Mix Examples-Alternating Current Questions in English

(D) The dimensions of the given quantities are:
$L = [M L^2 T^{-2} A^{-2}]$
$C = [M^{-1} L^{-2} T^4 A^2]$
$R = [M L^2 T^{-3} A^{-2}]$
$1$. For $\frac{1}{RC}$: The time constant $\tau = RC$ has dimensions of time $[T]$. Thus,$\frac{1}{RC}$ has dimensions of $[T^{-1}]$,which is frequency.
$2$. For $\frac{R}{L}$: The ratio $\frac{R}{L}$ has dimensions $[M L^2 T^{-3} A^{-2}] / [M L^2 T^{-2} A^{-2}] = [T^{-1}]$,which is frequency.
$3$. For $\frac{1}{\sqrt{LC}}$: The resonant frequency $\omega = \frac{1}{\sqrt{LC}}$ has dimensions $[T^{-1}]$,which is frequency.
$4$. For $\frac{C}{L}$: The dimensions are $[M^{-1} L^{-2} T^4 A^2] / [M L^2 T^{-2} A^{-2}] = [M^{-2} L^{-4} T^6 A^4]$. This does not represent frequency.
Therefore,the correct option is $D$.

(B) The power transmitted is given by $P = VI$,where $V$ is the voltage and $I$ is the current. For a fixed power $P$,if the voltage $V$ is increased,the current $I$ decreases.
The power loss in the transmission lines due to heating is given by $P_{loss} = I^2 R$,where $R$ is the resistance of the wires.
Substituting $I = P/V$,we get $P_{loss} = (P/V)^2 R = P^2 R / V^2$.
Thus,$P_{loss} \propto 1/V^2$. By increasing the voltage $V$,the power loss is significantly reduced,making the transmission more economical and efficient. Therefore,option $B$ is correct.

(D) The power consumed by a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the bulb.
For a bulb,the resistance $R$ is a constant property of the filament.
When connected to a $DC$ source of voltage $V$,the power consumed is $P_{DC} = \frac{V^2}{R}$.
When connected to an $AC$ source of the same $RMS$ voltage $V$,the power consumed is $P_{AC} = \frac{V_{rms}^2}{R} = \frac{V^2}{R}$.
Since the power consumed in both cases is identical,the brightness of the bulb will be the same.

(A) In an $LCR$ series $ac$ circuit,the phase difference $\phi$ between the applied voltage $V$ and the current $i$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
If $X_L > X_C$,the circuit is inductive,and the phase angle $\phi$ lies between $0$ and $\pi/2$ (current lags behind voltage).
If $X_L < X_C$,the circuit is capacitive,and the phase angle $\phi$ lies between $-\pi/2$ and $0$ (current leads the voltage).
If $X_L = X_C$,the circuit is at resonance,and $\phi = 0$.
Therefore,the magnitude of the phase angle $|\phi|$ ranges from $0$ to $\pi/2$ depending on the values of $X_L$ and $X_C$.

A choke coil consists of an inductor with high inductance and negligible resistance. In an $AC$ circuit, the average power consumed is given by $P = V_{rms} I_{rms} \cos \phi$. For a pure inductor, the phase angle $\phi = 90^{\circ}$, so the power factor $\cos 90^{\circ} = 0$. Consequently, the power consumed by an ideal choke coil is zero. In contrast, a $Rheostat$ is a resistor that dissipates energy as heat ($I^2R$ loss). Therefore, a choke coil is preferred to control current in $AC$ circuits without significant power loss.

(D) The dimensions of the given quantities are as follows:
$1$. For an $RC$ circuit,the time constant is $\tau = RC$. Thus,the dimension of $\frac{1}{RC}$ is $[T^{-1}]$,which is the dimension of frequency.
$2$. For an $RL$ circuit,the time constant is $\tau = \frac{L}{R}$. Thus,the dimension of $\frac{R}{L}$ is $[T^{-1}]$,which is the dimension of frequency.
$3$. For an $LC$ circuit,the resonant frequency is given by $\omega = \frac{1}{\sqrt{LC}}$. Thus,the dimension of $\frac{1}{\sqrt{LC}}$ is $[T^{-1}]$,which is the dimension of frequency.
$4$. For the combination $\frac{C}{L}$,we know that the resonant frequency $\omega = \frac{1}{\sqrt{LC}}$. Therefore,$\omega^2 = \frac{1}{LC}$,which implies $LC = \frac{1}{\omega^2}$. The dimension of $LC$ is $[T^2]$. Thus,the dimension of $\frac{C}{L}$ is $\frac{[C]}{[L]} = \frac{[C]^2}{[LC]} = \frac{[C]^2}{[T^2]}$. This does not have the dimensions of frequency.
Therefore,the correct option is $D$.

(D) In an $A.C.$ circuit,the phase relationship between current and voltage depends on the components present in the circuit.
If the circuit is purely resistive,the current is in phase with the voltage.
If the circuit is inductive,the current lags behind the voltage.
If the circuit is capacitive,the current leads the voltage.
Therefore,the current may lead,lag,or be in phase with the voltage depending on the circuit configuration.

(C) From the circuit diagram,we have an $LCR$ series circuit connected to an $AC$ source of $220 \, V$.
The voltmeter is connected across the resistor,so its reading is the voltage across the resistor $(V_R)$.
In an $LCR$ series circuit,the total voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
From the diagram,the voltage across the inductor $V_L = 300 \, V$ and the voltage across the capacitor $V_C = 300 \, V$.
Substituting these values: $220 = \sqrt{V_R^2 + (300 - 300)^2}$.
$220 = \sqrt{V_R^2 + 0} = V_R$.
Thus,the voltmeter reading is $V_R = 220 \, V$.
The current $i$ in the circuit is given by $i = \frac{V_R}{R}$.
Given $R = 100 \, \Omega$,we have $i = \frac{220}{100} = 2.2 \, A$.
Therefore,the voltmeter reading is $220 \, V$ and the ammeter reading is $2.2 \, A$.

(D) The circuit consists of a resistor $R_1 = 6\,\Omega$ in series with an $LCR$ branch containing $L = 5\,mH$,$R_2 = 4\,\Omega$,and $C = 50\,\mu F$.
Given $\omega = 2000\,rad/s$.
Inductive reactance $X_L = \omega L = 2000 \times 5 \times 10^{-3} = 10\,\Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 50 \times 10^{-6}} = 10\,\Omega$.
Total resistance $R_{total} = R_1 + R_2 = 6 + 4 = 10\,\Omega$.
Since $X_L = X_C$,the circuit is in resonance.
Total impedance $Z = \sqrt{R_{total}^2 + (X_L - X_C)^2} = \sqrt{10^2 + 0^2} = 10\,\Omega$.
Peak current $I_0 = \frac{V_0}{Z} = \frac{20}{10} = 2\,A$.
$RMS$ current $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2}{1.414} \approx 1.41\,A$.
The voltmeter is connected across the $LCR$ branch. The impedance of this branch is $Z_{LCR} = \sqrt{R_2^2 + (X_L - X_C)^2} = \sqrt{4^2 + 0^2} = 4\,\Omega$.
The voltage across this branch is $V_{LCR} = I_{rms} \times Z_{LCR} = 1.41 \times 4 = 5.64\,V \approx 5.6\,V$.

(A) Yes, it is possible in an $AC$ circuit. Kirchhoff's Current Law $(KCL)$ states that the algebraic sum of currents at a junction is zero for $DC$ circuits. However, in $AC$ circuits, the currents are phasors. The sum of the currents at the junction $B$ is given by the vector sum of the phasors: $\vec{I}_{in} = \vec{I}_{out1} + \vec{I}_{out2}$. If the phase angles between the currents are such that the vector sum of $15 \ A$ and $5 \ A$ results in $10 \ A$, then this distribution is possible.

(C) From the given phasor diagram,it is observed that the current $i_{rms}$ is lagging behind the voltage $E_{rms}$ by a phase angle of $\phi = 45^\circ$.
In an $AC$ circuit,current lags behind voltage when the circuit is inductive in nature.
An $LR$ circuit is inherently inductive,so the current will always lag behind the voltage.
An $LCR$ circuit can also be inductive if the inductive reactance $X_L$ is greater than the capacitive reactance $X_C$ (i.e.,$X_L > X_C$).
Therefore,the circuit could be either an $LR$ circuit or an $LCR$ circuit with $X_L > X_C$.
Thus,the correct option is $C$.

(B) In an anti-resonant circuit (parallel $LC$ circuit),the impedance is maximum at the resonant frequency $\nu_0$.
Since the current $i = \frac{V}{Z}$,the current $i$ is minimum at the resonant frequency.
At frequencies other than the resonant frequency,the impedance $Z$ decreases,causing the current $i$ to increase.
Therefore,the $i -
u$ curve shows a minimum at the resonant frequency,which corresponds to the graph shown in option $B$.

(C) The circuit consists of an inductor $L$ and a bulb $B$ in series with an $AC$ source. The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb and $X_L = \omega L$ is the inductive reactance.
When an iron rod is inserted into the coil,the permeability of the core increases,which significantly increases the self-inductance $L$ of the coil.
Since $X_L = \omega L$,an increase in $L$ leads to an increase in the inductive reactance $X_L$.
As the total impedance $Z = \sqrt{R^2 + X_L^2}$ increases,the total current in the circuit $I = \frac{V}{Z}$ decreases.
The brightness of the bulb depends on the power dissipated,$P = I^2 R$. Since the current $I$ decreases,the power dissipated by the bulb decreases,and thus the brightness of the bulb decreases.

(C) Given: Resistance $R = 100 \, \Omega$,Capacitive reactance $X_C = 100 \, \Omega$,and $RMS$ voltage $V_{rms} = 220 \, V$.
The total impedance of the $RC$ series circuit is $Z = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \, \Omega$.
The peak voltage of the source is $V_0 = V_{rms} \sqrt{2} = 220\sqrt{2} \, V$.
The peak conduction current in the circuit is $I_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{100\sqrt{2}} = 2.2 \, A$.
According to Maxwell's modification of Ampere's law,the displacement current $I_d$ between the plates of a capacitor is equal to the conduction current $I_c$ in the connecting wires.
Therefore,the peak value of the displacement current is equal to the peak value of the conduction current,which is $2.2 \, A$.

(C) Given: $C = \frac{2.5}{\pi} \times 10^{-6} \, F$,$R = 3000 \, \Omega$,$V_{rms} = 200 \, V$,$\nu = 50 \, Hz$.
First,calculate the capacitive reactance $X_C = \frac{1}{2\pi \nu C} = \frac{1}{2\pi \times 50 \times (\frac{2.5}{\pi} \times 10^{-6})} = \frac{1}{250 \times 10^{-6}} = 4000 \, \Omega$.
The impedance of the circuit is $Z = \sqrt{R^2 + X_C^2} = \sqrt{3000^2 + 4000^2} = 5000 \, \Omega$.
The power factor is $\cos \phi = \frac{R}{Z} = \frac{3000}{5000} = 0.6$.
The power dissipated is $P = \frac{V_{rms}^2 \cos \phi}{Z} = \frac{200^2 \times 0.6}{5000} = \frac{40000 \times 0.6}{5000} = 8 \times 0.6 = 4.8 \, W$.

(B) When $C$ is removed,the circuit becomes an $RL$ circuit. The phase angle $\phi$ is given by $\tan \phi = \frac{X_L}{R}$.
Given $\phi = \pi /3$,so $\tan(\pi /3) = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{R} \implies X_L = R\sqrt{3}$.
When $L$ is removed,the circuit becomes an $RC$ circuit. The phase angle $\phi$ is given by $\tan \phi = \frac{X_C}{R}$.
Given $\phi = \pi /3$,so $\tan(\pi /3) = \frac{X_C}{R} \implies \sqrt{3} = \frac{X_C}{R} \implies X_C = R\sqrt{3}$.
Since $X_L = X_C$,the circuit is in a state of resonance.
In an $LCR$ circuit at resonance,the impedance $Z = R$.
Given $R = 100 \ \Omega$,therefore $Z = 100 \ \Omega$.

(B) Given: $e = 75 \sin(\omega t) \text{ V}$ and $i = 1.5 \sin(\omega t + 45^\circ) \text{ A}$.
Here,the current $i$ leads the voltage $e$ by a phase angle of $\phi = 45^\circ$.
Since the current leads the voltage,the circuit must be capacitive in nature. This implies that there must be a capacitor in the box.
Since there is a phase difference between voltage and current,there must be a resistance $R$ in the box to provide the phase shift.
Because the circuit is capacitive,the net reactance $X = X_C - X_L$ must be capacitive $(X_C > X_L)$. This does not strictly rule out the presence of an inductor,but it implies the net effect is capacitive. However,the statement 'There must be an inductor' is not necessarily true,making it the wrong statement.
The power factor is $\cos \phi = \cos(45^\circ) = 1/\sqrt{2} \approx 0.707$. Thus,option $(d)$ is correct.
Therefore,the wrong statement is $(b)$.

(A) The dimension of frequency is $[T^{-1}]$.
$1$. For an $RC$ circuit,the time constant is $\tau = RC$. Since $\tau$ has the dimension of time $[T]$,the quantity $\frac{1}{RC}$ has the dimension of frequency $[T^{-1}]$.
$2$. For an $RL$ circuit,the time constant is $\tau = \frac{L}{R}$. Since $\tau$ has the dimension of time $[T]$,the quantity $\frac{R}{L}$ has the dimension of frequency $[T^{-1}]$.
$3$. For an $LC$ circuit,the resonant frequency is given by $\omega = \frac{1}{\sqrt{LC}}$. Thus,$\frac{1}{\sqrt{LC}}$ also has the dimension of frequency $[T^{-1}]$.
Comparing these with the given options,the combination $\frac{1}{RC}$ and $\frac{R}{L}$ represents dimensions of frequency.

(C) In the upper branch containing the capacitor,the current $I_1$ leads the applied voltage $V$ by a phase angle of $\phi_1 = \frac{\pi}{2}$.
In the lower branch containing the resistor $R$ and inductor $X_L$,the current $I_2$ lags behind the applied voltage $V$ by a phase angle $\phi_2 = \tan^{-1} \left( \frac{X_L}{R} \right)$.
The total phase difference between the two currents $I_1$ and $I_2$ is the sum of their individual phase shifts relative to the voltage $V$.
Therefore,the phase difference $\Delta \phi = \phi_1 + \phi_2 = \frac{\pi}{2} + \tan^{-1} \left( \frac{X_L}{R} \right)$.

(D) The reactance of a circuit is given by $X = X_L - X_C$,where $X_L = \omega L$ is the inductive reactance and $X_C = 1/(\omega C)$ is the capacitive reactance.
For the net reactance to be zero $(X = 0)$:
$1$. If the circuit contains both an inductor and a capacitor,they must be in resonance such that $X_L = X_C$,which results in $X = 0$.
$2$. If the circuit contains only resistors,there are no inductors or capacitors present,so $X_L = 0$ and $X_C = 0$,leading to $X = 0$.
Therefore,the circuit can contain both an inductor and a capacitor (in resonance) or neither an inductor nor a capacitor (purely resistive circuit).
Thus,the correct option is $(D)$.

(A) In an $LCR$ circuit, the charge on the capacitor at any time $t$ is given by $Q(t) = Q_0 e^{-Rt/2L} \cos(\omega' t + \phi)$.
The maximum charge on the capacitor at any cycle is given by $Q_{max}(t) = Q_0 e^{-Rt/2L}$.
Squaring both sides, we get $Q^2_{max}(t) = Q^2_0 e^{-Rt/L}$.
This equation represents an exponential decay of the square of the maximum charge with time.
The rate of decay depends on the factor $R/L$. Since $L_1 > L_2$, the decay constant $R/L_1$ is smaller than $R/L_2$.
Therefore, the charge in the circuit with $L_2$ decays faster than the circuit with $L_1$.
Thus, the graph for $L_1$ will show a slower decay compared to the graph for $L_2$.

(A) When capacitance is removed,the circuit becomes an $LR$ circuit.
Given the phase angle $\phi = 30^\circ$,we have $\tan \phi = \frac{\omega L}{R}$.
$\omega L = R \tan 30^\circ = 200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, \Omega$.
When the inductor is removed,the circuit becomes a $CR$ circuit.
Given the phase angle $\phi = 30^\circ$,we have $\tan \phi = \frac{1}{\omega C R}$.
$\frac{1}{\omega C} = R \tan 30^\circ = 200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, \Omega$.
In the original $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
Substituting the values: $Z = \sqrt{200^2 + (\frac{200}{\sqrt{3}} - \frac{200}{\sqrt{3}})^2} = \sqrt{200^2 + 0} = 200 \, \Omega$.
The power dissipated in an $AC$ circuit is $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$.
Since $\omega L = \frac{1}{\omega C}$,the circuit is in resonance,so $\phi = 0^\circ$ and $\cos \phi = 1$.
$P = \frac{V_{rms}^2}{Z^2} \times R = \frac{220^2}{200^2} \times 200 = \frac{220 \times 220}{200} = 242 \, W$.

(A) Applying Kirchhoff's Voltage Law $(KVL)$ to the $LCR$ circuit at any time $t$:
$\frac{q}{C} - iR - L \frac{di}{dt} = 0$
Since $i = -\frac{dq}{dt}$,we have:
$\frac{q}{C} + R \frac{dq}{dt} + L \frac{d^2q}{dt^2} = 0$
Rearranging gives the differential equation for a damped harmonic oscillator:
$\frac{d^2q}{dt^2} + \frac{R}{L} \frac{dq}{dt} + \frac{q}{LC} = 0$
The charge on the capacitor decays as $q(t) = Q_0 e^{-\frac{Rt}{2L}} \cos(\omega' t + \phi)$.
The maximum charge at any cycle is given by the envelope $Q_{Max} = Q_0 e^{-\frac{Rt}{2L}}$.
Squaring this,we get $Q_{Max}^2 = Q_0^2 e^{-\frac{Rt}{L}}$.
Comparing the decay constant $\lambda = \frac{R}{L}$,we see that for a fixed $R$,a smaller $L$ results in a larger decay constant,meaning the charge decays faster.
Since $L_1 > L_2$,the decay constant for $L_2$ is larger than for $L_1$.
Therefore,the curve for $L_2$ decays faster than the curve for $L_1$.

(A) The given equations are $e = 100 \sin(20t)$ and $i = 20 \sin(30t - \frac{\pi}{4})$.
Note: The frequencies of the $e.m.f.$ and current are different $(20 \neq 30)$. In such a case,the average power over a complete cycle is zero because the phase difference $\phi$ is not constant.
However,if we assume the question implies a standard phase difference $\phi = 45^{\circ}$ (or $\frac{\pi}{4}$) for a single frequency circuit,the calculation is as follows:
Average power $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos \phi = \left(\frac{V_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos \phi$
$P_{\text{avg}} = \left(\frac{100}{\sqrt{2}}\right) \left(\frac{20}{\sqrt{2}}\right) \cos 45^{\circ} = \frac{2000}{2} \times \frac{1}{\sqrt{2}} = \frac{1000}{\sqrt{2}} \text{ W}$.
Wattless current $I_w = I_{\text{rms}} \sin \phi = \left(\frac{I_0}{\sqrt{2}}\right) \sin 45^{\circ} = \left(\frac{20}{\sqrt{2}}\right) \times \frac{1}{\sqrt{2}} = \frac{20}{2} = 10 \text{ A}$.
Thus,the values are $\frac{1000}{\sqrt{2}}$ and $10$.

(D) Given: $R = 100 \,\Omega$,$V = 200 \,V$,$\omega = 300 \,rad/s$.
When capacitance is removed,the circuit becomes an $LR$ circuit. The phase angle $\phi_1 = 60^o$ (current lags voltage). Wait,the problem states current leads by $60^o$ when capacitance is removed,which implies an $RC$ circuit. Let's re-evaluate: If $C$ is removed,we have $LR$ circuit,$\tan(60^o) = \frac{X_L}{R} \implies X_L = R \tan(60^o) = 100\sqrt{3} \,\Omega$.
If $L$ is removed,we have $RC$ circuit,$\tan(60^o) = \frac{X_C}{R} \implies X_C = R \tan(60^o) = 100\sqrt{3} \,\Omega$.
Since $X_L = X_C$,the circuit is at resonance when all components are connected.
At resonance,$Z = R = 100 \,\Omega$.
Current $I = \frac{V}{Z} = \frac{200}{100} = 2 \,A$.
Power dissipated $P = I^2 R = (2)^2 \times 100 = 400 \,W$.

(D) Statement-$I$ is true because a capacitor can be used to limit current in an $a.c.$ circuit without dissipating power, similar to a choke coil.
Statement-$II$ is true because the capacitive reactance $X_C = 1 / (2\pi f C)$. For $d.c.$, $f = 0$, so $X_C = \infty$ (blocking $d.c.$), while for $a.c.$, it offers finite reactance.
Since a capacitor provides reactance to limit current in $a.c.$ circuits without power loss, Statement-$II$ correctly explains why it can replace a choke coil.

(A) The net voltage $V_{net}$ across the resistor is the sum of the three sources: $V_{net} = V_1 + V_2 + V_3$.
Representing these as phasors:
$V_1 = 3 \angle 0^\circ = 3 + j0$
$V_2 = 5 \angle 30^\circ = 5(\cos 30^\circ + j \sin 30^\circ) = 5(\frac{\sqrt{3}}{2} + j \frac{1}{2}) = \frac{5\sqrt{3}}{2} + j 2.5$
$V_3 = 5 \angle -127^\circ = 5(\cos(-127^\circ) + j \sin(-127^\circ)) \approx 5(-0.6 - j 0.8) = -3 - j 4$
Summing the components:
$V_{real} = 3 + \frac{5\sqrt{3}}{2} - 3 = \frac{5\sqrt{3}}{2} \approx 4.33$
$V_{imag} = 0 + 2.5 - 4 = -1.5$
Peak voltage $V_{max} = \sqrt{V_{real}^2 + V_{imag}^2} = \sqrt{(\frac{5\sqrt{3}}{2})^2 + (-1.5)^2} = \sqrt{\frac{75}{4} + 2.25} = \sqrt{18.75 + 2.25} = \sqrt{21} \text{ V}$.
Peak current $I_{max} = \frac{V_{max}}{R} = \frac{\sqrt{21}}{\sqrt{7/3}} = \sqrt{\frac{21 \times 3}{7}} = \sqrt{9} = 3 \text{ A}$.

(B) The initial reactance of the circuit is $X_i = |X_{Ci} - X_{Li}| = |\frac{1}{\omega C_i} - \omega L_i|$.
Initially,the capacitor is filled with liquid $(C_i = \frac{K \varepsilon_0 A}{\ell})$ and the inductor is empty $(L_i = \mu_0 n^2 A \ell)$.
So,$X_i = |\frac{\ell}{\omega K \varepsilon_0 A} - \omega \mu_0 n^2 A \ell|$.
After the liquid fills the inductor,the capacitor becomes empty $(C_f = \frac{\varepsilon_0 A}{\ell})$ and the inductor is filled with liquid $(L_f = \mu_0 \mu_r n^2 A \ell)$.
So,$X_f = |\frac{\ell}{\omega \varepsilon_0 A} - \omega \mu_0 \mu_r n^2 A \ell|$.
Taking the ratio $\frac{X_i}{X_f} = \left| \frac{\frac{\ell}{\omega K \varepsilon_0 A} - \omega \mu_0 n^2 A \ell}{\frac{\ell}{\omega \varepsilon_0 A} - \omega \mu_0 \mu_r n^2 A \ell} \right|$.
Multiplying numerator and denominator by $\omega \varepsilon_0 A / \ell$,we get $\frac{X_i}{X_f} = \left| \frac{\frac{1}{K} - \omega^2 \mu_0 \varepsilon_0 n^2 A^2}{1 - \omega^2 \mu_0 \varepsilon_0 \mu_r n^2 A^2} \right|$.
Using $\mu_0 \varepsilon_0 = \frac{1}{c^2}$ and $\omega^2 A^2 n^2 = c^2$,the term $\omega^2 \mu_0 \varepsilon_0 n^2 A^2 = \frac{c^2}{c^2} = 1$.
Thus,$\frac{X_i}{X_f} = \left| \frac{\frac{1}{K} - 1}{1 - \mu_r} \right| = \left| \frac{1 - K}{K(1 - \mu_r)} \right| = \frac{1 - K}{K(1 - \mu_r)}$ (assuming $K > 1$ and $\mu_r > 1$ or vice versa to maintain magnitude).

(A) In position $1$,the circuit is an $LR$ series circuit. The phase angle $\phi$ is given by $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.
Given $\phi = \pi / 6$,$\omega = 1000 \text{ rad/s}$,and $L = \sqrt{3} \text{ mH} = \sqrt{3} \times 10^{-3} \text{ H}$.
$\tan(\pi / 6) = \frac{1000 \times \sqrt{3} \times 10^{-3}}{R} \implies \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{R} \implies R = 3 \, \Omega$.
In position $2$,the circuit is an $RC$ series circuit. The phase angle $\phi$ is given by $\tan \phi = \frac{X_C}{R} = \frac{1}{\omega C R}$.
Given $C = \frac{1000}{3} \, \mu\text{F} = \frac{1000}{3} \times 10^{-6} \text{ F}$.
$\tan \phi = \frac{1}{1000 \times (\frac{1000}{3} \times 10^{-6}) \times 3} = \frac{1}{1000 \times \frac{1}{3} \times 10^{-3} \times 3} = \frac{1}{1} = 1$.
Since $\tan \phi = 1$,$\phi = \pi / 4$. In an $RC$ circuit,current leads the source $emf$ by $\phi$. Therefore,current leads source $emf$ by $\pi / 4$.

(A) The heat produced by a steady $DC$ current $I$ in a resistor $R$ over time $t$ is given by $H_{DC} = I^2 R t$.
The heat produced by an $AC$ current with peak value $I$ in the same resistor $R$ over the same time $t$ is given by $H_{AC} = I_{rms}^2 R t$.
Since $I_{rms} = \frac{I}{\sqrt{2}}$,we have $H_{AC} = \left(\frac{I}{\sqrt{2}}\right)^2 R t = \frac{I^2 R t}{2}$.
Therefore,the ratio of heat produced is $\frac{H_{DC}}{H_{AC}} = \frac{I^2 R t}{I^2 R t / 2} = \frac{2}{1}$.

(C) In an $LCR$ series circuit,the maximum current occurs at resonance,where the impedance $Z$ is equal to the resistance $R$ of the coil.
Given $V_{rms} = 24 \,V$ and $I_{rms, max} = 6 \,A$.
At resonance,$Z = R = V_{rms} / I_{rms, max} = 24 / 6 = 4 \,\Omega$.
This $R = 4 \,\Omega$ is the resistance of the coil.
When the coil is connected to a $DC$ battery of $emf$ $E = 12 \,V$ and internal resistance $r = 4 \,\Omega$,the total resistance of the circuit is $R_{total} = R + r = 4 \,\Omega + 4 \,\Omega = 8 \,\Omega$.
The steady-state current $I$ is given by $I = E / R_{total} = 12 / 8 = 1.5 \,A$.

(D) In a series $LR$ circuit connected to an $AC$ voltage source $V(t) = V_0 \sin \omega t$,the current $I(t)$ is given by $I(t) = I_0 \sin(\omega t - \phi)$,where $I_0 = \frac{V_0}{Z}$ and $Z = \sqrt{R^2 + (\omega L)^2}$.
The transient part of the current,which involves the term $e^{-Rt/L}$,decays to zero as $t \to \infty$ because $t_0 \gg \frac{L}{R}$.
Therefore,after a very large time,only the steady-state sinusoidal current remains,which oscillates with the same frequency as the source voltage but with a phase lag $\phi = \tan^{-1}(\frac{\omega L}{R})$.
This corresponds to a steady-state sinusoidal oscillation.

(A) In a series $RLC$ circuit, the voltage across the inductor is $V_L = I X_L = I (2 \pi f L)$ and the voltage across the capacitor is $V_C = I X_C = I \left( \frac{1}{2 \pi f C} \right)$.
Given that the current $I$ is kept constant as the frequency $f$ is varied.
From the expressions, we can see that $V_L \propto f$, which represents a straight line passing through the origin.
Also, $V_C \propto \frac{1}{f}$, which represents a rectangular hyperbola.
Therefore, the graph where $V_L$ increases linearly with $f$ and $V_C$ decreases hyperbolically with $f$ is the correct representation.

(C) Given resistance $R = 100\,\Omega$,voltage $V = 200\,V$.
When the capacitor is removed,the circuit becomes an $L-R$ circuit. The phase angle $\phi = 60^o$ is given by $\tan 60^o = \frac{X_L}{R}$.
$\Rightarrow X_L = R \tan 60^o = 100 \times \sqrt{3} = 100\sqrt{3}\,\Omega$.
When the inductor is removed,the circuit becomes a $C-R$ circuit. The phase angle $\phi = 60^o$ is given by $\tan 60^o = \frac{X_C}{R}$.
$\Rightarrow X_C = R \tan 60^o = 100 \times \sqrt{3} = 100\sqrt{3}\,\Omega$.
Since $X_L = X_C$,the circuit is in resonance.
In resonance,the impedance $Z = R = 100\,\Omega$.
The current $I = \frac{V}{Z} = \frac{200}{100} = 2\,A$.

(A) The total resistance $R$ of the circuit is the sum of all individual resistances: $R = 0.1 \, \Omega + 5.9 \, \Omega + 4 \, \Omega = 10 \, \Omega$.
The inductive reactance is $X_L = \omega L = 2000 \, rad/s \times 5 \times 10^{-3} \, H = 10 \, \Omega$.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{2000 \, rad/s \times 50 \times 10^{-6} \, F} = \frac{1}{0.1} = 10 \, \Omega$.
The impedance $Z$ of the $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values, $Z = \sqrt{10^2 + (10 - 10)^2} = \sqrt{100} = 10 \, \Omega$.
The current in the circuit is $I = \frac{V}{Z} = \frac{20 \, V}{10 \, \Omega} = 2 \, A$.

(C) For $dc$ source: The inductor acts as a short circuit (resistance $R$ only). Power $P_{dc} = \frac{V^2}{R}$.
Given $V = 10\, V$ and $P_{dc} = 20\, W$,so $R = \frac{10^2}{20} = \frac{100}{20} = 5\, \Omega$.
For $ac$ source: The impedance $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi f L$.
Power $P_{ac} = \frac{V^2 R}{Z^2} = \frac{V^2 R}{R^2 + X_L^2}$.
Given $P_{ac} = 10\, W$,so $10 = \frac{10^2 \times 5}{5^2 + X_L^2} \Rightarrow 10 = \frac{500}{25 + X_L^2}$.
$25 + X_L^2 = 50 \Rightarrow X_L^2 = 25 \Rightarrow X_L = 5\, \Omega$.
Since $X_L = 2\pi f L$,we have $5 = 2 \times 3.14 \times f \times (10 \times 10^{-3})$.
$f = \frac{5}{2 \times 3.14 \times 0.01} = \frac{5}{0.0628} \approx 79.6\, Hz \approx 80\, Hz$.

(C) The given voltage source is $E = 150 \sin(100t)$,where the peak voltage $V_0 = 150 \, V$.
The device acts as a half-wave rectifier,allowing current to flow only during the positive half-cycle of the $AC$ input.
For a half-wave rectified current,the $r.m.s.$ value is given by $i_{rms} = \frac{i_0}{2}$,where $i_0$ is the peak current.
The peak current is $i_0 = \frac{V_0}{R} = \frac{150}{20} = 7.5 \, A$.
Therefore,$i_{rms} = \frac{7.5}{2} = 3.75 \, A$.

(A) 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 $\tan \phi = \frac{X_C}{R}$. Given $\phi = \pi/3$,we have $\tan(\pi/3) = \sqrt{3} = \frac{X_C}{R}$,so $X_C = R\sqrt{3}$.
When $C$ is removed,the circuit becomes an $R-L$ circuit. The phase difference is $\tan \phi = \frac{X_L}{R}$. Given $\phi = \pi/3$,we have $\tan(\pi/3) = \sqrt{3} = \frac{X_L}{R}$,so $X_L = R\sqrt{3}$.
Since $X_L = X_C$,the original $R-L-C$ circuit is in resonance.
At resonance,the impedance $Z = R$,and the phase difference $\phi = 0$.
The power factor is $\cos \phi = \cos(0) = 1$.

(D) The Assertion is incorrect because Ohm's law $(V = IR)$ can be applied to $a.c.$ circuits using the concept of impedance $(Z)$. For an $a.c.$ circuit, the relation is $V = IZ$, where $Z$ is the impedance.
The Reason is also incorrect because the resistance offered by a capacitor is called capacitive reactance $(X_C)$, not resistance. While it is true that $X_C = 1 / (2\pi fC)$ depends on frequency, calling it 'resistance' is technically inaccurate in the context of general circuit theory, and the statement as a whole does not justify the false assertion.

(A) The power transmitted through a line is given by $P = VI$, where $V$ is the voltage and $I$ is the current.
The power loss in the transmission line due to resistance $R$ is given by $P_{\text{loss}} = I^2R$.
Substituting $I = \frac{P}{V}$, we get $P_{\text{loss}} = (\frac{P}{V})^2 R = \frac{P^2 R}{V^2}$.
From this expression, it is clear that $P_{\text{loss}} \propto \frac{1}{V^2}$.
Therefore, by increasing the voltage $V$, the power loss $P_{\text{loss}}$ decreases significantly.
Thus, both the Assertion and the Reason are correct, and the Reason is the correct explanation of the Assertion.

(D) The peak current in a series $LCR$ circuit is given by $I_{\max} = \frac{\varepsilon_{\max}}{Z}$,where $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
In a purely resistive element,the current is independent of frequency,but in a series $LCR$ circuit,the impedance $Z$ depends on the angular frequency $\omega$.
As $\omega$ increases,the term $(\omega L - \frac{1}{\omega C})^2$ changes. Specifically,the current $I_{\max}$ reaches its maximum value at resonance $(\omega = \frac{1}{\sqrt{LC}})$ and decreases as we move away from resonance in either direction.
Therefore,the assertion that the current always increases with an increase in angular frequency is incorrect.
The reason provided is a correct formula for peak current,but it does not support the false assertion.
Thus,the Assertion is incorrect and the Reason is correct.

(N/A) Yes, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit due to Kirchhoff's Voltage Law. However, this is not true for $rms$ voltage because the voltages across different elements are generally not in phase.
$(b)$ $A$ capacitor is used in the primary circuit of an induction coil to prevent sparking at the contact breaker. When the circuit is broken, the rapid change in current induces a high back $emf$ in the primary coil, which would cause a spark across the contacts. The capacitor provides a path for the current, absorbing the energy and preventing the spark.
$(c)$ For a $dc$ signal, the frequency is $0$. The inductive reactance $X_L = 2\pi fL = 0$, while the capacitive reactance $X_C = 1/(2\pi fC) \to \infty$. Thus, the $dc$ voltage drops across $C$. For a high-frequency $ac$ signal, $X_L$ is very large and $X_C$ is very small. Thus, the $ac$ voltage drops across $L$.
$(d)$ If connected to an $ac$ line, the lamp will glow less brightly when an iron core is inserted. The iron core increases the inductance $L$ of the choke, which increases the inductive reactance $X_L = 2\pi fL$. This increases the total impedance of the circuit, reducing the current and thus the brightness of the lamp.
$(e)$ $A$ choke coil is used to limit the current in a fluorescent tube circuit without dissipating significant power, as it has a high reactance and low resistance. An ordinary resistor would dissipate power as heat $(I^2R)$, which is inefficient and wasteful.

(B) The primary reason for preferring $A.C.$ voltage over $D.C.$ voltage for long-distance power transmission is the ease of voltage transformation.
Using a transformer,$A.C.$ voltage can be easily stepped up to a very high value for transmission,which significantly reduces the current $(I)$ flowing through the lines.
Since power loss in transmission lines is given by $P_{loss} = I^2R$,reducing the current leads to a drastic reduction in energy loss due to heating ($I^2R$ loss).
$D.C.$ voltage cannot be easily stepped up or down using simple transformers,making it inefficient for long-distance transmission.

(A) In an $AC$ circuit,the voltage and current are time-varying quantities that can be represented as rotating vectors.
These rotating vectors are known as phasors.
$A$ phasor diagram is a graphical representation that shows the phase relationship between the alternating current and voltage in an $AC$ circuit.
It helps in visualizing the phase difference $\phi$ between the voltage and current vectors,which is essential for analyzing $AC$ circuits containing resistors,inductors,and capacitors.

(A) In a mechanical system undergoing forced oscillations,the differential equation is given by $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = F_0 \sin(\omega t)$.
In an $LCR$ series circuit,the differential equation for charge $q$ is given by $L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = V_0 \sin(\omega t)$.
Comparing these two equations,we can see that the charge $q$ in the $LCR$ circuit is analogous to the displacement $x$ in the mechanical system.
Therefore,the physical quantity that corresponds to displacement $x$ is charge $q$.

(A) In an $LCR$ series circuit,the differential equation for the charge $q$ is given by $L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = E(t)$.
Comparing this with the equation for a forced mechanical oscillator,$m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = F(t)$,we can identify the analogous quantities.
The term involving the second derivative of the displacement $x$ is $m \frac{d^2x}{dt^2}$,which corresponds to $L \frac{d^2q}{dt^2}$ in the electrical circuit.
Therefore,the physical quantity that corresponds to mass $m$ is the self-inductance $L$.

(A) In a mechanical system undergoing forced oscillations,the equation of motion is given by $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = F_0 \cos(\omega t)$,where $b$ is the damping constant.
In an $LCR$ series circuit,the differential equation for the charge $q$ is $L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = V_0 \cos(\omega t)$.
Comparing the two equations,we see that the term involving the first derivative of the variable (velocity $\frac{dx}{dt}$ in mechanics and current $I = \frac{dq}{dt}$ in circuits) represents the dissipative element.
Therefore,the resistance $R$ in an $LCR$ circuit corresponds to the damping constant $b$ in forced oscillations.

(A) Given: Power $P = 2 \, kW = 2000 \, W$,$V_{rms} = 223 \, V$,$\tan \phi = -\frac{3}{4}$.
Since the current lags,the circuit is inductive,meaning $X_L > X_C$. Thus,$\tan \phi = \frac{X_L - X_C}{R} = \frac{3}{4}$.
$Z = \frac{V^2}{P} = \frac{223^2}{2000} = \frac{50000}{2000} = 25 \, \Omega$.
Using $Z^2 = R^2 + (X_L - X_C)^2$,we have $25^2 = R^2 + (\frac{3}{4}R)^2 = R^2 + \frac{9}{16}R^2 = \frac{25}{16}R^2$.
$625 = \frac{25}{16}R^2 \Rightarrow R^2 = 400 \Rightarrow R = 20 \, \Omega$.
$X_L - X_C = \frac{3}{4} \times 20 = 15 \, \Omega$. Therefore,$X_C - X_L = -15 \, \Omega$.
$I_M = \frac{V_m}{Z} = \frac{\sqrt{2} \times 223}{25} \approx 12.6 \, A$.
If $R, X_L, X_C$ are doubled,$Z' = \sqrt{(2R)^2 + (2(X_L - X_C))^2} = 2Z = 50 \, \Omega$.
$I' = \frac{V}{Z'} = \frac{I}{2} = 6.3 \, A$. Power $P' = V I' \cos \phi = \frac{P}{2} = 1 \, kW$.