Only Inductor, Only Capacitor and Only Resistor Circuit Questions in English

(B) The power dissipated in a resistor in an $AC$ circuit is given by the formula: $P = \frac{V_{rms}^2}{R}$.
Given:
$V_{rms} = 30 \ V$
$R = 10 \ \Omega$
Substituting the values into the formula:
$P = \frac{(30)^2}{10} = \frac{900}{10} = 90 \ W$.
Therefore,the power dissipated is $90 \ W$.

(A) choke coil is designed to have high inductance and low resistance.
Since an inductor is an ideal non-resistive device,it does not consume power in the form of heat,unlike a resistor.
By using a coil with high inductance $(L)$ and very low resistance $(R)$,we can control the current in an $AC$ circuit while minimizing power loss $(P = I^2 R)$.
Therefore,the correct option is $A$.

(A) choke coil is an inductor with high inductance and negligible resistance. It is used to control current in $AC$ circuits.
In an $AC$ circuit,the inductive reactance is given by $X_L = \omega L = 2\pi f L$. Since $f > 0$,the choke coil offers significant impedance to the current.
In a $DC$ circuit,the frequency $f = 0$,so the inductive reactance $X_L = 0$. The coil acts as a simple wire with negligible resistance,failing to control the current effectively.

(B) For a purely capacitive circuit,the applied alternating e.m.f. is given by $e = e_0 \sin(\omega t)$.
In a capacitor,the current $i$ leads the voltage $e$ by a phase angle of $\pi / 2$.
Therefore,the expression for current is $i = i_0 \sin(\omega t + \pi / 2)$.
This indicates that the current is ahead of the e.m.f. by $\pi / 2$.

(C) In a purely resistive $AC$ circuit, the voltage $V$ and current $I$ are given by $V = V_m \sin(\omega t)$ and $I = I_m \sin(\omega t)$.
Since the phase angle for both is $\omega t$, there is no phase difference between them.
Therefore, the current and voltage are in the same phase.

(D) In a pure inductive circuit,the current lags behind the voltage by a phase angle of $90^{\circ}$.
The formula for average power dissipated in an $AC$ circuit is given by $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos \phi$,where $\phi$ is the phase difference between voltage and current.
For a pure inductor,the phase difference $\phi = 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the average power dissipated is $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos 90^{\circ} = 0$.
Therefore,no power is dissipated in a pure inductor.

(A) In a pure inductive circuit,the resistance $R = 0$.
For an inductor,the voltage $V$ leads the current $I$ by a phase angle of $\phi = 90^o$ or $\pi/2$ radians.
Therefore,the $e.m.f.$ of the applied $ac$ voltage leads the current by $90^o$.

(B) In a pure inductive circuit,the voltage $V$ across the inductor leads the current $I$ by a phase angle of $90^{\circ}$ (or $\frac{\pi}{2}$ radians).
Conversely,this means the current $I$ lags behind the electromotive force (e.m.f.) by $90^{\circ}$.
Mathematically,if $V = V_m \sin(\omega t)$,then $I = I_m \sin(\omega t - 90^{\circ})$.
Therefore,the current lags behind the e.m.f. by $90^{\circ}$.

(B) The given alternating voltage is $E = 200\sqrt{2} \sin(100t)$.
Comparing this with the standard equation $E = E_0 \sin(\omega t)$,we get the peak voltage $E_0 = 200\sqrt{2} \text{ V}$ and angular frequency $\omega = 100 \text{ rad/s}$.
The capacitance is $C = 1 \mu F = 1 \times 10^{-6} \text{ F}$.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = 10^4 \Omega$.
The $RMS$ voltage is $V_{rms} = \frac{E_0}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 \text{ V}$.
The reading of the $AC$ ammeter is the $RMS$ current $I_{rms} = \frac{V_{rms}}{X_C} = \frac{200}{10^4} = 2 \times 10^{-2} \text{ A}$.
Converting to milliamperes,$I_{rms} = 2 \times 10^{-2} \times 10^3 \text{ mA} = 20 \text{ mA}$.

(D) The average power dissipated in an $AC$ circuit is given by the formula $P = V_{rms} I_{rms} \cos \phi$,where $\phi$ is the phase difference between voltage and current.
In a pure capacitive circuit,the current leads the voltage by a phase angle of $\phi = 90^\circ$ (or $\pi/2$ radians).
The power factor is $\cos \phi = \cos 90^\circ = 0$.
Substituting this into the power formula,we get $P = V_{rms} I_{rms} \times 0 = 0$.
Therefore,the average power dissipation in a pure capacitor is zero.

(C) The inductive reactance $X_L$ of the inductor is given by the formula $X_L = 2\pi f L$.
Given: $V = 120 \, V$,$f = 60 \, Hz$,and $L = 0.70 \, H$.
Substituting the values: $X_L = 2 \times 3.1416 \times 60 \times 0.70 \approx 263.89 \, \Omega$.
The current $I$ flowing through the inductor is given by $I = \frac{V}{X_L}$.
$I = \frac{120}{263.89} \approx 0.455 \, A$.
Therefore,the correct option is $C$.

(B) The inductive reactance $X_L$ is given by the formula $X_L = 2\pi \nu L$.
Given: $X_L = 100\, \Omega$ and frequency $\nu = 50\, Hz$.
Substituting the values into the formula:
$100 = 2 \times \pi \times 50 \times L$
$100 = 100 \times \pi \times L$
$L = \frac{1}{\pi} = \frac{1}{3.14} \approx 0.318\, H$.
Rounding to two decimal places, we get $L \approx 0.32\, H$.

(A) The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2\pi \nu C}$.
Given values are frequency $\nu = 4000\, Hz$ and capacitance $C = 25\,\mu F = 25 \times 10^{-6}\, F$.
Substituting these values into the formula:
$X_C = \frac{1}{2 \times \pi \times 4000 \times 25 \times 10^{-6}}$
$X_C = \frac{1}{2 \times \pi \times 10^5 \times 10^{-6}}$
$X_C = \frac{1}{2 \times \pi \times 10^{-1}}$
$X_C = \frac{1}{0.2 \times \pi} = \frac{10}{2\pi} = \frac{5}{\pi}\,\Omega$.

(A) In an $AC$ circuit, the power dissipated is given by the formula $P = V_{rms} I_{rms} \cos \phi$, where $\phi$ is the phase angle between voltage and current.
For a purely inductive circuit, the phase difference between voltage and current is $\phi = 90^\circ$.
The power factor is $\cos 90^\circ = 0$.
Since the coil is made of a superconducting material, its resistance $R = 0$.
Therefore, the power dissipated in the coil is $P = I^2 R = I^2 \times 0 = 0$.

(B) In a purely resistive $ac$ circuit,the voltage $v = V_m \sin(\omega t)$ and the current $i = I_m \sin(\omega t)$ are given by Ohm's law,where $I_m = V_m / R$.
Since the phase angle for both voltage and current is $\omega t$,the phase difference between them is zero.
Therefore,the current is in phase with the $e.m.f.$

(A) Given: Inductance $L = 1 \, H$,Voltage $V = 200 \, V$,Frequency $f = 50 \, Hz$.
Inductive reactance is given by $X_L = \omega L = 2 \pi f L$.
Substituting the values: $X_L = 2 \times 3.1416 \times 50 \times 1 = 314.16 \, \Omega$.
The current $i$ is given by $i = \frac{V}{X_L}$.
$i = \frac{200}{314.16} \approx 0.637 \, A$.

(D) The inductive reactance $X_L$ is given by the formula: $X_L = 2 \pi \nu L$, where $\nu$ is the frequency and $L$ is the inductance.
Given: $X_L = 50 \, \Omega$ and $\nu = 50 \, Hz$.
Rearranging the formula for $L$: $L = \frac{X_L}{2 \pi \nu}$.
Substituting the values: $L = \frac{50}{2 \times 3.14 \times 50} = \frac{1}{2 \times 3.14} = \frac{1}{6.28} \approx 0.159 \, H$.
Rounding to two decimal places, we get $L \approx 0.16 \, H$.

(B) The capacitive reactance of a capacitor is given by the formula: ${X_C} = \frac{1}{{2\pi \nu C}}$.
From this relation,it is clear that ${X_C} \propto \frac{1}{\nu }$.
As the frequency $\nu$ increases,the capacitive reactance ${X_C}$ decreases.
Therefore,for high frequency,a capacitor offers less reactance.

(B) The correct answer is $(b)$.
In an $AC$ circuit,a choke coil is essentially an inductor with high inductance and negligible resistance.
It is used to limit or decrease the current in the circuit without significant loss of electrical energy.
Unlike a resistor,which dissipates energy as heat ($I^2R$ loss),an ideal choke coil consumes no power because the phase difference between voltage and current is $90^{\circ}$,resulting in a power factor of $\cos(90^{\circ}) = 0$.

(C) The formula for inductive reactance is given by $X_L = 2\pi \nu L$.
Given,inductance $L = \frac{1}{\pi } \, H$ and frequency $\nu = 50 \, Hz$.
Substituting these values into the formula:
$X_L = 2 \times \pi \times 50 \times \frac{1}{\pi }$.
$X_L = 2 \times 50 = 100 \, \Omega$.
Therefore,the correct option is $C$.

(A) The capacitive reactance $X_C$ is given by the formula: $X_C = \frac{1}{2\pi \nu C}$.
Given: $X_C = 25 \ \Omega$,$\nu = \frac{400}{\pi} \ Hz$.
Rearranging the formula for $C$: $C = \frac{1}{2\pi \nu X_C}$.
Substituting the values: $C = \frac{1}{2 \times \pi \times (\frac{400}{\pi}) \times 25}$.
$C = \frac{1}{2 \times 400 \times 25} = \frac{1}{20000} \ F$.
Converting to $\mu F$: $C = \frac{1}{20000} \times 10^6 \ \mu F = \frac{100}{2} \ \mu F = 50 \ \mu F$.

(C) The inductive reactance $X_L$ is given by $X_L = 2\pi fL$.
Given $f = 60\,Hz$ and $L = 0.7\,H$,we have:
$X_L = 2 \times 3.1416 \times 60 \times 0.7 = 263.89\,\Omega$.
The current $I$ in the inductor is given by $I = \frac{V}{X_L}$.
Substituting the values: $I = \frac{120}{263.89} \approx 0.455\,A$.

(A) In a purely capacitive $AC$ circuit,the voltage across the capacitor is given by $V = V_m \sin(\omega t)$.
The current in the circuit is given by $I = C \frac{dV}{dt} = C \frac{d}{dt}(V_m \sin(\omega t)) = \omega C V_m \cos(\omega t)$.
This can be written as $I = I_m \sin(\omega t + \frac{\pi}{2})$,where $I_m = \omega C V_m$.
Comparing the phases,the current leads the potential difference by a phase angle of $\frac{\pi}{2}$ radians (or $90^{\circ}$).
Therefore,the current is forward relative to the potential.

(B) In an $LCR$ circuit connected to an $ac$ supply,the resistor $(R)$ is the only component that dissipates energy in the form of heat,given by the formula $E = i^2 Rt$.
Inductors $(L)$ and capacitors $(C)$ do not dissipate energy; instead,they store energy in the form of magnetic fields and electric fields,respectively,and return it to the circuit.

(D) The instantaneous values of $e.m.f.$ $(E)$ and current $(i)$ in a purely inductive circuit are given by $E = E_0 \sin(\omega t)$ and $i = i_0 \sin(\omega t - \frac{\pi}{2})$.
The instantaneous power $P_{inst}$ is the product of $E$ and $i$:
$P_{inst} = E \cdot i = E_0 \sin(\omega t) \cdot i_0 \sin(\omega t - \frac{\pi}{2})$
Using the trigonometric identity $\sin(\omega t - \frac{\pi}{2}) = -\cos(\omega t)$:
$P_{inst} = E_0 i_0 \sin(\omega t) \cdot (-\cos(\omega t))$
$P_{inst} = -E_0 i_0 \sin(\omega t) \cos(\omega t)$
Using the double angle identity $\sin(2\theta) = 2 \sin \theta \cos \theta$:
$P_{inst} = -\frac{1}{2} E_0 i_0 \sin(2\omega t)$
The angular frequency of the instantaneous power is the coefficient of $t$ inside the sine function,which is $2\omega$.

(B) For a capacitive circuit,the capacitive reactance is given by $X_C = \frac{1}{\omega C}$.
The current $I$ in the circuit is given by $I = \frac{V}{X_C}$.
Substituting the expression for $X_C$,we get $I = \frac{V}{1/(\omega C)} = V \omega C$.
Since the voltage $V$ and capacitance $C$ are constant,the current $I$ is directly proportional to the angular frequency $\omega$ $(I \propto \omega)$.
This represents a straight line passing through the origin. Therefore,the graph in option $B$ correctly depicts the variation of current with frequency.

(C) In a purely inductive circuit,the voltage $V$ leads the current $i$ by a phase angle of $90^o$ (or $\pi/2$ radians).
If the voltage is represented as $V = V_m \sin(\omega t + \phi)$,then the current is represented as $i = I_m \sin(\omega t + \phi - 90^o) = -I_m \cos(\omega t + \phi)$.
Looking at the provided voltage graph,the voltage is at its maximum at $t = 0$. This corresponds to a cosine function $V = V_m \cos(\omega t)$.
Therefore,the current must be $i = I_m \sin(\omega t - 90^o) = -I_m \cos(\omega t)$.
At $t = 0$,$i = -I_m$,which means the current graph should start from the negative peak. Among the given options,the graph that represents a negative cosine wave is option $C$.

(C) The inductive reactance $X_L$ of an inductor is given by the formula: $X_L = 2\pi fL$, where $f$ is the frequency and $L$ is the inductance.
Taking the reciprocal of both sides, we get: $\frac{1}{X_L} = \frac{1}{2\pi fL}$.
This can be rewritten as: $\frac{1}{X_L} = \left( \frac{1}{2\pi L} \right) \cdot \frac{1}{f}$.
Let $y = \frac{1}{X_L}$ and $x = f$. Then the equation becomes $y = \frac{k}{x}$, where $k = \frac{1}{2\pi L}$ is a constant.
This is the equation of a rectangular hyperbola. Therefore, the graph between $\frac{1}{X_L}$ and $f$ is a rectangular hyperbola.

(C) The capacitive reactance is given by $X_C = \frac{1}{2\pi f C}$,which implies that $X_C \propto \frac{1}{f}$. This represents a rectangular hyperbola,which matches the graph of $X_1$.
The inductive reactance is given by $X_L = 2\pi f L$,which implies that $X_L \propto f$. This represents a straight line passing through the origin,which matches the graph of $X_2$.
Therefore,$X_1$ is a capacitor and $X_2$ is an inductor.

(B) The capacitive reactance $X_C$ is given by the formula:
$X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}$
This shows that $X_C$ is inversely proportional to the frequency $f$,i.e.,$X_C \propto \frac{1}{f}$.
As the frequency $f$ increases,the capacitive reactance $X_C$ decreases hyperbolically.
Therefore,the curve representing this relationship is a rectangular hyperbola,which corresponds to Graph $B$.

(D) The voltage across the inductor is $E = E_0 \sin(\omega t)$.
In a pure inductor,the current lags the voltage by $\pi/2$,so $i = i_0 \sin(\omega t - \pi/2) = -i_0 \cos(\omega t)$.
The instantaneous power $P$ is given by $P = E \cdot i$.
$P = (E_0 \sin(\omega t)) \cdot (-i_0 \cos(\omega t))$.
$P = -E_0 i_0 \sin(\omega t) \cos(\omega t)$.
Using the trigonometric identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$,we get:
$P = -\frac{E_0 i_0}{2} \sin(2\omega t)$.
The frequency of the instantaneous power is the coefficient of $t$ divided by $2\pi$. Since the angular frequency of the power is $2\omega$,the frequency is $2\omega$.

(B) Given,the alternating voltage is $e = 200 \sqrt{2} \sin(100 t) \; V$.
Comparing this with the standard equation $e = E_0 \sin(\omega t)$,we get peak voltage $E_0 = 200 \sqrt{2} \; V$ and angular frequency $\omega = 100 \; rad/s$.
The rms voltage is $E_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \; V$.
The capacitance is $C = 1 \; \mu F = 1 \times 10^{-6} \; F$.
The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \; \Omega$.
The rms current $I_{\text{rms}}$ is given by $I_{\text{rms}} = \frac{E_{\text{rms}}}{X_C} = \frac{200}{10^4} = 2 \times 10^{-2} \; A$.
Converting to milliamperes $(mA)$,$I_{\text{rms}} = 2 \times 10^{-2} \times 10^3 \; mA = 20 \; mA$.

(A) When an ideal capacitor is connected to an $AC$ voltage source,the current $I(t)$ leads the voltage $V(t)$ by a phase angle of $90^\circ$.
Since the energy stored in the capacitor during the charging phase is returned to the circuit during the discharging phase,the net energy consumed by an ideal capacitor over a complete cycle is zero.
Therefore,statement $A$ is the correct statement.

(A) Given: $f = 50\, Hz$,$C = 100\, \mu F = 10^{-4}\, F$,$I_m = 1.57\, A$.
Angular frequency $\omega = 2\pi f = 2 \times \pi \times 50 = 100\pi\, rad/s$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 10^{-4}} = \frac{100}{\pi} \approx \frac{100}{3.14} \approx 31.85\, \Omega$.
Peak voltage $E_m = I_m \times X_C = 1.57 \times 31.85 \approx 50\, V$.
In a purely capacitive circuit,the voltage lags behind the current by $\pi / 2$ radians. If current $I = I_m \sin(\omega t)$,then voltage $E = E_m \sin(\omega t - \pi / 2)$.
Substituting the values: $E = 50\, \sin(100\pi t - \pi / 2)$.

(D) The capacitive reactance $X$ is given by the formula $X = \frac{1}{2 \pi f C}$,where $f$ is the frequency and $C$ is the capacitance.
Given that the new capacitance $C^{\prime} = 2C$ and the new frequency $f^{\prime} = 2f$.
The new reactance $X^{\prime}$ is calculated as:
$X^{\prime} = \frac{1}{2 \pi f^{\prime} C^{\prime}} = \frac{1}{2 \pi (2f) (2C)} = \frac{1}{4 (2 \pi f C)}$.
Since $X = \frac{1}{2 \pi f C}$,we substitute this into the equation:
$X^{\prime} = \frac{X}{4}$.

(D) The inductive reactance is given by $X_L = \omega L = 2\pi f L$.
For $f_1 = 50\,Hz$ and $L = 2\,H$,the reactance is $X_{L1} = 2\pi \times 50 \times 2 = 200\pi\,\Omega$.
The initial current is $I = V / X_{L1} = V / (200\pi)$.
When the frequency is changed to $f_2 = 400\,Hz$,the new inductive reactance is $X_{L2} = 2\pi \times 400 \times 2 = 1600\pi\,\Omega$.
The new current $I'$ is $I' = V / X_{L2} = V / (1600\pi)$.
Comparing the two currents: $I' = (V / 200\pi) / 8 = I / 8$.
In a purely inductive circuit,the current lags the voltage by $90^o$.

(B) The given voltage is $V = V_{0} \sin(\omega t)$,where $V_{0} = 300\sqrt{2} \text{ V}$ and $\omega = 100 \text{ rad/s}$.
The capacitive reactance is $X_{C} = \frac{1}{\omega C} = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^{4} \ \Omega$.
The peak current is $i_{0} = \frac{V_{0}}{X_{C}} = \frac{300\sqrt{2}}{10^{4}} = 300\sqrt{2} \times 10^{-4} \text{ A} = 30\sqrt{2} \times 10^{-3} \text{ A} = 30\sqrt{2} \text{ mA}$.
The $AC$ ammeter measures the $rms$ value of the current.
$i_{rms} = \frac{i_{0}}{\sqrt{2}} = \frac{30\sqrt{2}}{\sqrt{2}} = 30 \text{ mA}$.

(D) In a pure inductor,the current lags behind the voltage by a phase angle of $\pi/2$ or $90^\circ$.
If the voltage is represented as $V = V_m \cos(\omega t)$,then the current is given by $i = I_m \cos(\omega t - \pi/2) = I_m \sin(\omega t)$.
The given voltage graph is a cosine wave starting at its maximum value at $t = 0$.
Since the current lags the voltage by $90^\circ$,the current graph should be a sine wave starting from zero at $t = 0$ and increasing in the positive direction.
Comparing this with the given options,the graph that represents a sine wave starting from zero at $t = 0$ is Option $D$.

(B) In a pure inductive circuit,the voltage across the inductor leads the current flowing through it by a phase angle of $\frac{\pi}{2}$ radians (or $90^{\circ}$).
This occurs because the induced electromotive force $(EMF)$ in the inductor opposes the change in current,causing the voltage to reach its peak before the current.

(C) For a pure inductor,the current $i$ lags behind the voltage $V$ by a phase angle of $\pi / 2$.
If the voltage is represented as $V = V_m \sin(\omega t)$,then the current is $i = I_m \sin(\omega t - \pi / 2) = -I_m \cos(\omega t)$.
In the given diagram,the voltage $V$ is shown as a sine wave starting from $0$ and increasing.
Therefore,the current $i$ must be a negative cosine wave,which starts at a negative maximum value at $t = 0$ and increases to $0$ at $t = \pi / (2\omega)$.
This corresponds to the graph shown in option $C$.

(B) In a pure inductive circuit,the current always lags behind the electromotive force (emf) by a phase angle of $\frac{\pi}{2}$.
Given the voltage $V(t) = V_0 \sin(\omega t)$,the current is given by $I(t) = I_0 \sin(\omega t - \frac{\pi}{2})$.
Here,$V_0 = 100 \, V$ and $\omega = 500 \, rad/s$.
The peak current $I_0$ is calculated as $I_0 = \frac{V_0}{\omega L} = \frac{100}{500 \times 0.02} = \frac{100}{10} = 10 \, A$.
Substituting the values into the current equation:
$I(t) = 10 \sin(500t - \frac{\pi}{2})$.
Using the trigonometric identity $\sin(\theta - \frac{\pi}{2}) = -\cos(\theta)$,we get:
$I(t) = -10 \cos(500t)$.

(B) The given alternating voltage is $v(t) = 220 \sin(100 \pi t)$.
Since the load is purely resistive,the current $i(t)$ is in phase with the voltage,given by $i(t) = I_0 \sin(100 \pi t)$,where $I_0$ is the peak current.
The current rises from zero to half of its peak value,so $i(t) = \frac{I_0}{2}$.
Substituting this into the equation: $\frac{I_0}{2} = I_0 \sin(100 \pi t) \implies \sin(100 \pi t) = \frac{1}{2}$.
This implies $100 \pi t = \frac{\pi}{6}$ (since $\sin(30^\circ) = 0.5$).
Solving for $t$: $t = \frac{\pi}{6 \times 100 \pi} = \frac{1}{600} \text{ s}$.
Converting to milliseconds: $t = \frac{1000}{600} \text{ ms} = 1.67 \text{ ms}$.
Wait,re-evaluating the question's intent based on the provided options and image: The image shows $\theta = \pi/6$ for $A/2$. If the question asks for the time to reach half the peak value,$t = 1.67 \text{ ms}$. If the question implies the time to reach the peak from half peak,it would be $t = \frac{\pi/2 - \pi/6}{100 \pi} = \frac{\pi/3}{100 \pi} = 3.33 \text{ ms}$. Given the options,$3.3 \text{ ms}$ is the intended answer.

(B) choke coil is an inductor used to control the current in an $AC$ circuit without significant power loss. The power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$ is the power factor. For an ideal choke coil,the resistance $R$ should be as small as possible compared to the inductive reactance $X_L$. Therefore,for an efficient choke coil,$X_L >> R$.

(C) In a pure resistive $AC$ circuit, the voltage $V = V_m \sin(\omega t)$ and the current $I = I_m \sin(\omega t)$ are in the same phase.
This means the phase difference between the voltage and the current is zero.
Therefore, the current and voltage reach their maximum and minimum values at the same time.

(B) The current through the inductor is given by $I = I_0 \sin(\omega t)$.
The voltage across an inductor is given by $V_L = L \frac{dI}{dt}$.
Substituting the expression for $I$:
$V_L = L \frac{d}{dt} (I_0 \sin(\omega t))$
$V_L = L I_0 \omega \cos(\omega t)$.
Using the trigonometric identity $\cos(\theta) = \sin(\theta + \pi/2)$:
$V_L = I_0 \omega L \sin(\omega t + \pi/2)$.
Thus,the voltage leads the current by a phase of $\pi/2$.

(C) The amplitude of the applied voltage $V_c$ is independent of the frequency of the $A.C.$ source,so it remains the same.
The capacitive reactance is given by $X_c = \frac{1}{2\pi f C}$.
As the frequency $f$ decreases,the capacitive reactance $X_c$ increases.
The amplitude of the current is given by $I_c = \frac{V_c}{X_c}$.
Since $V_c$ is constant and $X_c$ increases,the amplitude of the current $I_c$ decreases.

(A) The reactance of an inductor is given by $X_L = \omega L = 2 \pi f L$. As the frequency $f$ increases,$X_L$ increases.
The reactance of a capacitor is given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$. As the frequency $f$ increases,$X_C$ decreases.
The resistance of a resistor $R$ is independent of frequency.
Since the question asks what happens as the frequency is increased from a very low value,the inductive reactance $(X_L)$ increases directly with frequency.

(C) In an $AC$ circuit with a pure capacitor,the phase difference between voltage and current is $\phi = -\pi/2$.
The average power consumed is given by $P_{av} = E_v I_v \cos(\phi)$.
Substituting the value of $\phi$,we get $P_{av} = E_v I_v \cos(-\pi/2) = E_v I_v (0) = 0$.
Thus,there is no power loss in a pure capacitor circuit.
However,the Reason states that no current is flowing in the circuit,which is incorrect because an alternating current $(AC)$ does flow through the capacitor due to its capacitive reactance $X_C = 1/(2\pi f C)$.
Therefore,the Assertion is correct,but the Reason is incorrect.

(A) The inductive reactance $X_L$ is given by the formula $X_L = 2 \pi \nu L$.
Given: $L = 25.0 \; mH = 25 \times 10^{-3} \; H$,$\nu = 50 \; Hz$,and $V_{rms} = 220 \; V$.
Substituting the values: $X_L = 2 \times 3.14 \times 50 \times 25 \times 10^{-3} \; \Omega = 7.85 \; \Omega$.
The rms current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_L}$.
Substituting the values: $I_{rms} = \frac{220 \; V}{7.85 \; \Omega} \approx 28.03 \; A$.

(N/A) When a $dc$ source is connected to a capacitor, the capacitor gets charged and after charging, no current flows in the circuit; therefore, the lamp will not glow. There will be no change even if $C$ is reduced.
With an $ac$ source, the capacitor offers capacitive reactance $X_C = 1 / (\omega C)$ and current flows in the circuit. Consequently, the lamp will shine.
If the capacitance $C$ is reduced, the capacitive reactance $X_C = 1 / (\omega C)$ increases. As a result, the current in the circuit decreases, and the lamp will shine less brightly than before.