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 $A$.

(A) The given voltage is $V = V_m \sin(\omega t)$, where $V_m = 220 \text{ V}$ and $\omega = 2 \times 10^3 \text{ rad/s}$.
First, calculate the inductive reactance $X_L = \omega L = (2 \times 10^3) \times (10 \times 10^{-3}) = 20 \Omega$.
Next, calculate the capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{(2 \times 10^3) \times (25 \times 10^{-6})} = \frac{1}{0.05} = 20 \Omega$.
Since $X_L = X_C$, the circuit is in resonance.
At resonance, the impedance $Z = R = 100 \Omega$.
The current amplitude $I_m = \frac{V_m}{Z} = \frac{220}{100} = 2.2 \text{ A}$.

(D) In an $LCR$ series circuit, the current and voltage are in phase when the circuit is in resonance.
At resonance, the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$.
$X_L = X_C$
$2\pi f L = \frac{1}{2\pi f C}$
Given: $R = 100 \text{ } \Omega$, $L = (4/\pi^2) \text{ H}$, $f = 50 \text{ Hz}$.
Substituting the values into the resonance condition:
$2 \pi (50) \times (4/\pi^2) = \frac{1}{2 \pi (50) \times C}$
$100 \pi \times (4/\pi^2) = \frac{1}{100 \pi C}$
$400/\pi = \frac{1}{100 \pi C}$
$C = \frac{1}{100 \pi \times (400/\pi)}$
$C = \frac{1}{40000} \text{ F}$
$C = 0.25 \times 10^{-4} \text{ F} = 25 \times 10^{-6} \text{ F} = 25 \text{ } \mu\text{F}$.

(B) At resonance, the current $I$ in the circuit is given by $I = V_R / R$. Given $V_R = 200 \text{ V}$ and $R = 800 \text{ } \Omega$, we have $I = 200 / 800 = 0.25 \text{ A}$.
At resonance, the capacitive reactance $X_C$ is given by $X_C = 1 / (\omega C)$.
Given $\omega = 250 \text{ rad/s}$ and $C = 2 \times 10^{-6} \text{ F}$, we have $X_C = 1 / (250 \times 2 \times 10^{-6}) = 1 / (500 \times 10^{-6}) = 10^6 / 500 = 2000 \text{ } \Omega$.
The voltage across the capacitor $V_C$ is given by $V_C = I \times X_C$.
Substituting the values, $V_C = 0.25 \times 2000 = 500 \text{ V}$.

(A) In a series $LCR$ circuit, maximum current flows at resonance, where the inductive reactance equals the capacitive reactance $(X_L = X_C)$.
At resonance, the potential difference across the inductor $(V_L)$ and the capacitor $(V_C)$ are equal in magnitude but opposite in phase, so $V_L = V_C = I_{max} X_L = I_{max} X_C$.
The energy stored in the capacitor is $U_C = \frac{1}{2} C V_C^2$ and the energy stored in the inductor is $U_L = \frac{1}{2} L I_{max}^2$.
Since $V_C = I_{max} X_C$, we have $U_C = \frac{1}{2} C (I_{max} X_C)^2 = \frac{1}{2} C I_{max}^2 (\frac{1}{\omega C})^2 = \frac{1}{2} I_{max}^2 (\frac{1}{\omega^2 C})$.
Using $\omega^2 = \frac{1}{LC}$, we get $U_C = \frac{1}{2} I_{max}^2 (LC) = \frac{1}{2} L I_{max}^2$.
Thus, $U_C = U_L$, which means the ratio $U_C : U_L = 1 : 1$. However, checking the provided options, there seems to be a discrepancy. Re-evaluating the question context: if the question implies a specific frequency or state, but standard resonance is assumed, the ratio is $1:1$. Given the options provided, if we calculate $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$, at resonance $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-3} \times 2 \times 10^{-6}}} = \frac{1}{\sqrt{2 \times 10^{-9}}} \approx 22360 \text{ rad/s}$. The ratio of energies $U_C/U_L = (\frac{1}{2} C V_C^2) / (\frac{1}{2} L I^2) = (\frac{1}{2} C (I X_C)^2) / (\frac{1}{2} L I^2) = (C X_C^2) / L = C (\frac{1}{\omega C})^2 / L = \frac{1}{\omega^2 L C}$. Since $\omega^2 = \frac{1}{LC}$, the ratio is $1$. None of the options match $1:1$. Assuming a typo in the question's premise, if the ratio was intended to be calculated differently, based on standard physics, the answer is $1:1$.

(C) In an $LCR$ series circuit, the current is in phase with the emf when the circuit is in resonance.
At resonance, the inductive reactance $X_L$ equals the capacitive reactance $X_C$, and the impedance $Z$ of the circuit is equal to the resistance $R$.
Given: $R = 20 \text{ } \Omega$, $V_{rms} = 220 \text{ V}$.
The peak voltage (amplitude of emf) is $V_0 = V_{rms} \sqrt{2} = 220 \sqrt{2} \text{ V}$.
At resonance, the impedance $Z = R = 20 \text{ } \Omega$.
The amplitude of the current $I_0$ is given by $I_0 = \frac{V_0}{Z} = \frac{220 \sqrt{2}}{20} = 11 \sqrt{2} \text{ A}$.
We can write $I_0 = \sqrt{11^2 \times 2} = \sqrt{121 \times 2} = \sqrt{242} \text{ A}$.
Comparing this with the given form $[x]^{1/2} \text{ A}$, we get $x = 242$.

(A) The quality factor $(Q)$ for a series $LCR$ circuit is given by $Q = \frac{\omega_0 L}{R}$, where $\omega_0 = \frac{1}{\sqrt{LC}}$ is the resonant angular frequency.
Substituting $\omega_0$, we get $Q = \frac{1}{\sqrt{LC}} \cdot \frac{L}{R}$.
The band width $(BW)$ is given as $\frac{R}{L}$.
We need to find the ratio of the quality factor to the band width:
Ratio $= \frac{Q}{BW} = \frac{\frac{1}{\sqrt{LC}} \cdot \frac{L}{R}}{\frac{R}{L}} = \frac{1}{\sqrt{LC}} \cdot \frac{L}{R} \cdot \frac{L}{R} = \sqrt{\frac{1}{LC}} \cdot \frac{L^2}{R^2}$.
Thus, the correct option is $A$.

(C) The resonant frequency $f$ of a series $LCR$ circuit is given by the formula: $f = \frac{1}{2\pi\sqrt{LC}}$.
Given the initial frequency $f = \frac{1}{2\pi\sqrt{LC}}$.
The new inductance is $L' = L + 2L = 3L$.
The new capacitance is $C' = 9C$.
The new resonant frequency $f'$ is given by: $f' = \frac{1}{2\pi\sqrt{L'C'}} = \frac{1}{2\pi\sqrt{(3L)(9C)}} = \frac{1}{2\pi\sqrt{27LC}}$.
Simplifying the expression: $f' = \frac{1}{2\pi\sqrt{27}\sqrt{LC}} = \frac{1}{2\pi(3\sqrt{3})\sqrt{LC}}$.
Since $f = \frac{1}{2\pi\sqrt{LC}}$, we substitute this into the equation for $f'$:
$f' = \frac{f}{3\sqrt{3}}$.
Therefore, the new resonant frequency is $f/(3\sqrt{3})$.

(D) In an $LCR$ series circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$, i.e.,$X_L = X_C$.
Therefore, the impedance $Z$ becomes equal to the resistance $R$, which is the minimum possible value of impedance.
Since $Z$ is minimum, the current $I = \frac{V}{Z}$ is maximum.
At resonance, the phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R} = 0$, which means $\phi = 0$.
Thus, the current and voltage are in the same phase.

(A) परिपथ में अधिकतम धारा तब प्रवाहित होती है जब परिपथ अनुनाद (resonance) की स्थिति में होता है।
अनुनाद की स्थिति में, प्रेरणिक प्रतिघात $(X_L)$ धारितीय प्रतिघात $(X_C)$ के बराबर होता है।
$X_L = X_C \implies \omega L = \frac{1}{\omega C}$
यहाँ, $\omega = 2\pi f$ है।
अतः, $C = \frac{1}{\omega^2 L} = \frac{1}{(2\pi f)^2 L} = \frac{1}{4\pi^2 f^2 L}$
दिया गया है: $L = 2 \times 10^{-6} \text{ H}$, $f = 10 \text{ kHz} = 10^4 \text{ Hz}$, $\pi^2 = 10$
$C = \frac{1}{4 \times 10 \times (10^4)^2 \times 2 \times 10^{-6}}$
$C = \frac{1}{40 \times 10^8 \times 2 \times 10^{-6}} = \frac{1}{80 \times 10^2} = \frac{1}{8000}$
अतः, $x = 8000$।

(D) In an $LCR$ series circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance, the inductive reactance $X_L$ equals the capacitive reactance $X_C$, i.e.,$X_L = X_C$.
Therefore, the impedance $Z$ becomes $Z = \sqrt{R^2 + 0} = R$, which is the minimum possible value of impedance.
Since $Z$ is minimum, the current $I = \frac{V}{Z}$ is maximum.
Also, at resonance, the phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R} = 0$, which implies $\phi = 0$.
Since the phase angle is $0$, the current and voltage are in the same phase.

(A) At the resonant angular frequency $\omega_0$, the capacitive reactance $X_C$ and inductive reactance $X_L$ are equal, so $X_C = X_L$.
Given that $X_C = 1/(\omega_0 C)$ and $X_L = \omega_0 L$, at $\omega_0$ we have $1/(\omega_0 C) = \omega_0 L$.
When the angular frequency is doubled to $\omega' = 2\omega_0$, the new capacitive reactance is $X_C' = 1/(2\omega_0 C) = X_C / 2$.
The new inductive reactance is $X_L' = (2\omega_0) L = 2 X_L$.
Since $X_C = X_L$, the ratio of the new capacitive reactance to the new inductive reactance is $X_C' / X_L' = (X_C / 2) / (2 X_L) = 1/4$.

(A) In a series $LCR$ circuit, the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance, $X_L = X_C$, so $Z = R$ (minimum). Since $I = V/Z$, the current $I$ is maximum.
In a parallel $LCR$ circuit, the admittance $Y = \sqrt{(1/R)^2 + (1/X_C - 1/X_L)^2}$. At resonance, $1/X_C = 1/X_L$, so $Y = 1/R$ (minimum), which means impedance $Z = 1/Y = R$ is maximum. Since $I = V/Z$, the current $I$ is minimum.

(B) In a series $LCR$ circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$, where $X_L = 2\pi fL$ and $X_C = \frac{1}{2\pi fC}$.
The r.m.s. current is given by $i = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (2\pi fL - \frac{1}{2\pi fC})^2}}$.
At resonance frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$, the impedance $Z$ is minimum $(Z = R)$, and therefore the current $i$ is maximum.
As the frequency $f$ moves away from $f_0$ (either increasing or decreasing), the impedance $Z$ increases, causing the current $i$ to decrease.
This variation of current $i$ with frequency $f$ is represented by a bell-shaped curve, which corresponds to graph $(P)$.

(C) In a series $LCR$ circuit, the current $I$ is given by $I = V_R / R$. Given $V_R = 100 \text{ V}$ and $R = 1000 \Omega$, we have $I = 100 / 1000 = 0.1 \text{ A}$.
At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$.
The capacitive reactance is $X_C = 1 / (\omega C) = 1 / (200 \times 2 \times 10^{-6}) = 1 / (400 \times 10^{-6}) = 10^6 / 400 = 2500 \Omega$.
Since the circuit is at resonance, $X_L = X_C = 2500 \Omega$.
The voltage across the inductor $L$ is $V_L = I \times X_L = 0.1 \text{ A} \times 2500 \Omega = 250 \text{ V}$.