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

(D) The current in an inductive circuit is given by $I_{L} = \frac{V}{X_{L}}$,where $X_{L} = \omega L = 2\pi f L$. Thus,$I_{L} = \frac{V}{2\pi f L}$. As frequency $f$ increases,$I_{L}$ decreases.
The current in a capacitive circuit is given by $I_{C} = \frac{V}{X_{C}}$,where $X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. Thus,$I_{C} = V(2\pi f C)$. As frequency $f$ increases,$I_{C}$ increases.
Therefore,the current decreases in the first circuit and increases in the second circuit.

(A) The given alternating voltage is $E = E_0 \sin(\omega t)$, where $E_0 = 100 \sqrt{2} \text{ V}$ and $\omega = 50 \text{ rad/s}$.
The root mean square (rms) voltage is $E_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 \text{ V}$.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{50 \times 2 \times 10^{-6}} = \frac{1}{100 \times 10^{-6}} = 10^4 \Omega$.
The ammeter measures the rms current $I_{\text{rms}}$, which is given by $I_{\text{rms}} = \frac{E_{\text{rms}}}{X_C}$.
Substituting the values, $I_{\text{rms}} = \frac{100}{10^4} = 10^{-2} \text{ A} = 10 \text{ mA}$.
Thus, the ammeter reading is $10 \text{ mA}$.

(C) The inductive reactance of a coil is given by the formula $X_L = \omega L = 2\pi f L$,where $f$ is the frequency of the source.
For an $A.C.$ source,the frequency $f$ is non-zero,so the reactance $X_{AC} = 2\pi f L$.
For a $D.C.$ source,the frequency $f = 0$,so the reactance $X_{DC} = 2\pi (0) L = 0$.
The ratio of the reactance when connected to an $A.C.$ source to that when connected to a $D.C.$ source is $\frac{X_{AC}}{X_{DC}} = \frac{2\pi f L}{0} = \infty$.
Therefore,the ratio is $\infty$.

(B) In a pure inductor,the current lags behind the applied $e.m.f.$ (voltage) by a phase angle of $90^{\circ}$ or $\pi/2$ radians.
If the voltage is represented as $e_L = E_0 \sin(\omega t)$,then the current is $i_L = I_0 \sin(\omega t - \pi/2)$.
In phasor diagrams,this means the current vector $i_L$ is $90^{\circ}$ clockwise from the voltage vector $e_L$.
Looking at the given options:
Figure $(A)$ shows them in opposite directions $(180^{\circ})$.
Figure $(B)$ shows $i_L$ lagging $e_L$ by $90^{\circ}$ (clockwise).
Figure $(C)$ shows them in phase $(0^{\circ})$.
Figure $(D)$ shows $i_L$ leading $e_L$ by $90^{\circ}$ (counter-clockwise).
Therefore,figure $(B)$ correctly represents the phase relationship.

(D) In an $A.C.$ circuit containing only an inductor, the inductive reactance is given by $X_L = 2\pi fL$. The current is $I = V / X_L = V / (2\pi fL)$. As frequency $f$ increases, $X_L$ increases, so the current $I$ decreases.
In an $A.C.$ circuit containing only a capacitor, the capacitive reactance is given by $X_C = 1 / (2\pi fC)$. The current is $I = V / X_C = V \cdot (2\pi fC)$. As frequency $f$ increases, $X_C$ decreases, so the current $I$ increases.
Therefore, the current decreases in the first circuit and increases in the second circuit.

(C) When an $A.C.$ voltage $V = V_m \sin(\omega t)$ is applied to a pure inductor of inductance $L$,the induced electromotive force $(EMF)$ is given by $\varepsilon = -L \frac{di}{dt}$.
Applying Kirchhoff's voltage law,$V - L \frac{di}{dt} = 0$,which implies $V = L \frac{di}{dt}$.
Substituting $V$,we get $V_m \sin(\omega t) = L \frac{di}{dt}$,so $di = \frac{V_m}{L} \sin(\omega t) dt$.
Integrating both sides,$i = \int \frac{V_m}{L} \sin(\omega t) dt = -\frac{V_m}{\omega L} \cos(\omega t) = \frac{V_m}{\omega L} \sin(\omega t - \pi / 2)$.
Comparing the phase of voltage $(\omega t)$ and current $(\omega t - \pi / 2)$,it is clear that the current lags behind the voltage by a phase angle of $\pi / 2$ radians.

(C) The capacitive reactance $X_{C}$ is given by the formula $X_{C} = \frac{1}{2 \pi f C}$,where $f$ is the frequency and $C$ is the capacitance.
Let the initial reactance be $X_{C} = \frac{1}{2 \pi f C}$.
When the frequency $f$ is doubled $(f' = 2f)$ and the capacitance $C$ is doubled $(C' = 2C)$,the new reactance $X_{C}'$ is given by:
$X_{C}' = \frac{1}{2 \pi f' C'} = \frac{1}{2 \pi (2f) (2C)}$.
$X_{C}' = \frac{1}{4 (2 \pi f C)} = \frac{1}{4} X_{C}$.
Therefore,the new reactance is $\frac{X_{C}}{4}$.

(D) In a series $LCR$ circuit connected to an $A.C.$ source:
$1$. In a pure resistor $(R)$, the current and voltage are always in phase.
$2$. In a pure inductor $(L)$, the voltage leads the current by a phase angle of $90^{\circ}$ ($\pi/2$ radians), meaning they are out of phase.
$3$. In a pure capacitor $(C)$, the current leads the voltage by a phase angle of $90^{\circ}$ ($\pi/2$ radians), meaning they are out of phase.
Therefore, in both the inductor $(L)$ and the capacitor $(C)$, the current and voltage are out of phase with each other. Thus, option $D$ is correct.

(C) The power consumed in an $A$.$C$. circuit containing a resistor is given by the formula:
$P = I_{rms}^2 R$
Given:
$P = 108 \text{ W}$
$I_{rms} = 3 \text{ A}$
Substituting the values into the formula:
$108 = (3)^2 \times R$
$108 = 9 \times R$
$R = \frac{108}{9} = 12 \ \Omega$
Therefore,the resistance in the circuit is $12 \ \Omega$.

(A) Given: Inductance $L = 42 \, mH = 42 \times 10^{-3} \, H$, Voltage $V_{rms} = 200 \, V$, Frequency $f = 50 \, Hz$.
First, calculate the inductive reactance $X_L$ using the formula $X_L = \omega L = 2 \pi f L$.
Substituting the values: $X_L = 2 \times \frac{22}{7} \times 50 \times 42 \times 10^{-3} \, \Omega$.
$X_L = 2 \times 22 \times 50 \times 6 \times 10^{-3} \, \Omega = 4400 \times 6 \times 10^{-3} \, \Omega = 13.2 \, \Omega$.
The r.m.s. current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_L}$.
$I_{rms} = \frac{200}{13.2} \approx 15.15 \, A$.

(C) $1$. Ideal Inductor:
- The voltage and current in an ideal inductor are out of phase by $90^{\circ}$,with the current lagging the voltage by $90^{\circ}$.
- The instantaneous power is given by $p(t) = v(t) \times i(t)$.
- Over a complete cycle,the positive and negative power values cancel out,resulting in an average power of zero.
$2$. Ideal Capacitor:
- In an ideal capacitor,the voltage and current are also out of phase by $90^{\circ}$,but the current leads the voltage by $90^{\circ}$.
- Similar to the inductor case,the instantaneous power alternates between positive and negative,resulting in an average power of zero over a complete cycle.

(B) The capacitive reactance $X_C$ is given by the formula: $X_C = \frac{1}{2 \pi f C}$.
From this,we can see that $X_C \propto \frac{1}{f}$.
Given initial frequency $f_1 = 50 \ Hz$ and initial reactance $X_{C1} = 5 \ \Omega$.
New frequency $f_2 = 100 \ Hz$.
Since $f_2 = 2 f_1$,the new reactance $X_{C2}$ will be:
$X_{C2} = \frac{X_{C1}}{2} = \frac{5 \ \Omega}{2} = 2.5 \ \Omega$.

(D) The inductive reactance is given by the formula $X_{L} = \omega L = 2 \pi f L$.
Given that the new frequency $f' = 2f$ and the new inductance $L' = 3L$.
The new inductive reactance $X_{L}'$ is calculated as:
$X_{L}' = 2 \pi f' L'$
$X_{L}' = 2 \pi (2f) (3L)$
$X_{L}' = 6 \times (2 \pi f L)$
Since $X_{L} = 2 \pi f L$,we have $X_{L}' = 6 X_{L}$.

(A) The formula for capacitive reactance is $X_C = \frac{1}{2 \pi f C}$.
Given,the initial capacitive reactance is $X_C = X \ \Omega$.
So,$X = \frac{1}{2 \pi f C}$.
When the frequency is doubled $(f' = 2f)$ and the capacitance is doubled $(C' = 2C)$,the new capacitive reactance $X_C'$ is:
$X_C' = \frac{1}{2 \pi f' C'} = \frac{1}{2 \pi (2f) (2C)}$.
$X_C' = \frac{1}{4 (2 \pi f C)}$.
Since $X = \frac{1}{2 \pi f C}$,we substitute this into the equation:
$X_C' = \frac{X}{4} \ \Omega$.

(B) The given alternating voltage is $E = E_0 \sin(\omega t)$, where $E_0 = 200 \sqrt{2} \text{ V}$ and $\omega = 100 \text{ rad/s}$.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 1 \times 10^{-6}} = 10^4 \Omega$.
The peak current is $I_0 = \frac{E_0}{X_C} = \frac{200 \sqrt{2}}{10^4} = 2 \sqrt{2} \times 10^{-2} \text{ A}$.
The $A.C.$ ammeter measures the root mean square $(RMS)$ current, $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$.
$I_{\text{rms}} = \frac{2 \sqrt{2} \times 10^{-2}}{\sqrt{2}} = 2 \times 10^{-2} \text{ A} = 20 \text{ mA}$.

(C) In a pure resistive circuit,the voltage $(e_R)$ and current $(i_R)$ are in the same phase. This means that they reach their maximum and minimum values at the same time. The phase difference between them is $0$. Therefore,the phasor diagram representing this relationship shows both vectors pointing in the same direction,as depicted in figure $(A)$.

(A) The inductive reactance $X_L$ is given by the formula $X_L = \omega L$, where $\omega = 2 \pi f$.
Substituting the value of $\omega$, we get $X_L = 2 \pi f L$.
Given $X_L = 11 \, \Omega$ and $L = 0.07 \, H$, we can solve for frequency $f$:
$f = \frac{X_L}{2 \pi L} = \frac{11}{2 \times 3.14159 \times 0.07} \approx \frac{11}{0.4398} \approx 25 \, Hz$.
Thus, the frequency of the alternating voltage is $25 \, Hz$.

(C) The capacitive reactance $X_c$ is given by the formula: $X_c = \frac{1}{2 \pi f C}$.
Given values are $X_c = \frac{1}{1000} \Omega = 10^{-3} \Omega$ and $C = 5 \mu F = 5 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$10^{-3} = \frac{1}{2 \pi f (5 \times 10^{-6})}$
$10^{-3} = \frac{1}{10 \pi f \times 10^{-6}}$
$10^{-3} = \frac{1}{\pi f \times 10^{-5}}$
Rearranging for frequency $f$:
$f = \frac{1}{10^{-3} \times \pi \times 10^{-5}} = \frac{1}{\pi \times 10^{-8}} = \frac{10^8}{\pi} \text{ Hz}$.
Since $1 \text{ MHz} = 10^6 \text{ Hz}$,we have $f = \frac{100 \times 10^6}{\pi} \text{ Hz} = \frac{100}{\pi} \text{ MHz}$.

(A) The inductive reactance $X_L$ is given by the formula $X_L = 2 \pi f L$,where $f$ is the frequency and $L$ is the inductance.
Rearranging for frequency,we get $f = \frac{X_L}{2 \pi L}$.
Given values: $X_L = 330 \Omega$ and $L = 0.25 mH = 0.25 \times 10^{-3} H$.
Substituting the values: $f = \frac{330}{2 \times (22/7) \times 0.25 \times 10^{-3}}$.
$f = \frac{330 \times 7}{2 \times 22 \times 0.25 \times 10^{-3}} = \frac{2310}{11 \times 10^{-3}} = 210 \times 10^3 Hz = 210 kHz$.

(C) The expression for inductive reactance $(X_L)$ is given by:
$X_L = 2 \pi f L$
Where $f$ is the frequency of the alternating current and $L$ is the inductance.
From the formula,it is clear that $X_L$ is directly proportional to both the frequency $(f)$ and the inductance $(L)$.
Therefore,$X_L \propto f$ and $X_L \propto L$.

(D) Given the voltage equation $e = 220 \sin(50t)$.
Comparing this with the standard form $e = E_0 \sin(\omega t)$,we get the peak voltage $E_0 = 220 \text{ V}$ and angular frequency $\omega = 50 \text{ rad/s}$.
The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C}$.
Substituting the values: $X_C = \frac{1}{50 \times 50 \times 10^{-6}} = \frac{1}{2500 \times 10^{-6}} = \frac{10^6}{2500} = 400 \Omega$.
The peak current $I_0$ is given by $I_0 = \frac{E_0}{X_C}$.
$I_0 = \frac{220}{400} = \frac{22}{40} = 0.55 \text{ A}$.

(D) The formula for inductive reactance $(X_L)$ is given by $X_L = 2 \pi f L$,where $f$ is the frequency and $L$ is the inductance.
Given that the initial inductive reactance is $R = 2 \pi f L$.
If the new inductance $L' = 2L$ and the new frequency $f' = 2f$,the new inductive reactance $X_L'$ will be:
$X_L' = 2 \pi f' L' = 2 \pi (2f) (2L) = 4 (2 \pi f L) = 4R$.
Therefore,the new inductive reactance is $4R$.

(A) The r.m.s. current $I_{rms}$ in a capacitive circuit is given by $I_{rms} = \frac{V_{rms}}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
Substituting $X_C$,we get $I_{rms} = V_{rms} \cdot \omega \cdot C$.
Given values are $V_{rms} = 230 \text{ V}$,$\omega = 300 \text{ rad/s}$,and $C = 100 \text{ pF} = 100 \times 10^{-12} \text{ F}$.
$I_{rms} = 230 \times 300 \times 100 \times 10^{-12} \text{ A}$.
$I_{rms} = 230 \times 3 \times 10^4 \times 10^{-12} \text{ A}$.
$I_{rms} = 690 \times 10^{-8} \text{ A} = 6.9 \times 10^{-6} \text{ A}$.

(B) The inductive reactance $X_{L}$ of an inductor is given by the formula $X_{L} = 2 \pi f L$,where $f$ is the frequency of the $a.c.$ supply and $L$ is the self-inductance.
From this relation,it is clear that $X_{L} \propto f$.
Therefore,if the frequency $f$ increases,the inductive reactance $X_{L}$ also increases linearly.

(B) The capacitive reactance $X_{C}$ is given by the formula: $X_{C} = \frac{1}{2 \pi f C}$.
From this formula,it is clear that the capacitive reactance is inversely proportional to the frequency,i.e.,$X_{C} \propto \frac{1}{f}$.
Let the initial frequency be $f$ and the initial reactance be $X_{C}$.
Let the new frequency be $f' = 4f$ and the new reactance be $X_{C}'$.
Using the proportionality $X_{C} \cdot f = \text{constant}$,we get:
$X_{C}' \cdot f' = X_{C} \cdot f$
$X_{C}' \cdot (4f) = X_{C} \cdot f$
$X_{C}' = \frac{X_{C}}{4}$.

(D) The given alternating voltage is $e = e_0 \sin(\omega t)$, where $e_0 = 200 \sqrt{2} \text{ V}$ 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 $RMS$ voltage is $V_{rms} = \frac{e_0}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \text{ V}$.
The reading of the $AC$ ammeter gives the $RMS$ current $I_{rms} = \frac{V_{rms}}{X_C}$.
Substituting the values, $I_{rms} = \frac{200}{10^4} = 2 \times 10^{-2} \text{ A} = 20 \text{ mA}$.

(A) choke coil is an inductor with high inductance and low resistance. It is used to control the current in $AC$ circuits without dissipating significant power as heat. In $DC$ circuits,an inductor acts as a simple wire with negligible resistance,so it cannot be used to control current effectively. Therefore,a choke coil is used as a resistance (impedance) in $AC$ circuits.

(C) In $AC$ circuits,a resistor dissipates energy in the form of heat due to the $I^2R$ loss,regardless of the phase difference.
However,an ideal choke coil is an inductor with negligible resistance.
In an ideal inductor,the phase difference between voltage and current is $90^{\circ}$,which makes the power factor $\cos(90^{\circ}) = 0$.
Therefore,the average power consumed by an ideal choke coil is $P = V_{rms} I_{rms} \cos(90^{\circ}) = 0$.
Thus,a choke coil controls the current in an $AC$ circuit without wasting electrical energy as heat.

(A) The applied voltage is $e = e_0 \sin \omega t$.
In an inductor,the current lags the voltage by $\frac{\pi}{2}$. The inductive reactance is $X_L = \omega L$. Thus,the instantaneous current $i_L$ is:
$i_L = \frac{e_0}{\omega L} \sin \left(\omega t - \frac{\pi}{2}\right) = -\frac{e_0}{\omega L} \cos \omega t$
In a capacitor,the current leads the voltage by $\frac{\pi}{2}$. The capacitive reactance is $X_C = \frac{1}{\omega C}$. Thus,the instantaneous current $i_C$ is:
$i_C = \frac{e_0}{X_C} \sin \left(\omega t + \frac{\pi}{2}\right) = e_0 \omega C \cos \omega t$
Therefore,the currents are $-\frac{e_0}{\omega L} \cos \omega t$ and $e_0 \omega C \cos \omega t$.

(A) The power factor of an $AC$ circuit is defined as $\cos \phi$,where $\phi$ is the phase difference between the voltage and the current.
In a purely inductive circuit,the current lags behind the voltage by a phase angle of $\phi = 90^\circ$ (or $\pi/2$ radians).
In a purely capacitive circuit,the current leads the voltage by a phase angle of $\phi = 90^\circ$ (or $\pi/2$ radians).
Therefore,the power factor is $\cos(90^\circ) = 0$ for both cases.

(D) In an alternating current circuit,the phase relationship between voltage and current depends on the components present.
For a purely inductive circuit (containing only an inductor $L$),the voltage leads the current by a phase angle of $\pi / 2$ $(90^{\circ})$,which is equivalent to saying the current lags behind the voltage by $\pi / 2$.
For a purely capacitive circuit (containing only a capacitor $C$),the current leads the voltage by a phase angle of $\pi / 2$ $(90^{\circ})$.
For a purely resistive circuit (containing only a resistor $R$),the current and voltage are in the same phase (phase difference is $0$).
Therefore,since the current lags behind the voltage by $\pi / 2$,the circuit must contain only an inductor $L$.

(A) The reactance of an inductor in an $A.C.$ circuit is given by $X_L = \omega L = 2\pi f L$,where $f$ is the frequency of the $A.C.$ source.
For a $D.C.$ source,the frequency $f = 0$.
Therefore,the inductive reactance in a $D.C.$ circuit is $X_L = 2\pi (0) L = 0$.
The ratio of reactance in an $A.C.$ source to that in a $D.C.$ source is $\frac{X_{L(ac)}}{X_{L(dc)}} = \frac{\omega L}{0} = \infty$.

(C) Step $1$: Given values:
$V = 220 \ V$
$L = 50 \ mH = 50 \times 10^{-3} \ H$
$\nu = 50 \ Hz$
Step $2$: The inductive reactance $X_L$ is given by $X_L = \omega L = 2 \pi \nu L$.
Step $3$: The rms current $I$ is given by $I = \frac{V}{X_L} = \frac{V}{2 \pi \nu L}$.
Step $4$: Substituting the values:
$I = \frac{220}{2 \times 3.14159 \times 50 \times 50 \times 10^{-3}}$
$I = \frac{220}{15.70795} \approx 14.005 \ A$.
Thus,the rms current is approximately $14 \ A$.

(C) Wattless current is defined as the current in an $AC$ circuit for which the average power consumed is zero.
Power in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$, where $\phi$ is the phase angle between voltage and current.
For $P$ to be zero, $\cos \phi$ must be zero, which implies $\phi = 90^{\circ}$.
In a purely inductive circuit (Only $L$) or a purely capacitive circuit (Only $C$), the phase difference between voltage and current is $90^{\circ}$.
Therefore, in a circuit containing only an ideal inductor (Only $L$) or only an ideal capacitor, the power factor $\cos 90^{\circ} = 0$, resulting in wattless current.
Among the given options, Only $L$ is the correct choice.

(D) Given: Capacitance $C = 50 \mu F = 50 \times 10^{-6} F$,Voltage $V_{rms} = 110 V$,Frequency $\nu = 60 Hz$.
First,calculate the capacitive reactance $X_C$ using the formula $X_C = \frac{1}{2 \pi \nu C}$.
$X_C = \frac{1}{2 \times 3.1416 \times 60 \times 50 \times 10^{-6}} \approx 53.05 \Omega$.
The rms current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_C}$.
$I_{rms} = \frac{110}{53.05} \approx 2.07 A$.
Rounding to one decimal place,we get $I_{rms} = 2.1 A$.

(D) The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2 \pi \nu C}$,where $\nu$ is the frequency of the source.
For a $DC$ source,the frequency $\nu = 0 \ Hz$.
Substituting this value into the formula,we get $X_C = \frac{1}{2 \pi (0) C} = \frac{1}{0}$.
Therefore,the reactance of the capacitor in a $DC$ circuit is infinite.

(A) The given alternating voltage is $V = 100 \sqrt{2} \sin(100t) \text{ V}$.
Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the peak voltage $V_m = 100 \sqrt{2} \text{ V}$ and angular frequency $\omega = 100 \text{ rad/s}$.
The capacitive reactance is given by $X_C = \frac{1}{\omega C}$.
Substituting the values,$X_C = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \text{ } \Omega$.
The root mean square $(RMS)$ voltage is $V_{\text{rms}} = \frac{V_m}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 \text{ V}$.
The $RMS$ current $I_{\text{rms}}$ is given by $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$.
$I_{\text{rms}} = \frac{100}{10^4} = 10^{-2} \text{ A}$.
Converting to milliamperes,$I_{\text{rms}} = 10^{-2} \times 10^3 \text{ mA} = 10 \text{ mA}$.

(B) In a purely capacitive $AC$ circuit,the voltage $V$ is given by $V = V_m \sin(\omega t)$ and the current $I$ is given by $I = I_m \sin(\omega t + \frac{\pi}{2})$.
This shows that the current leads the voltage by a phase angle of $\frac{\pi}{2}$ radians.
Therefore,the correct option is $B$.

(C) The correct option is $C$.
In a $DC$ circuit,the frequency $v = 0 \ Hz$.
The angular frequency is given by $\omega = 2 \pi v = 2 \pi \times 0 = 0 \ rad/s$.
The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{\omega C}$.
Substituting the values,we get $X_C = \frac{1}{0 \times 1} = \frac{1}{0} = \infty$.
Therefore,the capacitive reactance of a capacitor in a $DC$ circuit is infinite,meaning it acts as an open circuit.

(D) The average power dissipation $P$ 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.
For an ideal capacitor,the current leads the voltage by a phase angle of $\phi = \frac{\pi}{2}$ rad.
Substituting this into the power formula:
$P = V_{rms} I_{rms} \cos(\frac{\pi}{2})$
Since $\cos(\frac{\pi}{2}) = 0$,the average power dissipation is $P = 0$.

(A) The correct answer is $A$. Inductive reactance $(X_L)$ is defined as the opposition offered by an inductor to the flow of alternating current.
$X_L = \omega L = 2 \pi f L$
where $f$ is the frequency of the $AC$ source and $L$ is the inductance.
Since $X_L$ is directly proportional to the frequency $(f)$,it effectively limits the flow of alternating current in a circuit.
For $DC$ circuits,the frequency $f = 0$,which means $X_L = 0$. Therefore,a pure inductor offers no resistance to $DC$ current.

(A) The inductive reactance is given by the formula $X_{L} = \omega L$.
Here,$\omega = 2\pi\nu$,where $\nu$ is the frequency of the voltage source.
Substituting this into the formula,we get $X_{L} = 2\pi\nu L$.
Since $2$,$\pi$,and $L$ are constants,we have $X_{L} \propto \nu$.
This represents a linear relationship between inductive reactance and frequency,which is a straight line passing through the origin.
Therefore,the graph in option $A$ correctly represents this variation.

(A) Given: Capacitance $C = 10 \mu F = 10 \times 10^{-6} \text{ F}$.
Source voltage $V = 50 \sqrt{2} \sin 100 t$.
Comparing this with $V = V_{\max} \sin \omega t$, we get $V_{\max} = 50 \sqrt{2} \text{ V}$ and $\omega = 100 \text{ rad/s}$.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10 \times 10^{-6}} = \frac{1}{10^{-3}} = 1000 \Omega$.
The $RMS$ voltage is $V_{\text{rms}} = \frac{V_{\max}}{\sqrt{2}} = \frac{50 \sqrt{2}}{\sqrt{2}} = 50 \text{ V}$.
The reading of the $AC$ ammeter gives the $RMS$ current $I_{\text{rms}}$, which is $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{50}{1000} = 0.05 \text{ A}$.
Converting to milliamperes, $I_{\text{rms}} = 0.05 \times 1000 \text{ mA} = 50 \text{ mA}$.

(D) The average power dissipated in an $AC$ circuit is given by the formula $ P_{av} = VI \cos \phi $,where $ V $ is the $RMS$ voltage,$ I $ is the $RMS$ current,and $ \phi $ is the phase angle between voltage and current.
In a pure inductor,the current lags behind the voltage by a phase angle of $ \phi = \pi / 2 $.
Substituting this into the power formula: $ P_{av} = VI \cos(\pi / 2) $.
Since $ \cos(\pi / 2) = 0 $,we get $ P_{av} = VI \times 0 = 0 $.
Therefore,the average power dissipated in a pure inductor is zero.

(A) The capacitive reactance $X_C$ of a capacitor is given by the formula: $X_C = \frac{1}{2 \pi f C}$.
Here,$f$ is the frequency of the $AC$ source and $C$ is the capacitance.
From the formula,it is clear that $X_C \propto \frac{1}{f}$.
Given,initial capacitive reactance $X_{C1} = 3 \ k\Omega$ at frequency $f_1 = f$.
When the frequency is doubled,$f_2 = 2f$.
The new capacitive reactance $X_{C2}$ will be: $X_{C2} = \frac{1}{2 \pi (2f) C} = \frac{1}{2} \times \left( \frac{1}{2 \pi f C} \right) = \frac{X_{C1}}{2}$.
Substituting the value: $X_{C2} = \frac{3 \ k\Omega}{2} = 1.5 \ k\Omega$.

(B) Given: Inductance $L = 70 mH = 70 \times 10^{-3} H$,Voltage $V_{rms} = 220 V$,Frequency $f = 50 Hz$.
Inductive reactance is given by $\chi_{L} = \omega L = 2 \pi f L$.
Substituting the values: $\chi_{L} = 2 \times \pi \times 50 \times 70 \times 10^{-3} = 7 \pi \Omega$.
The rms current is $i_{rms} = \frac{V_{rms}}{\chi_{L}} = \frac{220}{7 \pi} \approx \frac{10}{\pi} A$ (Note: Based on standard textbook values,if $L = 70 mH$,the result is $\frac{220}{7 \pi} A$. However,checking the options,if $L = 70 mH$ and $V = 220 V$,the calculation yields $10 A$ only if $\pi \approx 3.14$ and $L$ was adjusted. Re-evaluating: $i_{rms} = \frac{220}{2 \pi \times 50 \times 70 \times 10^{-3}} = \frac{220}{7 \pi} \approx 10 A$ is not exact. Given the options,$10 A$ is the intended answer assuming approximation or specific values).

(A) Given: Capacitance $C = \frac{50}{\pi} \mu F = \frac{50}{\pi} \times 10^{-6} \, F$.
Source voltage $V_{rms} = 250 \, V$ and frequency $f = 50 \, Hz$.
The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2 \pi f C}$.
Substituting the values: $X_C = \frac{1}{2 \pi \times 50 \times (\frac{50}{\pi} \times 10^{-6})} = \frac{1}{100 \times 50 \times 10^{-6}} = \frac{1}{5000 \times 10^{-6}} = \frac{1}{0.005} = 200 \, \Omega$.
The rms current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_C}$.
$I_{rms} = \frac{250}{200} = 1.25 \, A$.

(A) Given: $V_{rms} = 100 \text{ V}$, $f = 50 \text{ Hz}$, $C = 100 \mu F = 100 \times 10^{-6} \text{ F}$.
First, calculate the capacitive reactance $X_C$ using the formula $X_C = \frac{1}{2 \pi f C}$.
$X_C = \frac{1}{2 \times \pi \times 50 \times 100 \times 10^{-6}} = \frac{1}{100 \pi \times 10^{-4}} = \frac{1}{0.01 \pi} = \frac{100}{\pi} \Omega$.
The rms value of the current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_C}$.
$I_{rms} = \frac{100}{100 / \pi} = \pi \text{ A}$.
Since $\pi \approx 3.14$, the current is $3.14 \text{ A}$.