Inductance, Capacitance and Resistance in Series and Parallel Questions in English

(A) The capacitive reactance is given by $X_C = \frac{1}{\omega C}$. As $\omega$ increases,$X_C$ decreases continuously. Therefore,$(A)-(iv)$.
The inductive reactance is given by $X_L = \omega L$. As $\omega$ increases,$X_L$ increases continuously. Therefore,$(B)-(i)$.
The resistance $R$ is independent of the angular frequency $\omega$. Therefore,$(C)-(ii)$.
The total impedance $Z$ in an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance,$X_L = X_C$,so $Z$ is minimum. As $\omega$ increases from zero,$Z$ first decreases until resonance and then increases. Therefore,$(D)-(iii)$.

(D) In a series $LCR$ circuit, the total applied voltage $V$ is the phasor sum of the voltages across the individual components.
The formula for the total voltage is given by:
$V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given:
$V_L = 50 \,V$
$V_C = 20 \,V$
$V_R = 40 \,V$
Substituting these values into the formula:
$V = \sqrt{40^2 + (50 - 20)^2}$
$V = \sqrt{1600 + (30)^2}$
$V = \sqrt{1600 + 900}$
$V = \sqrt{2500}$
$V = 50 \,V$
Therefore, the applied $A.C.$ voltage is $50 \,V$.

(B) The impedance of a series $LR$ circuit is given by $Z_1 = \sqrt{R^2 + X_L^2}$.
When a capacitor is connected in series with this circuit,the new impedance becomes $Z_2 = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $(X_L - X_C)^2 < X_L^2$ (assuming $X_C$ is not zero),the new impedance $Z_2$ is less than the original impedance $Z_1$.
According to Ohm's law for $AC$ circuits,$I = \frac{V}{Z}$.
Since the impedance $Z$ decreases,the current $I$ flowing in the circuit increases.

(D) For the inductor coil at $50 \, Hz$:
$X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.
For the resistor:
$R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.
At the new frequency $f' = 40 \, Hz$, the new inductive reactance $X_L'$ is:
$X_L' = X_L \times \frac{f'}{f} = 12.5 \times \frac{40}{50} = 10 \, \Omega$.
When connected in series, the impedance $Z$ is:
$Z = \sqrt{R^2 + (X_L')^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.
The current $I$ in the series circuit is:
$I = \frac{V}{Z} = \frac{100}{10\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, A$.

(C) The $r.m.s.$ current $I$ in the circuit is given by $I = \frac{V}{Z}$.
Given $V = 220 \ V$ and $Z = 33 \ \Omega$,we have $I = \frac{220}{33} = \frac{20}{3} \ A$.
The true power $P$ consumed in an $a.c.$ circuit is given by $P = I^2 R$.
Substituting the values,$P = \left(\frac{20}{3}\right)^2 \times 18$.
$P = \frac{400}{9} \times 18 = 400 \times 2 = 800 \ W$.

(B) Given: $R=200 \Omega$, $L=663 \text{ mH}$, $C=26.5 \mu F$, $V_{0}=50 \text{ V}$, $X_{L}=250 \Omega$, $X_{C}=100 \Omega$.
The impedance $Z$ of a series $LCR$ circuit is given by $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.
Substituting the values: $Z = \sqrt{200^{2} + (250 - 100)^{2}} = \sqrt{40000 + 150^{2}}$.
$Z = \sqrt{40000 + 22500} = \sqrt{62500} = 250 \Omega$.
The peak current $i_{0}$ is given by $i_{0} = \frac{V_{0}}{Z}$.
$i_{0} = \frac{50}{250} = 0.2 \text{ A}$.

(C) The true power $P$ consumed in an $LCR$ series circuit is given by the formula: $P = V_{rms} \cdot I_{rms} \cdot \cos \phi$,where $\cos \phi$ is the power factor.
We know that the power factor $\cos \phi = \frac{R}{Z}$.
Also,the $r.m.s.$ current is given by $I_{rms} = \frac{V_{rms}}{Z}$.
Substituting these into the power formula: $P = V_{rms} \cdot \left( \frac{V_{rms}}{Z} \right) \cdot \left( \frac{R}{Z} \right) = \frac{V_{rms}^2 \cdot R}{Z^2}$.
Given values: $V_{rms} = 220 \ V$,$R = 18 \ \Omega$,$Z = 33 \ \Omega$.
$P = \frac{220^2 \cdot 18}{33^2} = \frac{48400 \cdot 18}{1089}$.
$P = \frac{871200}{1089} = 800 \ W$.
Therefore,the true power consumed is $800 \ W$.

(C) The given circuit is a series $LCR$ circuit where the bulb acts as a resistor $R$.
The current in the circuit is given by $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$.
At resonance,the angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$. At this frequency,the impedance $Z$ is minimum $(Z = R)$,and the current $I$ is maximum $(I = \frac{V}{R})$.
As the frequency increases from a low value,the impedance $Z$ decreases until it reaches the resonance frequency,causing the current $I$ and the brightness of the bulb to increase to a maximum.
As the frequency increases further beyond the resonance frequency,the impedance $Z$ increases again,causing the current $I$ and the brightness of the bulb to decrease.
Therefore,the brightness increases,reaches a maximum at resonance,and then decreases.

(A) rejector circuit,also known as a parallel resonant circuit,is a circuit in which an inductor $(L)$ and a capacitor $(C)$ are connected in parallel. At the resonant frequency,the impedance of this parallel combination becomes maximum,which effectively rejects or blocks the current at that specific frequency. Therefore,$L-C-R$ components are connected in parallel to form a rejector circuit.

(C) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.
Given $X_L = 3R$,the power factor $X_1$ is:
$X_1 = \frac{R}{\sqrt{R^2 + (3R)^2}} = \frac{R}{\sqrt{R^2 + 9R^2}} = \frac{R}{\sqrt{10R^2}} = \frac{1}{\sqrt{10}}$.
When a capacitor with $X_C = R$ is added in series,the circuit becomes an $LCR$ circuit.
The impedance $Z'$ of the $LCR$ circuit is $Z' = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $X_L = 3R$ and $X_C = R$,we have $X_L - X_C = 3R - R = 2R$.
Thus,$Z' = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5}$.
The new power factor $X_2$ is:
$X_2 = \frac{R}{Z'} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}}$.
The ratio $X_1 : X_2$ is:
$\frac{X_1}{X_2} = \frac{1/\sqrt{10}}{1/\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{10}} = \frac{1}{\sqrt{2}}$.
Therefore,the ratio is $1: \sqrt{2}$.

(D) The apparent power in an a.c. circuit is given by $P_{app} = V_{rms} I_{rms} = I_{rms}^2 Z$.
The true power (or average power) in an a.c. circuit is given by $P_{true} = V_{rms} I_{rms} \cos \phi = I_{rms}^2 R$.
The ratio of apparent power to true power is $\frac{P_{app}}{P_{true}} = \frac{I_{rms}^2 Z}{I_{rms}^2 R} = \frac{Z}{R}$.
Since $\cos \phi = \frac{R}{Z}$,we have $\frac{Z}{R} = \frac{1}{\cos \phi} = \sec \phi$.

(A) The given $AC$ voltage source is $V = 300 \sqrt{2} \sin(\omega t)$.
The peak voltage is $V_m = 300 \sqrt{2} \text{ V}$.
The $RMS$ voltage is $V_{rms} = \frac{V_m}{\sqrt{2}} = 300 \text{ V}$.
The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = 30 \Omega$, $X_L = 10 \Omega$, and $X_C = 10 \Omega$.
$Z = \sqrt{30^2 + (10 - 10)^2} = \sqrt{30^2 + 0^2} = 30 \Omega$.
The $RMS$ current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{Z}$.
$I_{rms} = \frac{300}{30} = 10 \text{ A}$.

(D) For an $L-C-R$ series circuit,the phase difference $\phi$ is given by $\tan \phi = \frac{|X_C - X_L|}{R}$.
Given $\phi = 45^{\circ}$ and $R = 100 \ \Omega$,we have $\tan 45^{\circ} = 1 = \frac{|X_C - X_L|}{R}$,which implies $|X_C - X_L| = R = 100 \ \Omega$.
The impedance $Z$ of the circuit is $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{R^2 + R^2} = R\sqrt{2} = 100\sqrt{2} \ \Omega$.
The peak voltage $V_m = 240 \ V$. The $rms$ voltage is $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{240}{\sqrt{2}} \ V$.
The $rms$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{240 / \sqrt{2}}{100\sqrt{2}} = \frac{240}{100 \times 2} = \frac{240}{200} = 1.2 \ A$.

(A) Given: $X_L = 24 \ \Omega$,$X_C = 16 \ \Omega$,$R = 6 \ \Omega$,$V = 100 \ V$.
In a series $LCR$ circuit,the impedance $Z$ is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
$Z = \sqrt{6^2 + (24 - 16)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ \Omega$.
The current $i$ in the circuit is:
$i = \frac{V}{Z} = \frac{100}{10} = 10 \ A$.
The potential difference across the series combination of the inductor and capacitor is given by:
$V_{LC} = i |X_L - X_C|$
$V_{LC} = 10 \times |24 - 16| = 10 \times 8 = 80 \ V$.

(C) In a series $L-C-R$ circuit,the circuit is capacitive when the capacitive reactance $X_C$ is greater than the inductive reactance $X_L$ (i.e.,$X_C > X_L$).
From the given graph,the intersection point $B$ represents the resonance frequency where $X_L = X_C$.
For frequencies less than the resonance frequency (i.e.,to the left of point $B$),the curve for $X_C$ lies above the curve for $X_L$,meaning $X_C > X_L$.
Among the given options,point $A$ lies to the left of point $B$,where $X_C > X_L$. Therefore,the circuit is capacitive at point $A$.

(C) Given,peak voltages are $V_{C}=30 \,V$,$V_{L}=110 \,V$,and $V_{R}=60 \,V$.
The peak voltage $(V_{0})$ across a series $L-C-R$ circuit is given by the formula:
$V_{0} = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$
Substituting the given values:
$V_{0} = \sqrt{(60)^{2} + (110 - 30)^{2}}$
$V_{0} = \sqrt{60^{2} + 80^{2}}$
$V_{0} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \,V$
The rms value of the applied voltage is related to the peak voltage by $V_{\text{rms}} = \frac{V_{0}}{\sqrt{2}}$.
$V_{\text{rms}} = \frac{100}{\sqrt{2}} \approx 70.7 \,V$.

(A) The instantaneous power dissipated in an electrical circuit is given by $P = I^2 R$.
In a series $LCR$ circuit,the inductor $(L)$ and the capacitor $(C)$ are reactive components that store energy in magnetic and electric fields,respectively,but they do not dissipate energy as heat.
The resistance $(R)$ is the only component that dissipates electrical energy into heat.
Therefore,the power dissipation in a series $LCR$ circuit occurs only through the resistance $R$.

(A) Initially, the potential difference across each element is $V_R = V_L = V_C = 20 \, V$. Since $V_L = V_C$, the circuit is at resonance, meaning the impedance $Z = R$. The source voltage is $V = V_R = 20 \, V$.
When the resistance $R$ is doubled to $2R$, the new impedance becomes $Z' = \sqrt{(2R)^2 + (X_L - X_C)^2}$. Since $X_L = X_C$ at resonance, $Z' = 2R$.
The new current in the circuit is $I' = V / Z' = V / (2R) = I / 2$, where $I$ is the original current.
The new potential difference across the resistor is $V_R' = I' \times (2R) = (I/2) \times (2R) = IR = 20 \, V$.
The new potential difference across the inductor is $V_L' = I' X_L = (I/2) X_L = V_L / 2 = 20 / 2 = 10 \, V$.
The new potential difference across the capacitor is $V_C' = I' X_C = (I/2) X_C = V_C / 2 = 20 / 2 = 10 \, V$.
Thus, the new potential differences are $20 \, V, 10 \, V, 10 \, V$.

(B) The phase angle $\theta$ between current $I$ and voltage $V$ in an $LCR$ series circuit is given by the formula:
$\tan \theta = \frac{X_L - X_C}{R}$
First,calculate the inductive reactance $X_L$:
$X_L = 2 \pi f L = 2 \pi \times 50 \times \left( \frac{200}{\pi} \times 10^{-3} \right) = 20 \, \Omega$
Next,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 50 \times (10^{-3} / \pi)} = \frac{1}{0.1} = 10 \, \Omega$
Given resistance $R = 10 \, \Omega$.
Substituting these values into the phase angle formula:
$\tan \theta = \frac{20 - 10}{10} = \frac{10}{10} = 1$
Since $\tan \theta = 1$,we have $\theta = \tan^{-1}(1) = \frac{\pi}{4}$ radians.
Thus,the phase angle of the circuit is $\frac{\pi}{4}$.

(D) Given: Inductance $L = 1 \text{ H}$,Capacitance $C = 20 \mu\text{F} = 20 \times 10^{-6} \text{ F}$,Resistance $R = 300 \Omega$,Frequency $f = \frac{50}{\pi} \text{ Hz}$.
First,calculate the inductive reactance $X_{L}$:
$X_{L} = 2 \pi f L = 2 \pi \left(\frac{50}{\pi}\right) \times 1 = 100 \Omega$.
Next,calculate the capacitive reactance $X_{C}$:
$X_{C} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \left(\frac{50}{\pi}\right) \times 20 \times 10^{-6}} = \frac{1}{100 \times 20 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega$.
The impedance $Z$ of a series $L-C-R$ circuit is given by:
$Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$.
Substituting the values:
$Z = \sqrt{(300)^{2} + (500 - 100)^{2}} = \sqrt{300^{2} + 400^{2}} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \Omega$.

(D) The current in an $AC$ circuit containing a capacitor is given by $i = \frac{V}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
When a dielectric is introduced into the gap between the plates of the capacitor,the capacitance $C$ increases $(C = K C_0)$.
As $C$ increases,the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.
Since the impedance $Z = \sqrt{R^2 + X_C^2}$ decreases,the current $i$ in the circuit increases.
Consequently,the brightness of the bulb,which depends on the power dissipated $(P = i^2 R)$,increases.

(C) In a series $LCR$ circuit,the total alternating emf $(V)$ is given by the phasor sum of the individual voltages across the resistor $(V_R)$,inductor $(V_L)$,and capacitor $(V_C)$.
The formula is:
$V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given values:
$V_R = 80 \ V$
$V_L = 40 \ V$
$V_C = 100 \ V$
Substituting these values into the formula:
$V = \sqrt{(80)^2 + (40 - 100)^2}$
$V = \sqrt{6400 + (-60)^2}$
$V = \sqrt{6400 + 3600}$
$V = \sqrt{10000}$
$V = 100 \ V$
Therefore,the value of the alternating emf is $100 \ V$.

(D) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given: $R = 40 \ \Omega$,$X_C = 20 \ \Omega$,$X_L = 50 \ \Omega$,and $V = 100 \ V$.
Substituting the values:
$Z = \sqrt{40^2 + (50 - 20)^2}$
$Z = \sqrt{1600 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \ \Omega$.
The current $i$ in the circuit is given by $i = \frac{V}{Z}$.
$i = \frac{100 \ V}{50 \ \Omega} = 2 \ A$.

(C) In an $L-C-R$ circuit,the resonant frequency $f_0$ is given by the formula:
$f_0 = \frac{1}{2 \pi \sqrt{L C}}$
For the resonant frequency to remain the same,the product $LC$ must remain constant.
Let the new inductance be $L^{\prime}$ and the new capacitance be $C^{\prime} = 4C$.
Since $f_0 = f_0^{\prime}$,we have:
$L C = L^{\prime} C^{\prime}$
Substituting $C^{\prime} = 4C$ into the equation:
$L C = L^{\prime} (4 C)$
Dividing both sides by $4C$:
$L^{\prime} = \frac{L}{4}$
Therefore,the inductance should be changed from $L$ to $\frac{L}{4}$.

(D) In an $R-L-C$ series circuit,the given values are:
$R = 150 \Omega$
$C = 20 \mu F = 20 \times 10^{-6} F = 2 \times 10^{-5} F$
$L = 500 mH = 0.5 H$
$\omega = 400 rad s^{-1}$
First,calculate the inductive reactance $X_L$:
$X_L = \omega L = 400 \times 0.5 = 200 \Omega$
Next,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{\omega C} = \frac{1}{400 \times 2 \times 10^{-5}} = \frac{1}{800 \times 10^{-5}} = \frac{10^5}{800} = \frac{1000}{8} = 125 \Omega$
The phase angle $\phi$ is given by the formula:
$\tan \phi = \frac{X_L - X_C}{R}$
Substituting the values:
$\tan \phi = \frac{200 - 125}{150} = \frac{75}{150} = 0.5$
Therefore,the phase angle is $\phi = \tan^{-1}(0.5)$.

(A) In an $LCR$ series circuit, the applied voltage $V$ is given by the phasor sum of the potential differences across the components.
The formula for the applied voltage is $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Given values are:
$V_R = 40 \,V$
$V_L = 60 \,V$
$V_C = 30 \,V$
Substituting these values into the formula:
$V = \sqrt{40^2 + (60 - 30)^2}$
$V = \sqrt{40^2 + 30^2}$
$V = \sqrt{1600 + 900}$
$V = \sqrt{2500}$
$V = 50 \,V$.
Therefore, the applied $AC$ voltage is $50 \,V$.

(C) In an $L-C-R$ circuit,we are given:
$X_L = 2R$ and $X_C = \frac{X_L}{3}$.
Substituting $X_L = 2R$ into the expression for $X_C$,we get $X_C = \frac{2R}{3}$.
The impedance $Z$ of the $L-C-R$ circuit is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values of $X_L$ and $X_C$:
$Z = \sqrt{R^2 + (2R - \frac{2R}{3})^2} = \sqrt{R^2 + (\frac{4R}{3})^2} = \sqrt{R^2 + \frac{16R^2}{9}} = \sqrt{\frac{25R^2}{9}} = \frac{5R}{3}$.
The power factor is defined as $\cos \phi = \frac{R}{Z}$.
Substituting the value of $Z$:
$\cos \phi = \frac{R}{5R/3} = \frac{3}{5} = 0.6$.

(C) The power factor $(\cos \phi)$ of an $LR$ series circuit is given by $\cos \phi = \frac{R}{Z}$, where $R$ is the resistance and $Z$ is the impedance.
Given, $\cos \phi = 0.5 = \frac{1}{2}$.
We know that $Z = \sqrt{R^2 + X_L^2}$, where $X_L$ is the inductive reactance.
So, $\frac{R}{\sqrt{R^2 + X_L^2}} = \frac{1}{2}$.
Squaring both sides, we get $\frac{R^2}{R^2 + X_L^2} = \frac{1}{4}$.
$4R^2 = R^2 + X_L^2$.
$3R^2 = X_L^2$.
Taking the square root, $\sqrt{3}R = X_L$.
Therefore, the ratio of resistance to reactance is $\frac{R}{X_L} = \frac{1}{\sqrt{3}}$.

(A) The phase difference $\phi$ between the current and the voltage in a series $LCR$ circuit is given by the relation:
$\tan \phi = \frac{X_L - X_C}{R}$
In a series $LCR$ circuit,the current leads the source voltage when the circuit is capacitive in nature.
This happens when the capacitive reactance $X_C$ is greater than the inductive reactance $X_L$,i.e.,$X_C > X_L$.
In this condition,the phase angle $\phi$ becomes negative,indicating that the current leads the voltage.

(B) In an $AC$ circuit,the average power dissipated is given by the formula:
$P = V_{rms} I_{rms} \cos \phi$
where $\phi$ is the phase difference between voltage and current.
For an ideal inductor $(L)$ and an ideal capacitor $(C)$,the phase difference between voltage and current is $90^{\circ}$.
Therefore,the power dissipated by $L$ or $C$ is:
$P_{L \text{ or } C} = V I \cos 90^{\circ} = 0$ (since $\cos 90^{\circ} = 0$).
For a resistor $(R)$,the voltage and current are in phase,so the phase difference $\phi = 0^{\circ}$.
Thus,the power dissipated by the resistor is $P_R = V I \cos 0^{\circ} = V I$.
Hence,only the resistor $(R)$ dissipates energy in an $L-C-R$ circuit.

(A) Given,the applied electromotive force (emf) is $E = 30 \sin 200 t$.
Comparing this with $E = E_{\max} \sin \omega t$,we get $E_{\max} = 30 \text{ V}$ and $\omega = 200 \text{ rad s}^{-1}$.
From the circuit diagram,we have resistance $R = 10 \Omega$,inductance $L = 0.05 \text{ H}$,and capacitance $C = 500 \mu\text{F} = 500 \times 10^{-6} \text{ F}$.
First,calculate the inductive reactance: $X_L = L \omega = 0.05 \times 200 = 10 \Omega$.
Next,calculate the capacitive reactance: $X_C = \frac{1}{C \omega} = \frac{1}{500 \times 10^{-6} \times 200} = \frac{1}{0.1} = 10 \Omega$.
Since $X_L = X_C$,the circuit is in a state of electrical resonance.
In resonance,the impedance of the circuit is $Z = R = 10 \Omega$.
The amplitude of the current is given by $I_{\max} = \frac{E_{\max}}{Z} = \frac{30}{10} = 3 \text{ A}$.

(A) The circuit consists of a resistor $R$ in series with a parallel combination of an inductor $L$ and a capacitor $C$.
At the given frequency $f = \frac{1}{2 \pi \sqrt{L C}}$,the angular frequency is $\omega = 2 \pi f = \frac{1}{\sqrt{L C}}$.
This is the resonant frequency of the $L-C$ parallel circuit.
At this frequency,the inductive reactance $X_L = \omega L$ and capacitive reactance $X_C = \frac{1}{\omega C}$ are equal,i.e.,$X_L = X_C$.
For a parallel $L-C$ circuit,the equivalent impedance $Z_{LC}$ is given by $\frac{1}{Z_{LC}} = \sqrt{(\frac{1}{X_L})^2 + (\frac{1}{X_C} - \frac{1}{X_L})^2}$ is not correct here; rather,the admittance $Y = \sqrt{(\frac{1}{R_{LC}})^2 + (\frac{1}{X_L} - \frac{1}{X_C})^2}$. Since $X_L = X_C$,the net reactance of the parallel $L-C$ combination becomes infinite $(Z_{LC} \to \infty)$.
Therefore,the parallel $L-C$ combination acts as an open circuit.
Consequently,the current through the entire circuit,including the resistor $R$,becomes zero.

(A) Given: $R = 10 \, \Omega$, $L = 6.0 \, \mu H = 6.0 \times 10^{-6} \, H$, $C = 10 \, \mu F = 10 \times 10^{-6} \, F$, $f = 50 \, Hz$, $V_m = 280 \, V$.
The angular frequency is $\omega = 2 \pi f = 2 \times 3.14 \times 50 = 314 \, rad/s$.
Capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{314 \times 10 \times 10^{-6}} = \frac{1}{3.14 \times 10^{-3}} \approx 318.47 \, \Omega$.
Inductive reactance: $X_L = \omega L = 314 \times 6.0 \times 10^{-6} = 1.884 \times 10^{-3} \, \Omega$.
Impedance of the circuit: $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{10^2 + (318.47 - 0.001884)^2} \approx \sqrt{100 + (318.47)^2} \approx 318.63 \, \Omega$.
$RMS$ voltage: $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{280}{1.414} \approx 198 \, V$.
$RMS$ current: $I_{rms} = \frac{V_{rms}}{Z} = \frac{198}{318.63} \approx 0.621 \, A$.
Average power dissipated as heat in the resistor: $P = I_{rms}^2 R = (0.621)^2 \times 10 = 0.3856 \times 10 \approx 3.86 \, W$.
Rounding to the nearest given option, the power is $3.8 \, W$.

(A) In an $LCR$ series circuit,the phase difference $\phi$ between voltage and current is given by $\tan \phi = \frac{X_C - X_L}{R}$.
Since the current leads the voltage,the phase angle is $\phi = -45^{\circ}$.
Therefore,$\tan(-45^{\circ}) = \frac{X_C - X_L}{R} = -1$.
This implies $X_L - X_C = R$,or $X_C - X_L = -R$.
However,the standard convention for current leading is $\tan \phi = \frac{X_L - X_C}{R}$ where $\phi$ is the phase of voltage relative to current. If current leads,$\phi = -45^{\circ}$,so $\tan(-45^{\circ}) = -1 = \frac{X_L - X_C}{R}$,which gives $X_C - X_L = R$.
Substituting $X_C = \frac{1}{2 \pi f C}$ and $X_L = 2 \pi f L$:
$\frac{1}{2 \pi f C} - 2 \pi f L = R$
$\frac{1}{2 \pi f C} = R + 2 \pi f L$
$C = \frac{1}{2 \pi f (R + 2 \pi f L)}$.

(C) Given: Capacitance $C = \frac{10^{-3}}{2 \pi} F$,Inductance $L = \frac{100}{\pi} mH = \frac{100}{\pi} \times 10^{-3} H$,Resistance $R = 10 \Omega$,and Frequency $f = 50 Hz$.
First,calculate the inductive reactance $X_L = \omega L = 2 \pi f L = 2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} = 10 \Omega$.
Next,calculate the capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 50 \times \frac{10^{-3}}{2 \pi}} = \frac{1}{50 \times 10^{-3}} = \frac{1000}{50} = 20 \Omega$.
The phase angle $\phi$ is given by $\tan \phi = \frac{X_C - X_L}{R}$.
Substituting the values: $\tan \phi = \frac{20 - 10}{10} = \frac{10}{10} = 1$.
Therefore,$\phi = \tan^{-1}(1) = 45^{\circ}$.

(D) Given: Peak emf, $V_0 = 8 \, V$, Frequency $f = \frac{30}{\pi} \, Hz$, Resistance $R = 8 \, \Omega$, Average power $P_{avg} = 0.4 \, W$.
$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{8}{\sqrt{2}} \, V$.
Average power is given by $P_{avg} = V_{rms} I_{rms} \cos \phi = V_{rms} \left(\frac{V_{rms}}{Z}\right) \left(\frac{R}{Z}\right) = \frac{V_{rms}^2 R}{Z^2}$.
Substituting the values: $0.4 = \frac{(8/\sqrt{2})^2 \times 8}{Z^2} = \frac{32 \times 8}{Z^2} = \frac{256}{Z^2}$.
$Z^2 = \frac{256}{0.4} = 640 \, \Omega^2$.
Since $Z^2 = R^2 + X_L^2$, we have $640 = 8^2 + X_L^2 = 64 + X_L^2$.
$X_L^2 = 640 - 64 = 576$, so $X_L = 24 \, \Omega$.
Using $X_L = 2 \pi f L$, we get $24 = 2 \pi (\frac{30}{\pi}) L = 60 L$.
$L = \frac{24}{60} = 0.4 \, H$.

(D) The impedance $Z$ for a series $LCR$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Here,$R = 5 \ \Omega$,$L = 20 \ mH = 20 \times 10^{-3} \ H$,and $C = 500 \ \mu F = 500 \times 10^{-6} \ F$.
The angular frequency is $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \ rad/s$.
Inductive reactance $X_L = \omega L = 100 \pi \times 20 \times 10^{-3} = 2 \pi \approx 6.28 \ \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times 500 \times 10^{-6}} = \frac{1}{0.05 \pi} = \frac{20}{\pi} \approx 6.37 \ \Omega$.
Now,$Z = \sqrt{5^2 + (6.28 - 6.37)^2} = \sqrt{25 + (-0.09)^2} = \sqrt{25 + 0.0081} = \sqrt{25.0081} \approx 5 \ \Omega$.

(B) The power factor is given by $\cos \phi = \frac{R}{Z} = 0.5$.
Since $R = 100 \text{ } \Omega$,we have $Z = \frac{R}{0.5} = \frac{100}{0.5} = 200 \text{ } \Omega$.
The impedance of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Squaring both sides,$Z^2 = R^2 + (X_L - X_C)^2$.
Substituting the values,$200^2 = 100^2 + (X_L - X_C)^2$.
$40000 = 10000 + (X_L - X_C)^2$.
$(X_L - X_C)^2 = 30000$.
$|X_L - X_C| = \sqrt{30000} = 100\sqrt{3} \text{ } \Omega$.
Given that the difference is $\sqrt{3}\alpha \text{ } \Omega$,we equate $\sqrt{3}\alpha = 100\sqrt{3}$.
Therefore,$\alpha = 100$.