Half Power Frequency , Quality Factor ,Resonance in AC Circuit Questions in English

(D) The resonance frequency $f$ of a series $LCR$ circuit is given by the formula: $f = \frac{1}{2 \pi \sqrt{LC}}$.
Initially,$f_1 = \frac{1}{2 \pi \sqrt{LC}}$.
Finally,the resonance frequency becomes $f_2 = 2f_1$ by changing $C$ to $C^{\prime}$.
Thus,$f_2 = \frac{1}{2 \pi \sqrt{LC^{\prime}}}$.
Substituting $f_2 = 2f_1$ into the equation:
$\frac{1}{2 \pi \sqrt{LC^{\prime}}} = 2 \times \frac{1}{2 \pi \sqrt{LC}}$.
Canceling common terms $2 \pi$ and $\sqrt{L}$ from both sides:
$\frac{1}{\sqrt{C^{\prime}}} = \frac{2}{\sqrt{C}}$.
Squaring both sides:
$\frac{1}{C^{\prime}} = \frac{4}{C}$.
Therefore,the ratio $\frac{C}{C^{\prime}} = 4$.
Note: The change in resistance $R$ does not affect the resonance frequency of an $LCR$ circuit.

(D) The resonant frequency of a series $LCR$ circuit is given by the formula: $f = \frac{1}{2\pi\sqrt{LC}}$.
From this formula,we can see that the resonant frequency $f$ is inversely proportional to the square root of the inductance $L$,provided the capacitance $C$ remains constant: $f \propto \frac{1}{\sqrt{L}}$.
Let the initial resonant frequency be $f_0$ with inductance $L$. Thus,$f_0 = \frac{1}{2\pi\sqrt{LC}}$.
When the inductance is increased by $\sqrt{3}$ times,the new inductance $L' = \sqrt{3}L$. The resistance $R$ does not affect the resonant frequency.
The new resonant frequency $f'$ is given by: $f' = \frac{1}{2\pi\sqrt{L'C}} = \frac{1}{2\pi\sqrt{(\sqrt{3}L)C}}$.
Dividing $f'$ by $f_0$: $\frac{f'}{f_0} = \frac{\frac{1}{2\pi\sqrt{\sqrt{3}LC}}}{\frac{1}{2\pi\sqrt{LC}}} = \frac{1}{\sqrt{\sqrt{3}}} = \left(\frac{1}{3^{1/2}}\right)^{1/2} = \left(\frac{1}{3}\right)^{1/4}$.
Therefore,the new resonant frequency is $f' = \left(\frac{1}{3}\right)^{1/4} f_0$.

(D) The quality factor $Q$ of an $R-L-C$ series circuit is defined as the ratio of the voltage drop across the inductor (or capacitor) to the voltage drop across the resistor at resonance.
$Q = \frac{V_L}{V_R} = \frac{I X_L}{I R} = \frac{\omega_0 L}{R}$.
At resonance,the resonant frequency is given by $\omega_0 = \frac{1}{\sqrt{LC}}$.
Substituting $\sqrt{LC} = \frac{1}{\omega_0}$,we can also write $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.
Comparing this with the given options,the expression for the quality factor is $\frac{\omega_0 L}{R}$.

(C) In a series $R-L-C$ circuit,the resonance frequency is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.
Given: $R = 100 \Omega$,$L = 20 \text{ mH} = 20 \times 10^{-3} \text{ H}$,$f_0 = \frac{1250}{\pi} \text{ Hz}$.
Substituting the values into the resonance formula:
$\frac{1250}{\pi} = \frac{1}{2\pi\sqrt{20 \times 10^{-3} \times C}}$
$1250 = \frac{1}{2\sqrt{0.02 \times C}}$
$2500 = \frac{1}{\sqrt{0.02 \times C}}$
Squaring both sides:
$6.25 \times 10^6 = \frac{1}{0.02 \times C}$
$C = \frac{1}{0.02 \times 6.25 \times 10^6} = \frac{1}{0.125 \times 10^6} = 8 \times 10^{-6} \text{ F} = 8 \mu\text{F}$.
When charged by a $DC$ source of $V = 25 \text{ V}$,the charge $Q$ stored is:
$Q = C \times V = 8 \times 10^{-6} \text{ F} \times 25 \text{ V} = 200 \times 10^{-6} \text{ C} = 0.2 \times 10^{-3} \text{ C} = 0.2 \text{ mC}$.

(D) In a series $L-C-R$ circuit,the current is the same at two different frequencies $f_1$ and $f_2$ if these frequencies are equidistant from the resonance frequency $f_0$ in a geometric sense.
The resonance frequency $f_0$ is given by the geometric mean of the two frequencies at which the current is equal:
$f_0 = \sqrt{f_1 \times f_2}$
Given:
$f_1 = 200 \ Hz$
$f_2 = 800 \ Hz$
Substituting the values:
$f_0 = \sqrt{200 \times 800}$
$f_0 = \sqrt{160000}$
$f_0 = 400 \ Hz$
Thus,the resonance frequency of the circuit is $400 \ Hz$.

(C) The impedance $Z$ of a series $LCR$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
At very low frequencies $(\omega \to 0)$,the capacitive reactance $X_C = \frac{1}{\omega C}$ is very large,making the impedance $Z$ very high.
As the frequency $\omega$ increases,the term $(\omega L - \frac{1}{\omega C})^2$ decreases until it reaches zero at the resonance frequency $\omega_0 = \frac{1}{\sqrt{LC}}$. At this point,$Z = R$,which is the minimum value.
As the frequency increases further beyond $\omega_0$,the inductive reactance $X_L = \omega L$ becomes dominant,and the term $(\omega L - \frac{1}{\omega C})^2$ increases again,causing the impedance $Z$ to increase.
Therefore,the impedance first decreases and then increases.

(B) In an $LCR$ circuit,the current is maximum at resonance.
At resonance,the angular frequency $\omega$ is given by $\omega = \frac{1}{\sqrt{LC}}$.
From the given voltage equation $V = 5 \sin(100t)$,we have $\omega = 100 \text{ rad/s}$.
Given $L = 2 \text{ H}$,we can substitute these values into the resonance formula:
$100 = \frac{1}{\sqrt{2 \times C}}$
Squaring both sides:
$10000 = \frac{1}{2C}$
$C = \frac{1}{2 \times 10000} = \frac{1}{20000} \text{ F}$
$C = 0.5 \times 10^{-4} \text{ F} = 50 \times 10^{-6} \text{ F} = 50 \mu\text{F}$.

(A) In an $LCR$ series circuit,resonance occurs when the inductive reactance $X_L$ equals the capacitive reactance $X_C$ $(X_L = X_C)$.
At this condition,the total impedance $Z$ is equal to the resistance $R$,meaning the circuit behaves purely resistively.
The power factor is given by $\cos \phi = \frac{R}{Z}$.
At resonance,$Z = R$,so $\cos \phi = \frac{R}{R} = 1$.

(D) At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$.
The resonant angular frequency is given by $\omega_0 = \frac{1}{\sqrt{LC}}$.
The inductive reactance at resonance is $X_L = \omega_0 L = \frac{1}{\sqrt{LC}} \times L = \sqrt{\frac{L}{C}}$.
Given: $L = 1.6 \ \text{H}$ and $C = 40 \ \mu\text{F} = 40 \times 10^{-6} \ \text{F}$.
Substituting these values into the formula:
$X_L = \sqrt{\frac{1.6}{40 \times 10^{-6}}} = \sqrt{\frac{1.6 \times 10^6}{40}} = \sqrt{0.04 \times 10^6} = \sqrt{40000} = 200 \ \Omega$.
Therefore,the inductive reactance at resonant frequency is $200 \ \Omega$.

(A) The resonant frequency $f_r$ of an $LCR$ series circuit is given by the formula: $f_r = \frac{1}{2 \pi \sqrt{LC}}$.
Given values are $L = 1 \text{ mH} = 10^{-3} \text{ H}$ and $C = 0.1 \mu\text{F} = 0.1 \times 10^{-6} \text{ F} = 10^{-7} \text{ F}$.
Substituting these values into the formula:
$\sqrt{LC} = \sqrt{10^{-3} \times 10^{-7}} = \sqrt{10^{-10}} = 10^{-5}$.
Now,$f_r = \frac{1}{2 \pi \times 10^{-5}} = \frac{10^5}{2 \times 3.14159} \approx \frac{100000}{6.283} \approx 15915.5 \text{ Hz}$.
Converting to kHz,$f_r \approx 15.9 \text{ kHz}$.
Therefore,the correct option is $A$.