-4} \ J$.
When the capacitor is connected to the inductor,the total energy is conserved and oscillates between the electric field of the capacitor and the magnetic field of the inductor.
The maximum energy in the inductor is given by $U_{max} = \frac{1}{2} L I_{max}^2$.
Equating the energies: $\frac{1}{2} L I_{max}^2 = \frac{1}{2} C V^2$.
$I_{max}^2 = \frac{C V^2}{L} = \frac{4 \times 10^{-6} \times 100}{10 \times 10^{-3}} = \frac{4 \times 10^{-4}}{10^{-2}} = 4 \times 10^{-2} = 0.04$.
Therefore,$I_{max} = \sqrt{0.04} = 0.2 \ A$.

(D) When $S_1$ is closed,the capacitor $C$ charges to a potential $V$. The energy stored in the capacitor is $U_E = \frac{1}{2} CV^2$.
When $S_1$ is opened and $S_2$ is closed,the capacitor discharges through the inductor $L$,forming an $LC$ oscillation circuit.
At $t = 0$ (the moment $S_2$ is closed),the energy is entirely electric (stored in the capacitor).
As the capacitor discharges,the energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.
By the law of conservation of energy,the maximum electric energy equals the maximum magnetic energy:
$\frac{1}{2} CV^2 = \frac{1}{2} LI_{max}^2$
$I_{max}^2 = \frac{C}{L} V^2$
$I_{max} = V \sqrt{\frac{C}{L}}$
Thus,the instantaneous current in the circuit can reach a maximum value of $V \sqrt{\frac{C}{L}}$.

(A) The energy stored in a charged capacitor is given by $U = \frac{1}{2} CV^2$.
When the capacitor discharges through an inductor,the energy is transferred from the electric field of the capacitor to the magnetic field of the inductor.
At the moment of maximum current $(I_0)$,all the energy is stored in the inductor as magnetic energy,given by $U = \frac{1}{2} LI_0^2$.
By the law of conservation of energy: $\frac{1}{2} CV^2 = \frac{1}{2} LI_0^2$.
Solving for $I_0$: $I_0^2 = \frac{CV^2}{L}$.
Given: $C = 1 \mu F = 10^{-6} \ F$,$V = 50 \ V$,$L = 10 \ mH = 10 \times 10^{-3} \ H = 10^{-2} \ H$.
Substituting the values: $I_0^2 = \frac{10^{-6} \times (50)^2}{10^{-2}} = \frac{10^{-6} \times 2500}{10^{-2}} = 2500 \times 10^{-4} = 0.25$.
Therefore,$I_0 = \sqrt{0.25} = 0.5 \ A$.

(A) The magnetic energy $U_B$ in an inductor is given by $U_B = \frac{1}{2} L I^2$. Since the current $I$ in an $A.C.$ circuit varies as $I = I_0 \sin(\omega t)$,the energy varies as $U_B \propto \sin^2(\omega t)$.
Energy changes from maximum to minimum in $\frac{1}{4}$ of a time period $T$.
Given,$\frac{T}{4} = 5 \ ms = 5 \times 10^{-3} \ s$.
Therefore,$T = 20 \times 10^{-3} \ s = 0.02 \ s$.
The frequency $f$ is given by $f = \frac{1}{T} = \frac{1}{0.02} = 50 \ Hz$.

(B) When $S_1$ is closed,the capacitor $C$ charges to a potential difference $V$.
When $S_1$ is opened and $S_2$ is closed,the capacitor $C$ and inductor $L$ form an $LC$ oscillating circuit.
The energy initially stored in the capacitor is $U_E = \frac{1}{2} C V^2$.
As the capacitor discharges,the energy is transferred to the inductor as magnetic energy $U_B = \frac{1}{2} L I^2$.
By the law of conservation of energy,the maximum current $I_{max}$ occurs when all electrostatic energy is converted to magnetic energy:
$\frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2$
$I_{max}^2 = \frac{C V^2}{L}$
$I_{max} = V \sqrt{\frac{C}{L}}$.
Thus,the instantaneous current in the circuit can reach this maximum value.

(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 $q = \frac{Q_{0}}{\sqrt{2}}$.

(D) The angular frequency of free oscillations for an $LC$ circuit is given by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$.
Given values are $L = 16 \text{ mH} = 16 \times 10^{-3} \text{ H}$ and $C = 10 \mu F = 10 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$\omega_{0} = \frac{1}{\sqrt{16 \times 10^{-3} \times 10 \times 10^{-6}}}$
$\omega_{0} = \frac{1}{\sqrt{16 \times 10^{-8}}}$
$\omega_{0} = \frac{1}{4 \times 10^{-4}}$
$\omega_{0} = \frac{10^{4}}{4} = 2500 \text{ rad s}^{-1}$.

(B) Given, capacitance, $C = 10^{-6} \,F$.
Inductance, $L = 10^{-4} \,H$.
The frequency of electrical oscillations in an $L-C$ circuit is given by the formula:
$f = \frac{1}{2 \pi \sqrt{LC}}$
Substituting the values:
$f = \frac{1}{2 \pi \sqrt{10^{-4} \times 10^{-6}}}$
$f = \frac{1}{2 \pi \sqrt{10^{-10}}}$
$f = \frac{1}{2 \pi \times 10^{-5}}$
$f = \frac{10^5}{2 \pi} \,Hz$

(B) The charge $q$ as a function of time $t$ is given by $q = q_0 \sin(\omega t)$,where $q_0$ is the maximum charge and $\omega$ is the angular frequency.
Differentiating with respect to $t$,the current $I$ is given by $I = \frac{dq}{dt} = \omega q_0 \cos(\omega t)$.
At $t = 0$,the current is $I = \omega q_0 \cos(0) = \omega q_0$.
Given $I = 2 \ A$ at $t = 0$,we have $2 = \omega q_0$.
Since $\omega = \frac{1}{\sqrt{LC}}$,we substitute this into the equation: $q_0 = I \sqrt{LC}$.
Given $L = 3 \times 10^{-3} \ H$ and $C = 2.7 \times 10^{-6} \ F$,we calculate:
$q_0 = 2 \times \sqrt{3 \times 10^{-3} \times 2.7 \times 10^{-6}}$
$q_0 = 2 \times \sqrt{8.1 \times 10^{-9}} = 2 \times \sqrt{81 \times 10^{-10}}$
$q_0 = 2 \times 9 \times 10^{-5} = 18 \times 10^{-5} \ C$.

(C) Given, $L = 0.5 \, mH = 0.5 \times 10^{-3} \, H$ and $C = 20 \, \mu F = 20 \times 10^{-6} \, F$.
The resonant frequency of an $L-C$ circuit is given by the formula:
$f = \frac{1}{2 \pi \sqrt{L C}}$
Substituting the values:
$f = \frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}}$
$f = \frac{1}{2 \pi \sqrt{10 \times 10^{-9}}} = \frac{1}{2 \pi \sqrt{10^{-8}}}$
$f = \frac{1}{2 \pi \times 10^{-4}} = \frac{10^4}{2 \pi} \approx \frac{10000}{6.283} \approx 1592.3 \, Hz$
Thus, the resonant frequency is approximately $1592 \, Hz$.

(C) For $LC$-oscillations,the differential equation is given by $L \frac{di}{dt} + \frac{q}{C} = 0$.
Differentiating with respect to time $t$,we get $L \frac{d^2i}{dt^2} + \frac{1}{C} \frac{dq}{dt} = 0$. Since $i = \frac{dq}{dt}$,this becomes $L \frac{d^2i}{dt^2} + \frac{1}{C} i = 0$ ... $(i)$.
For a mechanical spring-block system,the equation of motion is $m \frac{d^2x}{dt^2} + kx = 0$ ... $(ii)$.
Comparing equation $(i)$ and $(ii)$,we observe that the mass $m$ is analogous to inductance $L$,and the force constant $k$ is analogous to the reciprocal of capacitance,i.e.,$k \propto \frac{1}{C}$.

(C) The resonant frequency of an $LC$ circuit is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.
When a dielectric slab of constant $K$ is inserted between the plates of the capacitor,the new capacitance becomes $C' = KC$.
The new resonant frequency $f'$ is given by $f' = \frac{1}{2\pi\sqrt{LC'}} = \frac{1}{2\pi\sqrt{L(KC)}} = \frac{1}{\sqrt{K}} \times \frac{1}{2\pi\sqrt{LC}}$.
Substituting the given values,$K = 16$ and $f_0 = \frac{1}{2\pi\sqrt{LC}}$,we get $f' = \frac{f_0}{\sqrt{16}} = \frac{f_0}{4}$.
Therefore,the correct option is $C$.

(D) Given: Capacitance $C = 196 \text{ pF} = 196 \times 10^{-12} \text{ F}$.
Inductance $L = 441 \text{ } \mu\text{H} = 441 \times 10^{-6} \text{ H}$.
The resonant frequency $f$ of an $L-C$ circuit is given by the formula $f = \frac{1}{2 \pi \sqrt{LC}}$.
Substituting the values:
$f = \frac{1}{2 \pi \sqrt{441 \times 10^{-6} \times 196 \times 10^{-12}}}$
$f = \frac{1}{2 \pi \sqrt{(21^2 \times 14^2) \times 10^{-18}}}$
$f = \frac{1}{2 \pi \times 21 \times 14 \times 10^{-9}}$
$f = \frac{1}{2 \times 3.14159 \times 294 \times 10^{-9}}$
$f \approx \frac{1}{1847.25 \times 10^{-9}} \approx 0.5413 \times 10^6 \text{ Hz} = 5.41 \times 10^5 \text{ Hz}$.

(A) Given: Capacitance,$C = 400 \ pF = 400 \times 10^{-12} \ F$.
Inductance,$L = 400 \ \mu H = 400 \times 10^{-6} \ H$.
The frequency of the resonating $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
The wavelength $\lambda$ of the radiated electromagnetic wave is related to the frequency by $\lambda = \frac{c}{f}$,where $c = 3 \times 10^8 \ m/s$ is the speed of light.
Substituting the expression for $f$,we get $\lambda = c \times 2 \pi \sqrt{LC}$.
Substituting the values: $\lambda = 3 \times 10^8 \times 2 \times 3.14 \times \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}}$.
$\lambda = 3 \times 10^8 \times 6.28 \times \sqrt{160000 \times 10^{-18}}$.
$\lambda = 3 \times 10^8 \times 6.28 \times 400 \times 10^{-9}$.
$\lambda = 3 \times 6.28 \times 400 \times 10^{-1} = 753.6 \ m \approx 754 \ m$.

(B) In an $L-C$ circuit,the total energy is conserved and oscillates between the magnetic field of the inductor and the electric field of the capacitor.
The maximum energy stored in the capacitor is $U_{max} = \frac{1}{2} C V_{max}^2$.
The maximum energy stored in the inductor is $U_{max} = \frac{1}{2} L I_{max}^2$.
Equating these two,we get $\frac{1}{2} C V_{max}^2 = \frac{1}{2} L I_{max}^2$.
Therefore,$I_{max} = V_{max} \sqrt{\frac{C}{L}}$.
Given: $C = 20 \mu F = 20 \times 10^{-6} \ F$,$L = 80 \mu H = 80 \times 10^{-6} \ H$,and $V_{max} = 80 \ V$.
Substituting the values: $I_{max} = 80 \times \sqrt{\frac{20 \times 10^{-6}}{80 \times 10^{-6}}} = 80 \times \sqrt{\frac{1}{4}} = 80 \times \frac{1}{2} = 40 \ A$.

(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
When the capacitor is filled with a dielectric material of constant $K$,the new capacitance becomes $C' = KC$.
The new frequency is $f' = \frac{1}{2 \pi \sqrt{L(KC)}} = \frac{1}{\sqrt{K}} f$.
Given $f = 125 \ kHz$ and the frequency decreases by $25 \ kHz$,the new frequency is $f' = 125 \ kHz - 25 \ kHz = 100 \ kHz$.
Substituting these values: $\frac{100}{125} = \frac{1}{\sqrt{K}}$.
$\frac{4}{5} = \frac{1}{\sqrt{K}} \Rightarrow \sqrt{K} = \frac{5}{4} = 1.25$.
Therefore,$K = (1.25)^2 = 1.5625 \approx 1.56$.

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

(D) For a tuned circuit of a transistor oscillator unit,the given values are:
$L = 5 mH = 5 \times 10^{-3} H$
$C = 5 pF = 5 \times 10^{-12} F$
The natural frequency $f$ of the oscillator is given by the formula:
$f = \frac{1}{2 \pi \sqrt{LC}}$
Substituting the values:
$f = \frac{1}{2 \pi \sqrt{5 \times 10^{-3} \times 5 \times 10^{-12}}}$
$f = \frac{1}{2 \pi \sqrt{25 \times 10^{-15}}}$
$f = \frac{1}{2 \pi \times 5 \times 10^{-7.5}}$
Wait,correcting the calculation:
$f = \frac{1}{2 \pi \sqrt{25 \times 10^{-15}}} = \frac{1}{2 \pi \times 5 \times 10^{-7} \times \sqrt{10^{-1}}} = \frac{1}{10 \pi \times 10^{-7} \times 0.316}$
Actually,using $LC = 25 \times 10^{-15} = 2.5 \times 10^{-14}$:
$f = \frac{1}{2 \pi \sqrt{2.5 \times 10^{-14}}} = \frac{1}{2 \pi \times 1.581 \times 10^{-7}}$
$f = \frac{10^7}{9.93} \approx 1.006 \times 10^6 Hz = 1 MHz$.