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

(A) In an $L-C$ circuit,resonance occurs when the inductive reactance $(X_L)$ equals the capacitive reactance $(X_C)$.
$X_L = X_C$
$\omega L = \frac{1}{\omega C}$
$\omega^2 = \frac{1}{LC}$
$\omega = \frac{1}{\sqrt{LC}}$
Since $\omega = 2 \pi f$,the resonant frequency $f$ is given by:
$f = \frac{1}{2 \pi \sqrt{LC}}$

(C) In an $L-C-R$ series circuit,resonance occurs when the inductive reactance $(X_L)$ equals the capacitive reactance $(X_C)$,i.e.,$X_L = X_C$.
The impedance of the circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance,$Z = \sqrt{R^2 + 0} = R$.
The power factor $(\cos \phi)$ is defined as the ratio of resistance to impedance: $\cos \phi = \frac{R}{Z}$.
Substituting $Z = R$,we get $\cos \phi = \frac{R}{R} = 1$.
Therefore,the power factor at resonance is unity.

(A) The current in an $LCR$ series circuit is given by $i = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$,where $V$ is the $rms$ voltage,$R$ is the resistance,$X_L = \omega L$ is the inductive reactance,and $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
For the current to be maximum,the impedance must be minimum,which occurs when $X_L = X_C$.
This condition is known as resonance,where $\omega L = \frac{1}{\omega C}$.
Rearranging for $L$,we get $L = \frac{1}{\omega^2 C}$.
Given $\omega = 1000 \ s^{-1}$ and $C = 10 \ \mu F = 10 \times 10^{-6} \ F$.
Substituting these values: $L = \frac{1}{(1000)^2 \times 10 \times 10^{-6}} = \frac{1}{10^6 \times 10^{-5}} = \frac{1}{10} = 0.1 \ H$.
Converting to millihenry: $0.1 \ H = 100 \ mH$.

(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 net reactance $X = X_L - X_C = 0$.
The phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R} = \frac{0}{R} = 0$.
Thus,$\phi = 0$,which means the current and voltage are in the same phase at resonance.

(D) When the current is in phase with the voltage,the circuit is in resonance. In this condition,the inductive reactance equals the capacitive reactance,and the impedance $Z$ of the circuit is equal to the resistance $R$.
Given $R = 20 \ \Omega$.
Therefore,$Z = R = 20 \ \Omega$.
The given voltage $V_{rms} = 220 \ V$. The peak voltage $e_0$ is given by $e_0 = V_{rms} \sqrt{2} = 220 \sqrt{2} \ V$.
The maximum current $i_0$ is given by $i_0 = \frac{e_0}{Z} = \frac{220 \sqrt{2}}{20} = 11 \sqrt{2} \ A$.
We can write $11 \sqrt{2} = \sqrt{11^2 \times 2} = \sqrt{121 \times 2} = \sqrt{242} \ A$.
Comparing this with $\sqrt{x} \ A$,we get $x = 242$.

(A) In a parallel resonant circuit,the capacitive reactance $(X_{C})$ and inductive reactance $(X_{L})$ are equal $(X_{L} = X_{C})$.
Since the inductor and capacitor are in parallel,the voltage across them is the same.
Therefore,the magnitudes of the currents are equal: $i_{L} = i_{C} = \frac{e}{X_{L}} = \frac{e}{X_{C}}$.
However,the current through the inductor lags the voltage by $90^{\circ}$,and the current through the capacitor leads the voltage by $90^{\circ}$.
Thus,the currents $i_{L}$ and $i_{C}$ are $180^{\circ}$ out of phase with each other.
The net current $i$ in the circuit is the phasor sum of $i_{L}$ and $i_{C}$,which is $i = |i_{L} - i_{C}| = 0$.
Hence,the condition is $i = 0$ and $i_{L} = i_{C} \neq 0$.

(D) In an $LC$ parallel resonant circuit,the impedance is maximum at the resonant frequency $f_r$,and the current is minimum at the resonant frequency $f_r$.
Graph $(4)$ shows that the current is minimum at the resonant frequency $f_r$,which is the correct characteristic for an $LC$ parallel resonant circuit.

(A) In an $LC$ parallel resonant circuit,at resonance,the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$.
This leads to a condition where the total impedance of the circuit becomes very high (theoretically infinite in an ideal circuit).
Because the impedance is very high,the current drawn from the source is minimal.
Therefore,the $LC$ parallel resonant circuit acts as a high-impedance circuit at resonance.

(D) In an $R-L-C$ series circuit,the phase difference $\phi$ is given by $\tan \phi = \frac{|X_L - X_C|}{R}$.
When $L$ is removed,the circuit becomes an $R-C$ circuit. The phase difference is $\tan \phi = \frac{X_C}{R} = \tan \frac{\pi}{3} = \sqrt{3}$. Thus,$X_C = \sqrt{3}R$.
When $C$ is removed,the circuit becomes an $R-L$ circuit. The phase difference is $\tan \phi = \frac{X_L}{R} = \tan \frac{\pi}{3} = \sqrt{3}$. Thus,$X_L = \sqrt{3}R$.
Since $X_L = X_C$,the circuit is in resonance.
At resonance,the impedance $Z = R$,and the phase difference $\phi = 0$.
The power factor is $\cos \phi = \cos 0 = 1$.

(A) Given: Resistance $R = 10 \ \Omega$,Inductance $L = 3 \text{ H}$,Capacitance $C = 27 \ \mu\text{F} = 27 \times 10^{-6} \text{ F}$.
Bandwidth $(\Delta \omega) = \frac{R}{L} = \frac{10}{3} \text{ rad/s}$.
Quality factor $(Q) = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{10} \sqrt{\frac{3}{27 \times 10^{-6}}} = \frac{1}{10} \sqrt{\frac{1}{9 \times 10^{-6}}} = \frac{1}{10} \times \frac{1}{3 \times 10^{-3}} = \frac{100}{3}$.
Ratio of Quality factor to Bandwidth = $\frac{Q}{\Delta \omega} = \frac{100/3}{10/3} = 10 \text{ s}$.

(B) The quality factor ($Q$-factor) of a resonant circuit is defined as the ratio of resonant frequency to the bandwidth of the circuit.
Mathematically,$Q = \frac{\omega_{0} L}{R}$.
At resonance,the angular frequency is given by $\omega_{0} = \frac{1}{\sqrt{LC}}$.
Substituting the value of $\omega_{0}$ into the $Q$-factor formula:
$Q = \frac{1}{\sqrt{LC}} \cdot \frac{L}{R} = \frac{1}{R} \cdot \frac{\sqrt{L}}{\sqrt{C}} = \frac{1}{R} \sqrt{\frac{L}{C}}$.

(B) Given: $L = \frac{3}{\pi^2} \ H$ and $f = 50 \ Hz$.
Since the voltage and current are in phase,the circuit is in resonance.
At resonance,the inductive reactance equals the capacitive reactance,so $X_C = X_L$.
$\frac{1}{\omega C} = \omega L \implies C = \frac{1}{\omega^2 L} = \frac{1}{(2\pi f)^2 L} = \frac{1}{4\pi^2 f^2 L}$.
Substituting the values:
$C = \frac{1}{4 \pi^2 \times (50)^2 \times \frac{3}{\pi^2}} = \frac{1}{4 \times 2500 \times 3} = \frac{1}{30000} \ F$.
$C = \frac{1}{3} \times 10^{-4} \ F \approx 0.33 \times 10^{-4} \ F$.

(B) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_C - X_L)^2}$.
At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,i.e.,$X_L = X_C$.
Substituting this into the impedance formula,we get $Z = \sqrt{R^2 + 0} = R$.
The power factor is defined as $\cos \phi = \frac{R}{Z}$.
Substituting $Z = R$,we get $\cos \phi = \frac{R}{R} = 1$.
Therefore,the power factor at resonance is $1$.

(D) The $Q$ factor of a series $LCR$ circuit is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values are $L = 2 \ H$,$C = 18 \ \mu F = 18 \times 10^{-6} \ F$,and $R = 10 \ \Omega$.
Substituting these values into the formula:
$Q = \frac{1}{10} \sqrt{\frac{2}{18 \times 10^{-6}}}$
$Q = \frac{1}{10} \sqrt{\frac{1}{9 \times 10^{-6}}}$
$Q = \frac{1}{10} \times \frac{1}{3 \times 10^{-3}}$
$Q = \frac{1}{10} \times \frac{1000}{3}$
$Q = \frac{100}{3} \approx 33.33$
Therefore,the correct option is $D$.

(C) The resonance frequency $v_{0}$ for an $LCR$ series circuit is given by the formula:
$v_{0} = \frac{1}{2 \pi \sqrt{LC}}$
Given values are $L = 25 \times 10^{-3} \, H$, $C = 62.5 \times 10^{-6} \, F$.
Substituting these values into the formula:
$v_{0} = \frac{1}{2 \times 3.14159 \times \sqrt{25 \times 10^{-3} \times 62.5 \times 10^{-6}}}$
$v_{0} = \frac{1}{2 \times 3.14159 \times \sqrt{1562.5 \times 10^{-9}}}$
$v_{0} = \frac{1}{2 \times 3.14159 \times \sqrt{1.5625 \times 10^{-6}}}$
$v_{0} = \frac{1}{2 \times 3.14159 \times 1.25 \times 10^{-3}}$
$v_{0} = \frac{1}{7.85398 \times 10^{-3}}$
$v_{0} \approx 127.39 \, Hz$
Thus, the correct option is $C$.

(C) At resonance condition,the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$.
Therefore,the net reactance $X = X_L - X_C = 0$.
The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the resonance condition $X_L = X_C$,we get $Z = \sqrt{R^2 + 0^2} = R$.
Given $R = 3 \ \Omega$,the impedance at resonance is $Z = 3 \ \Omega$.

(D) The $Q$-factor (Quality factor) of an $L-C-R$ series circuit is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values:
$L = 9 \ H$
$R = 10 \ \Omega$
$C = 100 \ \mu F = 100 \times 10^{-6} \ F = 10^{-4} \ F$
Substituting these values into the formula:
$Q = \frac{1}{10} \sqrt{\frac{9}{10^{-4}}}$
$Q = \frac{1}{10} \times \frac{3}{10^{-2}}$
$Q = \frac{1}{10} \times 300$
$Q = 30$
Therefore,the correct option is $D$.

(C) The $Q$-factor (Quality factor) of an $L-C-R$ circuit is defined as the ratio of the resonant frequency to the bandwidth.
$Q = \frac{f_0}{f_2 - f_1}$
Given:
Resonant frequency $f_0 = 5000 \ Hz$
Half-power frequencies $f_1 = 4950 \ Hz$ and $f_2 = 5050 \ Hz$
Bandwidth $\Delta f = f_2 - f_1 = 5050 \ Hz - 4950 \ Hz = 100 \ Hz$
Substituting the values:
$Q = \frac{5000}{100} = 50$
Therefore,the $Q$-factor is $50$.

(D) The $Q$ factor (Quality factor) of a series $LCR$ circuit is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values are $R = 6 \Omega$,$L = 1 \text{ H}$,and $C = 17.36 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$Q = \frac{1}{6} \sqrt{\frac{1}{17.36 \times 10^{-6}}}$
$Q = \frac{1}{6} \sqrt{\frac{10^6}{17.36}}$
$Q = \frac{1}{6} \times \sqrt{57603.68}$
$Q \approx \frac{1}{6} \times 240$
$Q \approx 40$
Therefore,the correct option is $D$.

(D) In a series $LCR$ circuit, the total applied voltage $V$ is given by the phasor sum of the individual voltages across the components:
$V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given values are $V_R = 5 \,V$, $V_L = 10 \,V$, and $V_C = 10 \,V$.
Substituting these values into the formula:
$V = \sqrt{5^2 + (10 - 10)^2}$
$V = \sqrt{25 + 0^2}$
$V = \sqrt{25}$
$V = 5 \,V$
Therefore, the applied $AC$ voltage is $5 \,V$.

(D) The resonance frequency of an $L-C-R$ circuit is given by the formula:
$\nu_{0} = \frac{1}{2 \pi \sqrt{LC}}$
From this formula,we can see that the resonance frequency is inversely proportional to the square root of the capacitance $C$:
$\nu_{0} \propto \frac{1}{\sqrt{C}}$
Let the initial frequency be $\nu_{0}$ with capacitance $C$,and the new frequency be $\nu_{0}'$ with capacitance $C' = 4C$.
Taking the ratio of the two frequencies:
$\frac{\nu_{0}}{\nu_{0}'} = \sqrt{\frac{C'}{C}}$
Substituting $C' = 4C$ into the equation:
$\frac{\nu_{0}}{\nu_{0}'} = \sqrt{\frac{4C}{C}} = \sqrt{4} = 2$
Therefore,the new resonance frequency is:
$\nu_{0}' = \frac{\nu_{0}}{2}$

(A) In an $L-C-R$ series $AC$ circuit,the impedance $Z$ is given by the formula: $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$,where $X_{L} = \omega L$ and $X_{C} = \frac{1}{\omega C}$.
At resonance,the inductive reactance equals the capacitive reactance,i.e.,$X_{L} = X_{C}$ or $\omega L = \frac{1}{\omega C}$.
Substituting this into the impedance formula: $Z = \sqrt{R^{2} + (0)^{2}} = \sqrt{R^{2}} = R$.
Therefore,at resonance,the impedance of the circuit is equal to the resistance $R$.

(C) The resonance frequency $f_0$ in an $L-C-R$ series circuit is given by the formula:
$f_0 = \frac{1}{2 \pi \sqrt{L C}}$
From this expression,we can see that the resonance frequency is inversely proportional to the square root of the capacitance:
$f_0 \propto \frac{1}{\sqrt{C}}$
As the value of capacitance $C$ increases,the resonance frequency $f_0$ decreases. This relationship represents a hyperbolic curve,which corresponds to the graph shown in option $C$.

(B) From the given graph,we can identify the resonant angular frequency $\omega_r$ and the half-power frequencies $\omega_1$ and $\omega_2$ where the impedance $Z = \sqrt{2} Z_{\text{min}}$.
$1$. The resonant frequency is $\omega_r = 500 \text{ rad/s}$.
$2$. The lower half-power frequency is $\omega_1 = 400 \text{ rad/s}$.
$3$. The upper half-power frequency is $\omega_2 = 600 \text{ rad/s}$.
The quality factor $Q$ is defined as the ratio of the resonant frequency to the bandwidth:
$Q = \frac{\omega_r}{\omega_2 - \omega_1}$
Substituting the values:
$Q = \frac{500}{600 - 400}$
$Q = \frac{500}{200}$
$Q = 2.5$

(A) In an $L-C-R$ series resonance circuit,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$.
Thus,$X_L = X_C$.
The phase difference $\phi$ is given by the formula $\tan \phi = \frac{X_L - X_C}{R}$.
Substituting $X_L = X_C$,we get $\tan \phi = \frac{0}{R} = 0$.
Therefore,$\phi = 0^{\circ}$.

(A) In the given circuit,$V_{1}$ measures the voltage across the resistor $R$,$V_{2}$ measures the voltage across the inductor $L$,and $V_{3}$ measures the voltage across the capacitor $C$.
At resonance,the inductive reactance $X_{L}$ is equal to the capacitive reactance $X_{C}$ $(X_{L} = X_{C})$.
Since the same current $I$ flows through the series combination of $R, L,$ and $C$,the voltage across the inductor is $V_{L} = I X_{L}$ and the voltage across the capacitor is $V_{C} = I X_{C}$.
Therefore,at resonance,$V_{L} = V_{C}$.
Since $V_{2}$ measures $V_{L}$ and $V_{3}$ measures $V_{C}$,the readings of $V_{2}$ and $V_{3}$ are equal.

(C) The current $i$ in the series $R-L-C$ circuit is given by $i = \frac{V_R}{R}$.
Substituting the given values: $i = \frac{100 \, V}{1000 \, \Omega} = 0.1 \, A$.
At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$, and the potential difference across the inductor $V_L$ is equal to the potential difference across the capacitor $V_C$.
The potential difference across the capacitor is $V_C = i X_C = i \left( \frac{1}{\omega C} \right)$.
Substituting the values: $V_C = 0.1 \times \left( \frac{1}{200 \times 2 \times 10^{-6}} \right)$.
$V_C = 0.1 \times \left( \frac{1}{400 \times 10^{-6}} \right) = 0.1 \times \left( \frac{10^6}{400} \right) = 0.1 \times 2500 = 250 \, V$.
Since $V_L = V_C$ at resonance, the potential difference across the inductance coil is $250 \, V$.

(D) In a series $LCR$ circuit at resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$ $(X_L = X_C)$.
At resonance,the phase difference $\phi$ between voltage and current is $0^{\circ}$.
$1$. The power factor is $\cos \phi = \cos 0^{\circ} = 1$. Thus,option $D$ is incorrect.
$2$. Average power $P_{avg} = V_{rms} I_{rms} \cos \phi = V_{rms} I_{rms} (1) = V_{rms} I_{rms}$,which is equal to the apparent power. Thus,option $B$ is true.
$3$. Peak energy stored in the capacitor is $\frac{1}{2} C V_0^2$ and in the inductor is $\frac{1}{2} L I_0^2$. At resonance,these values are equal. Thus,option $A$ is true.
$4$. Wattless current is $I_{rms} \sin \phi$. Since $\phi = 0^{\circ}$,$\sin 0^{\circ} = 0$,so wattless current is zero. Thus,option $C$ is true.
Therefore,the statement that is not true is that the power factor is zero.

(D) For better tuning of an $L-C-R$ circuit used in communication,the quality factor $Q$ must be high.
The formula for the quality factor is $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.
To maximize $Q$,we need a small resistance $R$,a large inductance $L$,and a small capacitance $C$.
Comparing the given options:
$A: R=20, L=1.5, C=35$
$B: R=25, L=2.5, C=45$
$C: R=25, L=1.5, C=45$
$D: R=15, L=3.5, C=30$
Option $D$ has the minimum resistance $(15 \Omega)$,the maximum inductance $(3.5 \text{ H})$,and the minimum capacitance $(30 \mu\text{F})$.
Therefore,the combination in option $D$ provides the highest quality factor,resulting in better tuning.

(D) Maximum power is dissipated in an $LCR$ circuit at the resonant frequency.
Given: Inductance $L = 5 \ mH = 5 \times 10^{-3} \ H$,Capacitance $C = 2 \ \mu F = 2 \times 10^{-6} \ F$.
The resonant frequency $f_R$ is given by the formula:
$f_R = \frac{1}{2 \pi \sqrt{LC}}$
Substituting the values:
$f_R = \frac{1}{2 \pi \sqrt{5 \times 10^{-3} \times 2 \times 10^{-6}}}$
$f_R = \frac{1}{2 \pi \sqrt{10 \times 10^{-9}}} = \frac{1}{2 \pi \sqrt{10^{-8}}}$
$f_R = \frac{1}{2 \pi \times 10^{-4}}$
$f_R = \frac{10^4}{2 \pi} = \frac{10 \times 10^3}{2 \pi} = \frac{5 \times 10^3}{\pi} \ Hz$.

(C) In the given $ LCR $ series circuit, the voltage across the inductor is $ V_L = 50 \, V $ and the voltage across the capacitor is $ V_C = 50 \, V $.
Since $ V_L = V_C $, the circuit is in resonance.
In a resonant $ LCR $ circuit, the net reactance is zero $( X_L - X_C = 0 )$, meaning the circuit behaves as a purely resistive circuit.
Therefore, the entire source voltage drops across the resistor $ R $.
Thus, the voltmeter reading is $ V_R = V_{source} = 220 \, V $.
The current in the circuit is given by $ I = \frac{V_R}{R} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A $.
Hence, the ammeter reading is $ 2.2 \, A $ and the voltmeter reading is $ 220 \, V $.

(D) The resonant frequency $ f $ of a tuned amplifier circuit ($LC$ circuit) is given by the formula: $ f = \frac{1}{2 \pi \sqrt{LC}} $.
Rearranging the formula to solve for $ \sqrt{LC} $,we get: $ \sqrt{LC} = \frac{1}{2 \pi f} $.
Given the carrier frequency $ f = 2 \text{ MHz} = 2 \times 10^{6} \text{ Hz} $.
Substituting the value of $ f $ into the equation: $ \sqrt{LC} = \frac{1}{2 \pi \times (2 \times 10^{6})} $.
Calculating the denominator: $ \sqrt{LC} = \frac{1}{4 \pi \times 10^{6}} $.

(A) In an $LCR$ circuit,resonance occurs when the inductive reactance $(X_L = \omega L)$ is equal to the capacitive reactance $(X_C = 1/(\omega C))$.
At this condition,the net reactance $X = X_L - X_C = 0$.
The impedance of the circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance,$Z = R$,which is the minimum possible impedance.
Since the net reactance is zero,the phase angle $\phi = \tan^{-1}((X_L - X_C)/R) = 0$.
Therefore,the current and voltage are in phase at resonance.

(B) For better tuning of a series $LCR$ circuit,the quality factor $(Q)$ should be as high as possible.
The quality factor is given by the formula: $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.
To maximize $Q$,we need a small resistance $(R)$ and a large ratio of inductance $(L)$ to capacitance $(C)$.
Let us calculate the factor $F = \frac{1}{R} \sqrt{\frac{L}{C}}$ for each option:
$A: F = \frac{1}{20} \sqrt{\frac{1.5}{35 \times 10^{-6}}} \approx 10.35$
$B: F = \frac{1}{15} \sqrt{\frac{3.5}{30 \times 10^{-6}}} \approx 22.79$
$C: F = \frac{1}{25} \sqrt{\frac{2.5}{45 \times 10^{-6}}} \approx 9.43$
$D: F = \frac{1}{15} \sqrt{\frac{2.5}{45 \times 10^{-6}}} \approx 15.71$
Comparing the values,option $B$ provides the highest quality factor,which indicates better tuning.

(A) The power dissipated in an $LCR$ circuit is given by $P = I_{rms}^2 R = \frac{1}{2} I_0^2 R$,where $I_0$ is the peak current.
At resonance,the power dissipated is maximum,given by $P_{max} = \frac{1}{2} I_{max}^2 R$.
We want the power to be half of the maximum power: $P = \frac{1}{2} P_{max}$.
Substituting the expressions,we get $\frac{1}{2} I^2 R = \frac{1}{2} (\frac{1}{2} I_{max}^2 R)$.
This simplifies to $I^2 = \frac{1}{2} I_{max}^2$.
Taking the square root of both sides,we get $I = \frac{1}{\sqrt{2}} I_{max}$.
Thus,the current amplitude must be $\frac{1}{\sqrt{2}}$ times its maximum value.

(A) In an $LCR$ series circuit,the given values are:
$L = 10 \ H$,$C = 40 \ \mu F$,$R = 60 \ \Omega$,and $V = 240 \ V$.
At resonance,the inductive reactance $(X_L)$ equals the capacitive reactance $(X_C)$,which means the net impedance $(Z)$ of the circuit is equal to the resistance $(R)$.
$Z = R = 60 \ \Omega$.
The current $(I)$ at resonance is given by Ohm's law:
$I = \frac{V}{Z} = \frac{240 \ V}{60 \ \Omega} = 4 \ A$.

(B) In an $LCR$ series circuit, resonance occurs when the inductive reactance equals the capacitive reactance.
Given, inductive reactance $X_L = 2R$.
At resonance, $X_C = X_L = 2R$.
The impedance $Z$ of the circuit at resonance is equal to the resistance $R$, as $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$.
The power factor is defined as $\cos \phi = \frac{R}{Z}$.
Substituting the values, $\cos \phi = \frac{R}{R} = 1$.
Therefore, the power factor is $1$ and $X_C = 2R$.

(B) For the power factor to be unity $(\cos \phi = 1)$,the circuit must be in resonance.
In an $LCR$ series circuit at resonance,the inductive reactance equals the capacitive reactance,i.e.,$X_L = X_C$.
This implies $\omega L = \frac{1}{\omega C}$,where $\omega = 2 \pi f$.
Substituting the values: $2 \pi f = \frac{1}{\sqrt{LC}}$.
Given $f = 50 \text{ Hz}$,$L = 10 \text{ mH} = 10 \times 10^{-3} \text{ H}$.
$2 \pi \times 50 = \frac{1}{\sqrt{10 \times 10^{-3} \times C}}$.
$100 \pi = \frac{1}{\sqrt{0.01 \times C}}$.
Squaring both sides: $(100 \pi)^2 = \frac{1}{0.01 \times C}$.
$10000 \times \pi^2 = \frac{1}{0.01 \times C}$.
$C = \frac{1}{10000 \times \pi^2 \times 0.01} = \frac{1}{100 \times \pi^2} \approx \frac{1}{100 \times 9.8696} \approx \frac{1}{986.96} \approx 1.0132 \times 10^{-3} \text{ F}$.
Thus,the required capacitance is approximately $1.014 \times 10^{-3} \text{ F}$.

(B) Given: Capacitance $C = 16 \ \mu F = 16 \times 10^{-6} \ F$.
Inductance $L = \frac{10}{\pi^2} \ mH = \frac{10}{\pi^2} \times 10^{-3} \ H$.
For the reactances to be equal,$X_L = X_C$.
Substituting the formulas for inductive and capacitive reactance: $L \omega = \frac{1}{C \omega}$.
Since $\omega = 2 \pi f$,we have $L(2 \pi f) = \frac{1}{C(2 \pi f)}$.
Rearranging for frequency $f$: $f^2 = \frac{1}{4 \pi^2 LC}$,which gives $f = \frac{1}{2 \pi \sqrt{LC}}$.
Substituting the values: $f = \frac{1}{2 \pi} \sqrt{\frac{1}{(\frac{10}{\pi^2} \times 10^{-3}) \times (16 \times 10^{-6})}}$.
$f = \frac{1}{2 \pi} \sqrt{\frac{\pi^2}{160 \times 10^{-9}}} = \frac{1}{2 \pi} \sqrt{\frac{\pi^2}{1.6 \times 10^{-7}}} = \frac{1}{2 \pi} \times \frac{\pi}{\sqrt{1.6 \times 10^{-7}}}$.
Calculating the value: $f = \frac{1}{2} \times \frac{1}{\sqrt{16 \times 10^{-8}}} = \frac{1}{2 \times 4 \times 10^{-4}} = \frac{10^4}{8} = 1250 \ Hz = 1.25 \ kHz$.

(B) The impedance of an $LCR$ series circuit is minimum at the resonant frequency, where the inductive reactance equals the capacitive reactance $(X_L = X_C)$.
The resonant angular frequency is given by $\omega = \frac{1}{\sqrt{LC}}$.
Given: $L = \frac{25}{\pi^2} \times 10^{-3} \text{ H}$, $C = 0.1 \times 10^{-6} \text{ F}$.
Substituting the values:
$\omega = \frac{1}{\sqrt{\frac{25}{\pi^2} \times 10^{-3} \times 0.1 \times 10^{-6}}} = \frac{1}{\sqrt{\frac{2.5}{\pi^2} \times 10^{-9}}} = \frac{1}{\sqrt{\frac{25}{\pi^2} \times 10^{-10}}} = \frac{\pi}{5 \times 10^{-5}} = \frac{\pi}{5} \times 10^5 \text{ rad/s}$.
Since $\omega = 2\pi f$, we have $2\pi f = \frac{\pi}{5} \times 10^5$.
$f = \frac{10^5}{10} = 10^4 \text{ Hz} = 10 \text{ kHz}$.

(B) The resonance frequency of a series $L-C-R$ circuit is given by the formula:
$f_0 = \frac{1}{2 \pi \sqrt{L C}} \quad \dots (i)$
When the inductance is reduced to $L^{\prime} = \frac{L}{4}$ and the capacitance is increased to $C^{\prime} = 16 C$,the new resonance frequency $f_0^{\prime}$ is:
$f_0^{\prime} = \frac{1}{2 \pi \sqrt{L^{\prime} C^{\prime}}}$
Substituting the new values:
$f_0^{\prime} = \frac{1}{2 \pi \sqrt{(\frac{L}{4}) \times (16 C)}}$
$f_0^{\prime} = \frac{1}{2 \pi \sqrt{4 L C}}$
$f_0^{\prime} = \frac{1}{2} \times \frac{1}{2 \pi \sqrt{L C}}$
Using equation $(i)$,we get:
$f_0^{\prime} = \frac{f_0}{2}$

(B) The resonant frequency of an $L-C$ circuit is given by the formula: $f = \frac{1}{2 \pi \sqrt{LC}}$.
From this relation,we can see that $f \propto \frac{1}{\sqrt{C}}$.
Let the initial frequency be $f_1 = f$ and the initial capacitance be $C_1 = C$.
Let the new frequency be $f_2$ and the new capacitance be $C_2 = 4C$.
Using the proportionality $f \propto \frac{1}{\sqrt{C}}$,we have the ratio: $\frac{f_2}{f_1} = \sqrt{\frac{C_1}{C_2}}$.
Substituting the values: $\frac{f_2}{f} = \sqrt{\frac{C}{4C}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the new resonant frequency is $f_2 = \frac{f}{2}$.

(A) In an $LCR$ series circuit,the amplitude of the current is given by $I_0 = \frac{E_0}{Z}$,where $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
For the current amplitude to be maximum,the impedance $Z$ must be minimum.
This occurs at resonance,where $X_L = X_C$.
Given the $emf$ equation $e = 200 \sin(100 \pi t) \ V$,the angular frequency is $\omega = 100 \pi \ rad/s$.
The condition for resonance is $\omega L = \frac{1}{\omega C}$.
Substituting the given values: $100 \pi \times L = \frac{1}{100 \pi \times 1 \times 10^{-6}}$.
$L = \frac{1}{(100 \pi)^2 \times 10^{-6}} = \frac{1}{10000 \pi^2 \times 10^{-6}} = \frac{1}{10^{-2} \pi^2} = \frac{100}{\pi^2} \ H$.

(A) The quality factor $(Q)$ of a series $LCR$ circuit is defined as the ratio of the resonant frequency to the bandwidth of the circuit.
Mathematically,it is given by the formula: $Q = \frac{\omega_0 L}{R}$,where $\omega_0$ is the resonant angular frequency.
For a series $LCR$ circuit,the resonant angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$.
Substituting this value into the formula for $Q$:
$Q = \frac{1}{R} \cdot \frac{1}{\sqrt{LC}} \cdot L$
$Q = \frac{1}{R} \sqrt{\frac{L^2}{LC}}$
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
$Q = \sqrt{\frac{L}{CR^2}}$
Therefore,the correct option is $A$.

(B) In an $LCR$ series circuit,the total impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
For the circuit to be purely resistive,the net reactance must be zero,which means the inductive reactance must equal the capacitive reactance.
Therefore,$X_L - X_C = 0$,which implies $X_L = X_C$.
At this condition,the circuit is said to be in resonance,and the impedance $Z$ becomes equal to $R$.

(A) The $Q$-factor (quality factor) of a series $LCR$ circuit is given by the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Given values are:
$L = 2 \ H$
$C = 32 \ \mu F = 32 \times 10^{-6} \ F$
$R = 20 \ \Omega$
Substituting these values into the formula:
$Q = \frac{1}{20} \sqrt{\frac{2}{32 \times 10^{-6}}}$
$Q = \frac{1}{20} \sqrt{\frac{1}{16 \times 10^{-6}}}$
$Q = \frac{1}{20} \times \frac{1}{4 \times 10^{-3}}$
$Q = \frac{1}{20} \times \frac{1000}{4}$
$Q = \frac{250}{20} = 12.5$
Thus,the $Q$-value is $12.5$.

(D) In an $L-C-R$ series circuit,the phase difference $\phi$ between the current and the voltage is given by the relation $\cos \phi = \frac{R}{Z}$,where $Z$ is the impedance of the circuit.
At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,i.e.,$X_L = X_C$.
The impedance $Z$ of the circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting $X_L = X_C$ into the impedance formula,we get $Z = \sqrt{R^2 + 0} = R$.
Now,substituting $Z = R$ into the phase difference formula,we get $\cos \phi = \frac{R}{R} = 1$.
Since $\cos \phi = 1$,it follows that $\phi = 0^{\circ}$.
Therefore,at resonance,the applied voltage and the current are in the same phase.

(C) In the $AC$ circuit, the current is in phase with the emf, which implies that the circuit is in resonance. At resonance, the impedance $Z$ of the circuit is equal to the resistance $R$ of the inductor coil $(Z = R)$.
Given, rms voltage $V_{\text{rms}} = 8 \, V$ and rms current $I_{\text{rms}} = 16 \, A$.
The resistance of the inductor coil is $R = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{8}{16} = 0.5 \, \Omega$.
When the inductor coil is connected to a $6 \, V$ $DC$ battery, the capacitor acts as an open circuit for $DC$, but the question asks for the steady current through the inductor coil itself. Since the inductor coil has a resistance $R = 0.5 \, \Omega$, the steady current $I$ is given by Ohm's law:
$I = \frac{V_{\text{DC}}}{R} = \frac{6 \, V}{0.5 \, \Omega} = 12 \, A$.

(A) Key Idea: The current amplitude in a series $LCR$ circuit is maximum at resonance, where the inductive reactance equals the capacitive reactance, i.e., $\omega L = \frac{1}{\omega C}$.
Given: $L = 0.5 \text{ H}$, $R = 10 \Omega$, $V_{rms} = 200 \text{ V}$, and $f = \frac{150}{\pi} \text{ Hz}$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times \frac{150}{\pi} = 300 \text{ rad/s}$.
At resonance, the impedance $Z = R = 10 \Omega$.
The maximum current $I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{10} = 20 \text{ A}$.
The $rms$ voltage across the inductor is $V_L = I_{rms} \times X_L$, where $X_L = \omega L$.
$V_L = 20 \times (300 \times 0.5) = 20 \times 150 = 3000 \text{ V}$.