LC Oscillations Questions in English

(B) The natural frequency of an $LC$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
Given,initial frequency $f = 120 \ kHz$.
When a dielectric material of constant $K$ is introduced,the capacitance becomes $C' = KC$.
The new frequency is $f' = f - 20 \ kHz = 120 - 20 = 100 \ kHz$.
The new frequency is $f' = \frac{1}{2 \pi \sqrt{L(KC)}} = \frac{1}{\sqrt{K}} \times \frac{1}{2 \pi \sqrt{LC}} = \frac{f}{\sqrt{K}}$.
Therefore,$\frac{f}{f'} = \sqrt{K}$.
Substituting the values: $\frac{120}{100} = \sqrt{K} \Rightarrow 1.2 = \sqrt{K}$.
Squaring both sides: $K = (1.2)^2 = 1.44$.

(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
This implies $f \propto \frac{1}{\sqrt{C}}$.
When the capacitor $C$ is replaced by a capacitor $C'$ with a dielectric medium of constant $K$,the new capacitance becomes $C' = KC$.
The new frequency $f'$ is given as $f' = f - 25 \text{ kHz} = 125 \text{ kHz} - 25 \text{ kHz} = 100 \text{ kHz}$.
Taking the ratio of the frequencies:
$\frac{f'}{f} = \sqrt{\frac{C}{C'}} = \sqrt{\frac{C}{KC}} = \frac{1}{\sqrt{K}}$.
Substituting the values:
$\frac{100}{125} = \frac{1}{\sqrt{K}}$.
$\frac{4}{5} = \frac{1}{\sqrt{K}} \implies \sqrt{K} = \frac{5}{4} = 1.25$.
Therefore,$K = (1.25)^2 = 1.5625 \approx 1.56$.

(B) $1$. Initially,the switch $S_1$ is closed and $S_2$ is open. The capacitor $C$ charges to the potential $V$ of the battery.
$2$. The energy stored in the capacitor is $U_C = \frac{1}{2} C V^2$.
$3$. When $S_1$ is opened and $S_2$ is closed,the capacitor discharges through the inductor $L$,forming an $LC$ oscillation circuit.
$4$. According to the law of conservation of energy,the maximum energy stored in the capacitor is transferred to the inductor as magnetic energy when the current is maximum $(i_{max})$.
$5$. $\frac{1}{2} C V^2 = \frac{1}{2} L i_{max}^2$
$6$. Solving for $i_{max}$,we get $i_{max}^2 = \frac{C V^2}{L}$,which implies $i_{max} = V \sqrt{\frac{C}{L}}$.

(B) The capacitor is charged to a potential difference of $V = 50 \ V$. When the battery is removed and the capacitor is connected to the inductor,the circuit forms an $LC$ oscillator.
At $t=0$,the charge on the capacitor is maximum,so the potential difference across it is $V_{max} = 50 \ V$.
The voltage across the inductor is given by $V_L = L \frac{dI}{dt}$.
In an $LC$ circuit,the sum of potential differences is zero: $V_C + V_L = 0$,which implies $|V_L| = |V_C|$.
At $t=0$,the current $I=0$,so the entire potential of the capacitor appears across the inductor.
Thus,$L \left( \frac{dI}{dt} \right)_{max} = V_{max}$.
Given $L = 10 \ mH = 10 \times 10^{-3} \ H$ and $V_{max} = 50 \ V$.
$\left( \frac{dI}{dt} \right)_{max} = \frac{50}{10 \times 10^{-3}} = \frac{50}{0.01} = 5000 \ A s^{-1}$.

(A) The resonant frequency of an $LC$ circuit is given by the formula $\omega = \frac{1}{\sqrt{LC}}$.
Here,$\omega$ represents the angular frequency,which has the dimensions of $[T^{-1}]$.
Therefore,$(LC)^{-\frac{1}{2}} = \frac{1}{\sqrt{LC}} = \omega$.
The dimensional formula for angular frequency is $[M^0 L^0 T^{-1}]$.
Thus,the correct option is $A$.

(D) When the switch is in position $a$,the capacitor is charged to a potential $E$. The charge on the capacitor is $q = CE$.
The energy stored in the capacitor is $U = \frac{q^2}{2C} = \frac{(CE)^2}{2C} = \frac{1}{2} CE^2$.
When the switch is thrown to position $b$,the circuit becomes an $LC$ oscillator. The total energy is conserved,oscillating between the electric field of the capacitor and the magnetic field of the inductor.
Let $I_0$ be the amplitude of the oscillating current. The maximum magnetic energy in the inductor is $\frac{1}{2} L I_0^2$.
By the law of conservation of energy,the maximum electrical energy equals the maximum magnetic energy:
$\frac{1}{2} CE^2 = \frac{1}{2} L I_0^2$
$CE^2 = L I_0^2$
$I_0^2 = \frac{C}{L} E^2$
$I_0 = E \sqrt{\frac{C}{L}}$

(B) The angular frequency $\omega$ of free oscillations in an $LC$ circuit is given by the formula $\omega = \frac{1}{\sqrt{LC}}$.
Given values are $L = 27 \text{ mH} = 27 \times 10^{-3} \text{ H}$ and $C = 30 \mu\text{F} = 30 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$\omega = \frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}}$
$\omega = \frac{1}{\sqrt{810 \times 10^{-9}}}$
$\omega = \frac{1}{\sqrt{81 \times 10^{-8}}}$
$\omega = \frac{1}{9 \times 10^{-4}}$
$\omega = \frac{10^4}{9} \approx 1111.11 \text{ rad/s}$.
The closest option is $1100 \text{ rad/s}$.