LC Oscillations MCQ Quiz in मल्याळम - Objective Question with Answer for LC Oscillations - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• Resonance Frequency of an LC Oscillator:
• The resonance frequency (ω) of an LC oscillator is given by the formula:
  • ω = 1 / √(LC), where:
  • L: Inductance (in henry, H)
  • C: Capacitance (in farads, F)
• If the values of inductance and capacitance change, the resonance frequency also changes according to the above formula.
• In this problem, the inductance is doubled, and the capacitance is increased by a factor of eight.

Calculation:

Given:

Initial inductance = L

Initial capacitance = C

The resonance frequency is:

ω₀ = 1 / √(LC)

After the changes:

New inductance = 2L

New capacitance = 8C

The new resonance frequency is:

ω = 1 / √((2L)(8C)) = 1 / √(16LC) = (1 / 4) × (1 / √(LC))

Thus, the new resonance frequency is 1/4 of the initial resonance frequency.

∴ The value of x is 1/4.

Concept:

In an L-C circuit, the current drawn by the inductor and capacitor are out of phase. The net current drawn from the generator is the difference between the current drawn by the inductor and the current drawn by the capacitor. The net current can be expressed as:

I = Iₗ - I_ᶜ

Where:

• Iₗ = Current through the inductor (0.9 A)
• I_ᶜ = Current through the capacitor (0.4 A)

Calculation:

Given:

• Iₗ = 0.9 A
• I_ᶜ = 0.4 A

The net current drawn by the generator is:

I = Iₗ - I_ᶜ = 0.9 A - 0.4 A = 0.5 A

∴ The current drawn from the source is 0.5 A.

Hence, the correct answer is option 3.

Explanation:

The time period of oscillation in an LC circuit, which contains only an inductor (L) and a capacitor (C), is derived from the second-order differential equation:

$L\frac{d^{2}q}{d{t}^2}+\frac{q}{C}=0$

Solving this gives a solution for charge  $q(t)=q_0\cos (\omega t+\varphi )$ , where$\omega =\frac{1}{\sqrt{LC}}$ is the angular frequency. The time period T is related to angular frequency by $T=\frac{2\pi }{\omega }$ , which simplifies to:

$T=2\pi \sqrt{LC}$

Thus, option '4' is correct.

$f={\displaystyle \frac{1}{2\pi \sqrt{LC}}}$

At resonance the impedance is maximum and frequency and current will be minimum.

In this condition at every instant potential difference across the Power supply will be the same as Potential difference across the inductor and capacitor.

In such case, there will no current flow through the $A_3$

Explanation:

1. Capacitance:
   
 $c=4\mu F$
   

2. Resonance Frequency:
   The resonance frequency $\omega$ is given by:

   
  $\omega =\frac{1}{\sqrt{LC}}⟹f_R=\frac{1}{2\pi \sqrt{LC}}$

   Thus, 
   $f_R=\frac{1}{2\pi \sqrt{25\times {10}^{-3}\times 4\times {10}^{-6}}}=503Hz$

   Therefore, Option 1 is correct.

   Since $f_R=503Hz$:
    For $f>f_R$, current lags the voltage.
    For $f<f_R$ , current leads the voltage.

    Option 3 is correct because at 200 Hz, voltage lags the current. However, option D is incorrect because at 700 Hz, current lags the voltage.

Impedance Calculation:

3. Impedance z:
   The impedance is given by:   
  $$\begin{gathered} z=j(\omega L-\frac{1}{\omega c})\text{and}\omega =2\pi f=2000\pi \\ z=j(2000\pi \times 25\times {10}^{-3}-\frac{1}{2000\pi \times 4\times {10}^{-6}}) \\ z=j\left(117\right) \end{gathered}$$
   The correct option are (1) and (3).

Concept:

RL Circuit with Inductor and Resistor:

• In an RL circuit, the resistor (R) and inductor (L) are connected in series, and when the switch is connected to a power source, current flows through the circuit.
• Initially, the current grows exponentially through the circuit until it reaches a constant value when the inductor becomes fully energized, behaving like a short circuit. The current then becomes $I_0=\frac{V}{R}$, where (V) is the supply voltage.
• When the switch is disconnected from the power source and connected to another path (here, from (A) to (B), the current begins to decay exponentially due to the inductor's energy dissipation. The current at any time (t) is given by:
  • $I(t)=I_0{e}^{-\frac{t}{\tau }}$, where $\tau =\frac{L}{R}$ is the time constant of the circuit.
• The voltage across the resistor at any time is given by Ohm’s law:
  • $V_R=I(t)R$
• The voltage across the inductor is given by:
  • $V_L=L\frac{dI}{dt}$
• We are asked to find the ratio of the voltage across the resistor to the voltage across the inductor at a specific time $t=\frac{L}{R}$

Calculation:

Given,

• Time constant,$\tau =\frac{L}{R}$
• At $t=\frac{L}{R},t=\tau$

⇒ $I\left(\frac{L}{R}\right)=I_0{e}^{-1}=\frac{I_0}{e}$

⇒ $V_R=I(t)R=\frac{I_0}{e}R$

⇒ $V_L=L\frac{dI}{dt}=-L\times \frac{I_{0}R}{L}{e}^{-1}=-\frac{I_{0}R}{e}$

⇒$\frac{V_R}{V_L}=\frac{\frac{I_0}{e}R}{-\frac{I_0}{e}R}=-1$

∴ The correct option is 1

Concept:

Maximum Charge and Current in an LC Circuit:
In an LC circuit, the maximum current I_max
​occurs when the charge on the capacitor is at its maximum, and the energy stored in the capacitor is completely transferred to the inductor as magnetic energy.

The total energy in the circuit is conserved and can be expressed as:

$\frac{1}{2}{Li}_{max}^2=\frac{1}{2}\frac{Q_{max}^2}{C}$

Calculation:

Maximum energy stored in capacitor = Maximum energy stored in inductor which is given by :

$\frac{1}{2}{Li}_{max}^2=\frac{1}{2}\frac{Q_{max}^2}{C}$

⇒ $i_{max}^2=\frac{Q_{max}^2}{\mathrm{LC}}$

⇒ $i_{max}=\sqrt{\frac{Q_{max}^2}{LC}}=\frac{Q_{max}}{\sqrt{\mathrm{LC}}}$

= $\frac{2.7\times {10}^{-6}}{\sqrt{75\times {10}^{-3}\times 1.2\times {10}^{-6}}}$ = 9 mA

∴ The correct answer is (9).

CONCEPT:

• Resonance frequency: The frequency of an AC circuit at which the impedance of the circuit becomes minimum or current in the circuit becomes maximum is called resonance frequency.

• The resonance frequency is given by:

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

Where L is the inductance and C is the capacitance of the circuit.

CALCULATION:

Given - C = 10-6F and L = 10-4H

• The resonance frequency is given by

$\Rightarrow f=\frac{1}{2\pi \sqrt{LC}}$

$\Rightarrow f=\frac{1}{2\pi \sqrt{{10}^{-6}\times {10}^{-4}}}=\frac{{10}^5}{2\pi }Hz$

CONCEPT:

LC Oscillations:

• We know that a capacitor and an inductor can store electrical and magnetic energy, respectively.
• When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems.
• Let a capacitor and an inductor are connected as shown in the figure.
• Let a capacitor be charged Qo at t = 0 sec.
• The moment the circuit is completed, the charge on the capacitor starts decreasing, giving rise to a current in the circuit.
• The angular frequency of the oscillation is given as,

$\Rightarrow {\omega }_o=\frac{1}{\sqrt{LC}}$

Where L = self-inductance and C = capacitance

• The charge on the capacitor varies sinusoidally with time as,

⇒ Q = Qocos(ωot)

• The current in the circuit at any time t is given as,

⇒ I = Iosin(ωot)

Where Io = maximum current in the circuit

• The relation between the maximum charge and the maximum current is given as,

⇒ Io = ωoQo

CALCULATION:

Given C = 40 μF = 40×10^-6 F and L = 100 mH = 10^-1 H

Where L = self-inductance and C = capacitance

We know that the angular frequency of the LC oscillation is given as,

$\Rightarrow {\omega }_o=\frac{1}{\sqrt{LC}}$

$\Rightarrow {\omega }_o=\frac{1}{\sqrt{{10}^{-1}\times 40\times {10}^{-6}}}$

⇒ ω_o = 0.5 × 10³ rad/sec

⇒ ωo = 0.5 × 103 rad/sec

⇒ ωo = 500 rad/sec

• Hence, option 2 is correct.