RL, RC and LC AC Circuits Questions in English

(B) In an $A.C.$ circuit,the relationship between current $(I)$ and voltage $(V)$ can be represented by a Lissajous figure in a $I-V$ plot.
For a purely capacitive circuit,current leads voltage by $90^o$.
For a purely inductive circuit,current lags voltage by $90^o$.
In the given graph,the $x$-axis represents current and the $y$-axis represents voltage.
As time progresses (indicated by the arrow),the curve traces an ellipse.
At the point where the current is maximum (on the $x$-axis),the voltage is zero and increasing.
This indicates that the current reaches its peak before the voltage reaches its peak.
Therefore,the current leads the voltage by approximately $90^o$.

(B) Given: Impedance $Z = 10\,\Omega$,frequency $f = 1000\,Hz$,phase angle $\phi = 45^\circ$.
The phase angle in an $RL$ circuit is given by $\tan \phi = \frac{\omega L}{R}$.
Since $\phi = 45^\circ$,$\tan 45^\circ = 1$,so $\omega L = R$.
The impedance is $Z = \sqrt{R^2 + (\omega L)^2} = 10\,\Omega$.
Substituting $R = \omega L$,we get $Z = \sqrt{(\omega L)^2 + (\omega L)^2} = \sqrt{2}(\omega L) = 10$.
Thus,$\omega L = \frac{10}{\sqrt{2}} = 5\sqrt{2}\,\Omega$.
We know $\omega = 2\pi f = 2\pi \times 1000 = 2000\pi\,rad/s$.
Therefore,$L = \frac{5\sqrt{2}}{\omega} = \frac{5\sqrt{2}}{2000\pi} = \frac{\sqrt{2}}{400\pi} = \frac{1}{200\sqrt{2}\pi}\,H$.
Wait,re-evaluating the calculation: $L = \frac{5\sqrt{2}}{2000\pi} = \frac{1}{400\pi/\sqrt{2}} = \frac{1}{200\sqrt{2}\pi}$.
Checking the options provided,the correct expression is $L = \frac{1}{\sqrt{2} \times 200\pi}$.

(D) In the initial circuit,the resistors $R$ and $6\,\Omega$ are in series. The potential difference across the $6\,\Omega$ resistor is $V_{6\Omega} = 3\,V$.
Using Ohm's law,the current in the circuit is $I = \frac{V_{6\Omega}}{6\,\Omega} = \frac{3}{6} = 0.5\,A$.
The total voltage is $V = 5\,V$. In a series circuit with $R$ and $6\,\Omega$,$V = I \sqrt{(R+6)^2} = I(R+6)$. Thus,$5 = 0.5(R+6)$,which gives $R+6 = 10$,so $R = 4\,\Omega$.
When $R$ is replaced by an inductor $L$ with reactance $X_L$,the circuit becomes an $L-R$ series circuit. The current remains $I = 0.5\,A$.
The impedance of the new circuit is $Z = \sqrt{X_L^2 + 6^2}$.
Using $V = IZ$,we have $5 = 0.5 \sqrt{X_L^2 + 36}$.
$10 = \sqrt{X_L^2 + 36} \implies 100 = X_L^2 + 36 \implies X_L^2 = 64 \implies X_L = 8\,\Omega$.
The potential difference across the inductor is $V_L = I \cdot X_L = 0.5 \times 8 = 4\,V$.

(A) The power consumed in an $AC$ circuit is given by the formula:
$P = V_{rms} I_{rms} \cos\phi$
Since $\cos\phi = \frac{R}{Z}$,we have:
$P = V_{rms} I_{rms} \left( \frac{R}{Z} \right)$ ..........$(1)$
First,calculate the total impedance $Z$ of the circuit:
$Z = \frac{V_{rms}}{I_{rms}} = \frac{120}{3} = 40\,\Omega$
Now,substitute the values into equation $(1)$:
$108 = (120)(3) \times \frac{R}{40}$
$108 = 360 \times \frac{R}{40}$
$108 = 9R$
$R = \frac{108}{9} = 12\,\Omega$
Therefore,the resistance in the circuit is $12\,\Omega$.

(A) The total voltage $V$ is given by $V^2 = V_R^2 + V_L^2$,where $V_R = 54 \, V$ and $V = 120 \, V$.
$V_L = \sqrt{V^2 - V_R^2} = \sqrt{120^2 - 54^2} = \sqrt{14400 - 2916} = \sqrt{11484} \approx 107.16 \, V$.
The power consumed by the resistor is $P = V_R \cdot I$,so $I = P / V_R = 16 / 54 \, A$.
The voltage across the inductor is $V_L = I \cdot X_L = I \cdot (2 \pi f L)$.
Substituting the values: $107.16 = (16 / 54) \cdot 2 \cdot \pi \cdot 60 \cdot L$.
$107.16 = (16 / 54) \cdot 377 \cdot L$.
$L = (107.16 \cdot 54) / (16 \cdot 377) \approx 5786.64 / 6032 \approx 0.96 \, H$.
Rounding to the nearest integer,$L \approx 1 \, H$.

(A) $1$. The brightness of the bulb depends on the power dissipated in the circuit,which is given by $P = I_{rms}^2 R_{bulb}$.
$2$. The current in the $RC$ circuit is given by $I_{rms} = \frac{V_{rms}}{Z}$,where $Z = \sqrt{R_{total}^2 + X_C^2}$ is the impedance of the circuit.
$3$. Here,$X_C = \frac{1}{\omega C}$ is the capacitive reactance.
$4$. When a dielectric slab is inserted into the capacitor,the capacitance $C$ increases ($C' = KC$,where $K > 1$).
$5$. As $C$ increases,the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.
$6$. Since $Z = \sqrt{R_{total}^2 + X_C^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$.
$7$. $A$ decrease in impedance $Z$ results in an increase in the current $I_{rms}$ flowing through the circuit.
$8$. Since $P = I_{rms}^2 R_{bulb}$,an increase in $I_{rms}$ leads to an increase in the power dissipated by the bulb,hence the brightness of the bulb will increase.

(B) In an $L-R$ series circuit,the impedance $Z$ is given by the formula:
$Z = \sqrt{R^2 + X_L^2}$
where $R$ is the resistance and $X_L$ is the inductive reactance.
The inductive reactance $X_L$ is defined as:
$X_L = \omega L = 2 \pi f L$
Substituting the value of $X_L$ into the impedance formula:
$Z = \sqrt{R^2 + (2 \pi f L)^2}$
$Z = \sqrt{R^2 + 4 \pi^2 f^2 L^2}$

(D) In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_L = 2\pi fL$ and $X_C = \frac{1}{2\pi fC}$.
At $f = 0$ ($DC$ condition),the capacitive reactance $X_C = \frac{1}{2\pi fC} = \infty$. Since $X_C$ is in series,the total impedance $Z = \infty$. Therefore,the current $I = \frac{E}{Z} = \frac{200}{\infty} = 0\,A$.
At $f = \infty$,the inductive reactance $X_L = 2\pi fL = \infty$. Since $X_L$ is in series,the total impedance $Z = \infty$. Therefore,the current $I = \frac{E}{Z} = \frac{200}{\infty} = 0\,A$.
Thus,the current at both $f = 0$ and $f = \infty$ is $0\,A$.

(B) The total voltage $V$ in an $RC$ series circuit is given by $V^2 = V_R^2 + V_C^2$.
First,calculate the potential difference across the resistor: $V_R = I \times R = 1.5 \, A \times 30 \, \Omega = 45 \, V$.
Now,substitute the values into the formula: $V_C = \sqrt{V^2 - V_R^2}$.
$V_C = \sqrt{(120)^2 - (45)^2}$.
$V_C = \sqrt{14400 - 2025} = \sqrt{12375}$.
$V_C \approx 111.24 \, V$.
Rounding to the nearest whole number,the potential difference across the capacitor is $111 \, V$.

(C) Given:
$I = 10 \sin \omega t$
$V = 10 \sin \left( \omega t + \pi / 4 \right)$
Comparing these with the general forms $I = I_0 \sin \omega t$ and $V = V_0 \sin (\omega t + \phi)$,we see that the phase difference is $\phi = \pi / 4$.
Since the voltage leads the current by a phase angle of $\pi / 4$ (positive phase angle),the circuit must be inductive in nature.
An $RL$ circuit (Resistor and Inductor) results in a voltage that leads the current.
An $LCR$ circuit can also result in a voltage that leads the current if the inductive reactance $X_L$ is greater than the capacitive reactance $X_C$ (i.e.,$X_L > X_C$).
Therefore,the circuit could contain both $L$ and $R$,or it could contain all three elements $L, C,$ and $R$ provided the net reactance is inductive.
Thus,options $(b)$ and $(d)$ are correct.

(A) The phase angle $\phi$ in an $LCR$ series circuit is given by $\tan \phi = \frac{X_C - X_L}{R}$.
Substituting the given values: $\tan \phi = \frac{8 - 5}{4} = \frac{3}{4}$.
Therefore,$\phi = \tan^{-1}(3/4)$.
Since $X_C > X_L$,the circuit is capacitive in nature,which means the current leads the voltage by a phase angle of $\tan^{-1}(3/4)$.

(B) For an $LR$ series circuit,the time constant is given by $\tau = \frac{L}{R}$.
When the battery is replaced by an $AC$ source,the impedance $Z$ of the $LR$ circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (\omega L)^2}$.
The power factor $\cos \phi$ is defined as $\cos \phi = \frac{R}{Z}$.
Substituting the expression for $Z$,we get $\cos \phi = \frac{R}{\sqrt{R^2 + \omega^2 L^2}}$.
Dividing the numerator and denominator by $R$,we get $\cos \phi = \frac{1}{\sqrt{1 + \omega^2 (\frac{L}{R})^2}}$.
Since $\tau = \frac{L}{R}$,substituting this into the equation gives $\cos \phi = \frac{1}{\sqrt{1 + \omega^2 \tau^2}}$.

(A) The phase difference $\phi$ in an $RL$ circuit is given by the formula $\tan \phi = \frac{X_L}{R}$.
Here,$X_L = \omega L = 2\pi f L$.
Given: $R = 4\,\Omega$,$L = 0.01\,H$,$f = 50\,Hz$.
Substituting the values: $\tan \phi = \frac{2\pi \times 50 \times 0.01}{4}$.
$\tan \phi = \frac{100\pi \times 0.01}{4} = \frac{\pi}{4}$.
Therefore,the phase difference is $\phi = \tan^{-1}\left(\frac{\pi}{4}\right)$.

(B) In an $AC$ series $LR$ circuit,the voltage across the resistor $(V_R)$ and the inductor $(V_L)$ are out of phase by $90^{\circ}$.
The total potential difference $(V)$ of the source is given by the phasor sum:
$V = \sqrt{V_R^2 + V_L^2}$
Given:
$V_L = 16 \, V$
$V_R = 20 \, V$
Substituting the values:
$V = \sqrt{(20)^2 + (16)^2}$
$V = \sqrt{400 + 256}$
$V = \sqrt{656}$
$V \approx 25.61 \, V$
Rounding to the nearest provided option,the total potential difference is $25.6 \, V$.

(B) Given: $L = 0.7 \, H$,$R = 220 \, \Omega$,$V_{rms} = 220 \, V$,$f = 50 \, Hz$.
First,calculate the inductive reactance $X_L$:
$X_L = 2 \pi f L = 2 \times \frac{22}{7} \times 50 \times 0.7 = 220 \, \Omega$.
Phase angle $\phi$ is given by:
$\tan \phi = \frac{X_L}{R} = \frac{220}{220} = 1 \implies \phi = 45^o$.
The total impedance $Z$ is:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{220^2 + 220^2} = 220\sqrt{2} \, \Omega$.
The $rms$ current $I_{rms}$ is:
$I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{220\sqrt{2}} = \frac{1}{\sqrt{2}} \, A$.
The wattless component of current is $I_{rms} \sin \phi$:
$I_{wattless} = I_{rms} \sin 45^o = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 0.5 \, A$.
Thus,the phase angle is $45^o$ and the wattless component is $0.5 \, A$.

(D) In a pure inductor,the current lags behind the potential difference across the inductor by a phase angle of $\pi/2$ radians (or $90^{\circ}$).
This phase relationship is independent of the frequency of the $A.C.$ source.
Therefore,the phase angle between the current and the potential difference across the inductor $L$ remains constant at $90^{\circ}$ regardless of the change in frequency.
However,if the question refers to the phase angle $\phi$ of the entire $LR$ series circuit,it is given by $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R} = \frac{2\pi f L}{R}$.
As the frequency $f$ increases,$\tan \phi$ increases,and thus the phase angle $\phi$ between the source voltage and current increases.
Given the specific phrasing 'phase angle between current and the potential difference across $L$',the answer is that it remains constant. Since this is not an option,the question likely intends to ask for the phase angle of the circuit,which increases as frequency increases.

(A) Given,resistance $R = 30\,\Omega$ and inductive reactance $X_L = 20\,\Omega$ at frequency $f = 50\,Hz$.
Since $X_L = 2\pi f L$,we have $20 = 2\pi(50)L$,which implies $L = \frac{20}{100\pi} = \frac{1}{5\pi}\,H$.
When the frequency is changed to $f' = 100\,Hz$,the new inductive reactance $X_L'$ is $X_L' = 2\pi f' L = 2\pi(100)(\frac{1}{5\pi}) = 40\,\Omega$.
The impedance $Z$ of the coil at $100\,Hz$ is $Z = \sqrt{R^2 + (X_L')^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\,\Omega$.
The current $I$ in the coil is given by $I = \frac{V}{Z} = \frac{200}{50} = 4\,A$.

(A) For the $A.C.$ source,the impedance $Z = \frac{V}{I_{AC}} = \frac{12 \; V}{0.2 \; A} = 60 \; \Omega$.
For the $D.C.$ source,the resistance $R = \frac{V}{I_{DC}} = \frac{12 \; V}{0.4 \; A} = 30 \; \Omega$.
Since $Z > R$,the circuit must contain an inductor $(L)$ in addition to the resistor $(R)$.
In a $D.C.$ circuit,an ideal capacitor acts as an open circuit (infinite resistance),but since current flows in the $D.C.$ case,the circuit cannot contain a capacitor in series. Therefore,the circuit is a series $LR$ circuit.

(A) Given: $L = 40 \times 10^{-3} \; H$,$C = 100 \times 10^{-6} \; F$,$V(t) = 10 \sin (314 t)$.
Here,$\omega = 314 \; rad/s$.
Inductive reactance $X_{L} = \omega L = 314 \times 40 \times 10^{-3} = 12.56 \; \Omega$.
Capacitive reactance $X_{C} = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = \frac{10^{4}}{314} \approx 31.85 \; \Omega$.
Since $X_{C} > X_{L}$,the circuit is capacitive.
The net reactance $X = X_{C} - X_{L} = 31.85 - 12.56 = 19.29 \; \Omega$.
The peak current $I_{m} = \frac{V_{m}}{X} = \frac{10}{19.29} \approx 0.52 \; A$.
In a capacitive circuit,the current leads the voltage by $\frac{\pi}{2}$.
Thus,$I(t) = I_{m} \sin (\omega t + \frac{\pi}{2}) = 0.52 \sin (314 t + \frac{\pi}{2}) = 0.52 \cos (314 t)$.

(B) When the iron rod is inserted into the inductor,the magnetic permeability of the core increases,which in turn increases the self-inductance $(L)$ of the coil.
The inductive reactance of the coil is given by $X_L = \omega L$. As $L$ increases,the inductive reactance $X_L$ also increases.
The total impedance of the series circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb. As $X_L$ increases,the total impedance $Z$ of the circuit increases.
Since the source voltage $V$ is constant,the current in the circuit $I = V/Z$ decreases.
The glow of the light bulb depends on the power dissipated,$P = I^2 R$. Since the current $I$ decreases,the power dissipated in the bulb decreases,and therefore,the glow of the light bulb decreases.

(A) $R = 200 \; \Omega, C = 15.0 \; \mu F = 15.0 \times 10^{-6} \; F$
$V = 220 \; V, \nu = 50 \; Hz$
$(a)$ To calculate the current,we need the impedance $Z$ of the circuit:
$X_C = \frac{1}{2 \pi \nu C} = \frac{1}{2 \times 3.14 \times 50 \times 15.0 \times 10^{-6}} \approx 212.3 \; \Omega$
$Z = \sqrt{R^2 + X_C^2} = \sqrt{200^2 + 212.3^2} \approx 291.67 \; \Omega$
$I = \frac{V}{Z} = \frac{220}{291.67} \approx 0.754 \; A$
$(b)$ The voltage across the resistor is $V_R = I R = 0.754 \times 200 = 150.8 \; V$.
The voltage across the capacitor is $V_C = I X_C = 0.754 \times 212.3 = 160.1 \; V$.
The algebraic sum is $150.8 + 160.1 = 310.9 \; V$,which is greater than the source voltage of $220 \; V$. This is not a paradox because $V_R$ and $V_C$ are not in phase. $V_R$ is in phase with the current,while $V_C$ lags the current by $90^{\circ}$. The total voltage is the phasor sum: $V = \sqrt{V_R^2 + V_C^2} = \sqrt{150.8^2 + 160.1^2} \approx 220 \; V$.

(A) The resonant frequency of an $LC$ circuit is given by $f = \frac{1}{2\pi \sqrt{LC}}$.
Rearranging for capacitance,we get $C = \frac{1}{4\pi^2 f^2 L}$.
Given $L = 200 \;\mu H = 200 \times 10^{-6} \;H$.
For $f_1 = 800 \;kHz = 800 \times 10^3 \;Hz$:
$C_1 = \frac{1}{4 \times (3.14)^2 \times (800 \times 10^3)^2 \times 200 \times 10^{-6}} \approx 198 \;pF$.
For $f_2 = 1200 \;kHz = 1200 \times 10^3 \;Hz$:
$C_2 = \frac{1}{4 \times (3.14)^2 \times (1200 \times 10^3)^2 \times 200 \times 10^{-6}} \approx 88 \;pF$.
Thus,the range of the variable capacitor is $88 \;pF$ to $198 \;pF$.

(A) Given:
Inductance $L = 0.50 \; H$
Resistance $R = 100 \; \Omega$
$RMS$ Voltage $V_{rms} = 240 \; V$
Frequency $f = 50 \; Hz$
$(a)$ Peak voltage $V_0 = \sqrt{2} \times V_{rms} = 1.414 \times 240 = 339.41 \; V$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \; rad/s$.
Impedance $Z = \sqrt{R^2 + (\omega L)^2} = \sqrt{100^2 + (100 \pi \times 0.5)^2} = \sqrt{10000 + 24674} = \sqrt{34674} \approx 186.21 \; \Omega$.
Maximum current $I_0 = \frac{V_0}{Z} = \frac{339.41}{186.21} \approx 1.82 \; A$.
$(b)$ The phase angle $\phi$ is given by $\tan \phi = \frac{\omega L}{R} = \frac{100 \pi \times 0.5}{100} = \frac{\pi}{2} \approx 1.57$.
$\phi = \tan^{-1}(1.57) \approx 57.5^{\circ}$.
Converting to radians: $\phi = 57.5 \times \frac{\pi}{180} \approx 1.003 \; rad$.
The time lag $\Delta t = \frac{\phi}{\omega} = \frac{1.003}{100 \pi} \approx 3.19 \times 10^{-3} \; s = 3.2 \; ms$.

(A) Given: Inductance $L = 0.50 \; H$,Resistance $R = 100 \; \Omega$,$RMS$ Voltage $V_{rms} = 240 \; V$,Frequency $\nu = 10 \; kHz = 10^4 \; Hz$.
$(a)$ Peak voltage $V_0 = V_{rms} \sqrt{2} = 240 \sqrt{2} \approx 339.4 \; V$.
Angular frequency $\omega = 2 \pi \nu = 2 \pi \times 10^4 \; rad/s$.
Inductive reactance $X_L = \omega L = 2 \pi \times 10^4 \times 0.5 = \pi \times 10^4 \approx 31416 \; \Omega$.
Impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{100^2 + (31416)^2} \approx 31416 \; \Omega$.
Maximum current $I_0 = \frac{V_0}{Z} = \frac{339.4}{31416} \approx 1.08 \times 10^{-2} \; A$.
$(b)$ Phase angle $\phi = \tan^{-1}(\frac{\omega L}{R}) = \tan^{-1}(\frac{31416}{100}) \approx \tan^{-1}(314.16) \approx 89.82^{\circ}$.
Time lag $\Delta t = \frac{\phi}{\omega} = \frac{89.82 \times \pi / 180}{2 \pi \times 10^4} \approx 2.5 \times 10^{-5} \; s = 25 \; \mu s$.
At very high frequencies,$X_L = \omega L$ becomes very large,making $Z$ very large,so $I_0 \to 0$,acting like an open circuit. In a $DC$ circuit at steady state,$\omega = 0$,so $X_L = 0$,and the inductor acts as a pure conductor (short circuit).

(A) Capacitance of the capacitor,$C = 100 \;\mu\,F = 100 \times 10^{-6} \;F$.
Resistance of the resistor,$R = 40 \;\Omega$.
Supply voltage $(RMS)$,$V_{rms} = 110 \;V$.
Frequency,$f = 60 \;Hz$.
$(a)$ Angular frequency,$\omega = 2\pi f = 120\pi \;rad/s$.
Impedance $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C} = \frac{1}{120\pi \times 100 \times 10^{-6}} \approx 26.53 \;\Omega$.
$Z = \sqrt{40^2 + 26.53^2} = \sqrt{1600 + 703.84} = \sqrt{2303.84} \approx 48 \;\Omega$.
Peak voltage $V_0 = V_{rms} \sqrt{2} = 110 \times 1.414 = 155.56 \;V$.
Maximum current $I_0 = \frac{V_0}{Z} = \frac{155.56}{48} \approx 3.24 \;A$.
$(b)$ In an $RC$ circuit,current leads voltage by phase angle $\phi$,where $\tan \phi = \frac{X_C}{R} = \frac{26.53}{40} = 0.66325$.
$\phi = \tan^{-1}(0.66325) \approx 33.55^{\circ} = 0.5856 \;rad$.
The time lag $\Delta t = \frac{\phi}{\omega} = \frac{0.5856}{120\pi} \approx 1.55 \times 10^{-3} \;s = 1.55 \;ms$.

(A) Given: $C = 100 \;\mu F = 100 \times 10^{-6} \;F$,$R = 40 \;\Omega$,$V_{rms} = 110 \;V$,$f = 12 \;kHz = 12 \times 10^3 \;Hz$.
Angular frequency $\omega = 2\pi f = 2 \times \pi \times 12 \times 10^3 = 24\pi \times 10^3 \;rad/s$.
Peak voltage $V_0 = V_{rms} \sqrt{2} = 110 \sqrt{2} \;V$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{24\pi \times 10^3 \times 100 \times 10^{-6}} = \frac{1}{2.4\pi} \approx 0.1326 \;\Omega$.
Impedance $Z = \sqrt{R^2 + X_C^2} = \sqrt{40^2 + (0.1326)^2} \approx 40 \;\Omega$.
$(a)$ Maximum current $I_0 = \frac{V_0}{Z} = \frac{110 \sqrt{2}}{40} \approx 3.89 \;A \approx 3.9 \;A$.
$(b)$ Phase angle $\phi = \tan^{-1}(\frac{X_C}{R}) = \tan^{-1}(\frac{0.1326}{40}) \approx 0.19^\circ$.
Time lag $\Delta t = \frac{\phi}{\omega} = \frac{0.19 \times \pi}{180 \times 24\pi \times 10^3} \approx 4.4 \times 10^{-8} \;s$.
At very high frequencies,$X_C = \frac{1}{\omega C} \to 0$,so the capacitor acts as a conductor. In a $dc$ circuit,$f=0$,so $\omega=0$ and $X_C \to \infty$,acting as an open circuit.

(A) Given: $L = 80 \, mH = 0.08 \, H$,$C = 60 \, \mu F = 60 \times 10^{-6} \, F$,$V_{rms} = 230 \, V$,$f = 50 \, Hz$.
Angular frequency $\omega = 2 \pi f = 100 \pi \approx 314.16 \, rad/s$.
Inductive reactance $X_L = \omega L = 100 \pi \times 0.08 = 8 \pi \approx 25.13 \, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times 60 \times 10^{-6}} = \frac{10^6}{6000 \pi} \approx 53.05 \, \Omega$.
Net reactance $X = |X_L - X_C| = |25.13 - 53.05| = 27.92 \, \Omega$.
$(a)$ $I_{rms} = \frac{V_{rms}}{X} = \frac{230}{27.92} \approx 8.24 \, A$. Current amplitude $I_0 = I_{rms} \sqrt{2} \approx 8.24 \times 1.414 \approx 11.65 \, A$.
$(b)$ $V_{L,rms} = I_{rms} X_L = 8.24 \times 25.13 \approx 207.07 \, V$. $V_{C,rms} = I_{rms} X_C = 8.24 \times 53.05 \approx 437.13 \, V$.
$(c)$ Average power in inductor $P_L = V_{L,rms} I_{rms} \cos(90^\circ) = 0 \, W$.
$(d)$ Average power in capacitor $P_C = V_{C,rms} I_{rms} \cos(90^\circ) = 0 \, W$.
$(e)$ Total average power $P = P_L + P_C = 0 \, W$.

(N/A) In an $L-C-R$ series $AC$ circuit,the current $i$ is the same through all components. Let the instantaneous current be $i = I_m \sin(\omega t)$.
$(i)$ The voltage across the resistance $(V_R)$ is given by $V_R = iR$.
$(ii)$ The voltage across the inductor $(V_L)$ is given by $V_L = iX_L$,where $X_L = \omega L$ is the inductive reactance. Thus,$V_L = i(\omega L)$.
$(iii)$ The voltage across the capacitor $(V_C)$ is given by $V_C = iX_C$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance. Thus,$V_C = i(\frac{1}{\omega C})$.

(B) In an $LCR$ series circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Case $1$: If ${X_L} > {X_C}$,then $\tan \phi > 0$,which means $\phi$ is positive. In this case,the voltage leads the current,or the current lags the voltage.
Case $2$: If ${X_C} > {X_L}$,then $\tan \phi < 0$,which means $\phi$ is negative. In this case,the current leads the voltage.
Therefore,when ${X_C} < {X_L}$,the current lags the voltage,and when ${X_C} > {X_L}$,the current leads the voltage.

(N/A) The inductive reactance is given by $X_{L} = 2 \pi \nu L$ and the capacitive reactance is given by $X_{C} = \frac{1}{2 \pi \nu C}$.
For very high frequencies,$\nu \rightarrow \infty$.
As $\nu \rightarrow \infty$,the inductive reactance $X_{L} \rightarrow \infty$,which acts as an open circuit. Conversely,the capacitive reactance $X_{C} \rightarrow 0$,which acts as a short circuit.
Therefore,in the given circuit,the capacitors $C_{1}$ and $C_{2}$ behave as short circuits (wires),and the inductors $L_{1}$ and $L_{2}$ behave as open circuits (breaks in the path).
By replacing the components with their high-frequency equivalents,the circuit simplifies to a series combination of resistors $R_{1}$ and $R_{3}$.
Thus,the effective impedance is $Z_{eq} = R_{1} + R_{3}$.

(A) Given: Inductance $L = 0.01 \, H$,Resistance $R = 1 \, \Omega$,Voltage $V = 200 \, V$,Frequency $f = 50 \, Hz$.
$1$. Calculate the inductive reactance $X_L$:
$X_L = 2 \pi f L = 2 \times 3.1416 \times 50 \times 0.01 = 3.1416 \, \Omega$.
$2$. Calculate the impedance $Z$:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{1^2 + (3.1416)^2} = \sqrt{1 + 9.8696} = \sqrt{10.8696} \approx 3.297 \, \Omega$.
$3$. Calculate the phase angle $\phi$:
$\tan \phi = \frac{X_L}{R} = \frac{3.1416}{1} = 3.1416$.
$\phi = \tan^{-1}(3.1416) \approx 72.34^{\circ}$.
$4$. Calculate the time lag $\Delta t$:
$\Delta t = \frac{\phi}{\omega} = \frac{\phi}{2 \pi f} = \frac{72.34 \times \pi / 180}{2 \times \pi \times 50} = \frac{72.34}{36000} \approx 0.00201 \, s$ or $2.01 \, ms$.

(N/A) The given equation for an $LCR$ circuit is:
$L \frac{di}{dt} + Ri + \frac{q}{C} = V_m \sin \omega t$
$(a)$ Multiplying by $i$ on both sides:
$Li \frac{di}{dt} + i^2 R + \frac{q}{C} i = V_m i \sin \omega t$
Since $i = \frac{dq}{dt}$,we can write:
$\frac{d}{dt} \left( \frac{1}{2} L i^2 \right) + i^2 R + \frac{d}{dt} \left( \frac{q^2}{2C} \right) = Vi$
$(b)$ Physical interpretation:
$Li \frac{di}{dt} = \frac{d}{dt} (\frac{1}{2} L i^2)$ is the rate of change of magnetic energy stored in the inductor.
$i^2 R$ is the rate of Joule heating (power dissipation) in the resistor.
$\frac{q}{C} i = \frac{d}{dt} (\frac{q^2}{2C})$ is the rate of change of electric energy stored in the capacitor.
$Vi$ is the instantaneous power supplied by the source.
$(c)$ The equation $\frac{d}{dt} (\frac{1}{2} L i^2 + \frac{q^2}{2C}) + i^2 R = Vi$ represents the conservation of energy,where the power supplied by the source equals the sum of the rate of change of stored energy and the power dissipated as heat.
$(d)$ Integrating over one cycle $T = \frac{2\pi}{\omega}$:
$\int_0^T \frac{d}{dt} (\frac{1}{2} L i^2 + \frac{q^2}{2C}) dt + \int_0^T i^2 R dt = \int_0^T Vi dt$
Since the energy stored in $L$ and $C$ is periodic,the integral of the derivative over one cycle is zero.
$0 + I_{rms}^2 R T = \int_0^T Vi dt$
Since $I_{rms}^2 R T > 0$,the average power $\int_0^T Vi dt$ must be positive. This implies $\cos \phi > 0$,meaning the phase difference $\phi$ must be acute.

(A) The impedance $Z$ of an $RL$ circuit is given by $Z = \sqrt{R^2 + X_L^2} = 100\, \Omega$.
Given that the voltage leads the current by $\phi = 45^{\circ}$,we have $\tan \phi = \frac{X_L}{R}$.
Since $\tan 45^{\circ} = 1$,it follows that $X_L = R$.
Substituting $X_L = R$ into the impedance equation: $\sqrt{X_L^2 + X_L^2} = 100 \Rightarrow \sqrt{2} X_L = 100$.
Thus,$X_L = \frac{100}{\sqrt{2}} = 50\sqrt{2}\, \Omega$.
Since $X_L = 2\pi f L$,we have $L = \frac{X_L}{2\pi f} = \frac{50\sqrt{2}}{2 \times \pi \times 1000}$.
$L = \frac{25\sqrt{2}}{1000\pi} \approx \frac{25 \times 1.414}{3141.59} \approx 0.01125\, H = 1.125 \times 10^{-2}\, H$.

(B) When an iron rod is inserted into the inductor,its self-inductance $L$ increases because the permeability of the core increases.
The inductive reactance is given by $X_L = \omega L$. As $L$ increases,$X_L$ also increases.
The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2}$. Since $X_L$ increases,the total impedance $Z$ of the circuit increases.
The current in the circuit is given by $I = \frac{V}{Z}$. As $Z$ increases,the current $I$ flowing through the circuit decreases.
The power dissipated in the bulb is given by $P = I^2 R$. Since the current $I$ decreases,the power dissipated in the bulb decreases,and therefore,the glow of the light bulb decreases.

(A) For a series $R-C$ circuit,the total applied voltage $V$ is given by the phasor sum of the voltage across the resistor $V_R$ and the voltage across the capacitor $V_C$:
$V^2 = V_R^2 + V_C^2$
Given $V = 10\, V$ and $V_C = 8\, V$,we have:
$10^2 = V_R^2 + 8^2$
$100 = V_R^2 + 64$
$V_R^2 = 100 - 64 = 36$
$V_R = 6\, V$
Now,the phase difference $\phi$ between the current and the applied voltage is given by:
$\tan \phi = \frac{V_C}{V_R}$
$\tan \phi = \frac{8}{6} = \frac{4}{3}$
$\phi = \tan^{-1}\left(\frac{4}{3}\right)$
Thus,the voltage across $R$ is $6\, V$ and the phase difference is $\tan^{-1}\left(\frac{4}{3}\right)$.

(B) In an $LCR$ series circuit,the relationship between the source voltage $V$ and the voltages across the components is given by:
$V^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2}$
Given:
$V = 120 \ V$
$V_{R} = 40 \ V$
$V_{L} = 50 \ V$
Substituting the values into the equation:
$120^{2} = 40^{2} + (50 - V_{C})^{2}$
$14400 = 1600 + (50 - V_{C})^{2}$
$(50 - V_{C})^{2} = 14400 - 1600 = 12800$
Taking the square root on both sides:
$50 - V_{C} = \pm \sqrt{12800} = \pm \sqrt{6400 \times 2} = \pm 80\sqrt{2}$
Case $1$: $50 - V_{C} = 80\sqrt{2} \implies V_{C} = 50 - 80\sqrt{2}$ (Negative value,not possible for magnitude)
Case $2$: $50 - V_{C} = -80\sqrt{2} \implies V_{C} = 50 + 80\sqrt{2}$
$V_{C} = 10(5 + 8\sqrt{2}) \ V$

(D) The phase difference $\phi$ between voltage and current is defined as follows:
$(a)$ In a purely resistive circuit,voltage and current are in the same phase,so the phase difference is $0$.
$(b)$ In a pure inductive circuit,voltage leads current by $\frac{\pi}{2}$,which means current lags voltage by $\frac{\pi}{2}$.
$(c)$ In a pure capacitive circuit,current leads voltage by $\frac{\pi}{2}$.
$(d)$ In an $LCR$ series circuit,the phase difference is given by $\tan \phi = \frac{X_L - X_C}{R}$,or equivalently $\tan \phi = \frac{-(X_C - X_L)}{R}$,which corresponds to $\tan^{-1}\left(\frac{X_C - X_L}{R}\right)$ depending on the sign convention.
Comparing these,we get $(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)$.

(B) The power factor of an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$.
For the initial circuit ($RL$ circuit):
$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (3R)^2} = \sqrt{R^2 + 9R^2} = \sqrt{10R^2} = R\sqrt{10}$.
Old power factor,$\cos \phi = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}}$.
For the new circuit ($RLC$ circuit):
$Z' = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (3R - 2R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.
New power factor,$\cos \phi' = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The ratio of the new power factor to the old power factor is:
$\frac{\cos \phi'}{\cos \phi} = \frac{1/\sqrt{2}}{1/\sqrt{10}} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} = \frac{\sqrt{5}}{1}$.
Comparing this with $\sqrt{5} : x$,we get $x = 1$.

(A) The phase angle $\phi$ in an $LCR$ circuit is given by $\tan \phi = \frac{X_C - X_L}{R}$.
Since the current leads the voltage,the circuit is capacitive,and the phase angle is $\phi = -45^{\circ}$.
Thus,$\tan(-45^{\circ}) = \frac{X_C - X_L}{R} \implies -1 = \frac{X_C - X_L}{R}$.
This gives $X_L - X_C = R$.
Given $L = 30 \, {mH} = 0.03 \, {H}$,$R = 1 \, \Omega$,and $\omega = 300 \, {rad/s}$.
$X_L = \omega L = 300 \times 0.03 = 9 \, \Omega$.
Substituting the values: $9 - X_C = 1 \implies X_C = 8 \, \Omega$.
Since $X_C = \frac{1}{\omega C}$,we have $\frac{1}{300 \times C} = 8$.
$C = \frac{1}{300 \times 8} = \frac{1}{2400} = \frac{1}{2.4} \times 10^{-3} \, {F}$.
Wait,re-evaluating the phase condition: If current leads voltage,$\phi$ is negative,so $\tan \phi = \frac{X_C - X_L}{R} = \tan(-45^{\circ}) = -1$.
$X_L - X_C = R \implies 9 - X_C = 1 \implies X_C = 8$. This leads to $C = \frac{1}{2400} \approx 0.416 \times 10^{-3} \, {F}$.
If the question implies the magnitude of the phase difference is $45^{\circ}$ and current leads,then $X_C > X_L$ is incorrect; it should be $X_L - X_C = R$ if current lags. For current to lead,$X_C - X_L = R$ is not possible if $X_C < X_L$. Let's re-check: $\tan \phi = \frac{X_L - X_C}{R}$. For current to lead,$\phi$ must be negative,so $\frac{X_C - X_L}{R} = \tan(45^{\circ}) = 1$ is wrong. The correct relation is $\tan \phi = \frac{X_L - X_C}{R}$. For current to lead,$\phi = -45^{\circ}$,so $-1 = \frac{X_L - X_C}{R} \implies X_C - X_L = R$.
$X_C - 9 = 1 \implies X_C = 10 \, \Omega$.
$\frac{1}{300 \times C} = 10 \implies C = \frac{1}{3000} = \frac{1}{3} \times 10^{-3} \, {F}$.
Thus,$x = 3$.

(D) Given: $R=100\,\Omega$,$L=0.5\,mH = 0.5 \times 10^{-3}\,H$,$C=0.1\,pF = 0.1 \times 10^{-12}\,F$,$f=50\,Hz$.
Angular frequency $\omega = 2\pi f = 2\pi \times 50 = 100\pi\,rad/s$.
Inductive reactance $X_L = \omega L = 100\pi \times 0.5 \times 10^{-3} = 0.05\pi\,\Omega \approx 0.157\,\Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 0.1 \times 10^{-12}} = \frac{10^{11}}{10\pi} = \frac{10^{10}}{\pi}\,\Omega \approx 3.18 \times 10^9\,\Omega$.
Since $X_C \gg X_L$ and $X_C \gg R$,the circuit is predominantly capacitive.
The phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Since $X_C$ is extremely large compared to $R$ and $X_L$,$\tan \phi \approx -\infty$,which implies $\phi \approx -90^{\circ}$.
The negative sign indicates that the current leads the voltage by approximately $90^{\circ}$,confirming the circuit is predominantly capacitive.

(C) Inductive reactance,$X_L = \omega L = 2\pi f L$.
Substituting the values,$X_L = 2 \times \frac{22}{7} \times 50 \times 0.07 = 100 \times \frac{22}{7} \times 0.07 = 22\,\Omega$.
Impedance of the circuit,$Z = \sqrt{R^2 + X_L^2} = \sqrt{12^2 + 22^2} = \sqrt{144 + 484} = \sqrt{628} \approx 25.06\,\Omega \approx 25\,\Omega$.
Current in the circuit,$I = \frac{V}{Z} = \frac{220}{25} = 8.8\,A$.
Phase angle $\phi$ is given by $\tan \phi = \frac{X_L}{R} = \frac{22}{12} = \frac{11}{6}$.
Therefore,$\phi = \tan^{-1}\left(\frac{11}{6}\right)$.

(A) Given: $V(t) = 210 \sin(3000t) \text{ V}$,$L = 10 \text{ mH} = 10^{-2} \text{ H}$,$C = 25 \mu\text{F} = 25 \times 10^{-6} \text{ F}$,$R = 100 \Omega$.
Angular frequency $\omega = 3000 \text{ rad/s}$.
Inductive reactance $X_L = \omega L = 3000 \times 10^{-2} = 30 \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{3000 \times 25 \times 10^{-6}} = \frac{1}{0.075} = \frac{1000}{75} = \frac{40}{3} \Omega \approx 13.33 \Omega$.
The phase difference $\Phi$ is given by $\tan \Phi = \frac{X_L - X_C}{R}$.
$\tan \Phi = \frac{30 - 13.33}{100} = \frac{16.67}{100} = 0.1667 \approx 0.17$.
Therefore,$\Phi = \tan^{-1}(0.17)$.

(B) The resistance of the lamp is $R = \frac{V_L^2}{P} = \frac{100^2}{50} = 200\,\Omega$.
In an $RC$ series circuit,the total voltage $V$ is given by $V^2 = V_R^2 + V_C^2$,where $V_R = 100\,V$ is the voltage across the lamp.
$200^2 = 100^2 + V_C^2 \Rightarrow V_C^2 = 40000 - 10000 = 30000 \Rightarrow V_C = 100\sqrt{3}\,V$.
The current in the circuit is $I = \frac{V_R}{R} = \frac{100}{200} = 0.5\,A$.
The capacitive reactance is $X_C = \frac{V_C}{I} = \frac{100\sqrt{3}}{0.5} = 200\sqrt{3}\,\Omega$.
Since $X_C = \frac{1}{2\pi f C}$,we have $200\sqrt{3} = \frac{1}{2 \times \pi \times 50 \times C} \Rightarrow C = \frac{1}{200\sqrt{3} \times 100\pi} = \frac{1}{20000\pi\sqrt{3}}\,F$.
Converting to $\mu F$: $C = \frac{10^6}{20000\pi\sqrt{3}} = \frac{50}{\pi\sqrt{3}}\,\mu F$.
Comparing this with $C = \frac{50}{\pi\sqrt{x}}\,\mu F$,we get $x = 3$.

(D) For element $X$,the current is in phase with the voltage,so $X$ is a resistor.
The resistance $R = \frac{V_0}{I_0} = \frac{100}{5} = 20\,\Omega$.
For element $Y$,the current lags behind the voltage by $\frac{\pi}{2}$,so $Y$ is an inductor.
The inductive reactance $X_L = \frac{V_0}{I_0} = \frac{100}{5} = 20\,\Omega$.
When $X$ and $Y$ are connected in series,the total impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\,\Omega$.
The peak current in the series circuit is $I_0 = \frac{V_0}{Z} = \frac{100}{20\sqrt{2}} = \frac{5}{\sqrt{2}}\,A$.
The rms current is $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{5/\sqrt{2}}{\sqrt{2}} = \frac{5}{2}\,A$.

(C) In a series $RLC$ circuit,the total voltage (source $e.m.f.$) is given by the phasor sum of the individual potential differences:
$V_{source} = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given:
$V_R = 30 \,V$
$V_L = 60 \,V$
$V_C = 100 \,V$
Substituting these values into the formula:
$V_{source} = \sqrt{(30)^2 + (60 - 100)^2}$
$V_{source} = \sqrt{(30)^2 + (-40)^2}$
$V_{source} = \sqrt{900 + 1600}$
$V_{source} = \sqrt{2500} = 50 \,V$
Thus,the $e.m.f.$ of the source is $50 \,V$.

(C) Given: $L = 10^{-3} \,H$,$C = 3 \times 10^{-7} \,F$,$f = 50 \,Hz$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \,rad/s$.
Inductive reactance $X_L = \omega L = (100 \pi) \times 10^{-3} = 0.1 \pi \, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times 3 \times 10^{-7}} = \frac{10^7}{300 \pi} = \frac{10^5}{3 \pi} \, \Omega$.
Since it is a series $L-C$ circuit,the impedance $Z = |X_L - X_C|$.
$Z = |0.1 \pi - \frac{10^5}{3 \pi}|$.
Since $\frac{10^5}{3 \pi} \approx 10610 \, \Omega$ and $0.1 \pi \approx 0.314 \, \Omega$,the impedance is $Z = \frac{10^5}{3 \pi} - 0.1 \pi = \frac{10^5}{3 \pi} - \frac{\pi}{10} \, \Omega$.

(C) The bulb is rated $P = 100 \,W$ and $V_b = 300 \,V$.
The current through the bulb is $I = \frac{P}{V_b} = \frac{100}{300} = \frac{1}{3} \,A$.
The resistance of the bulb is $R = \frac{V_b^2}{P} = \frac{300^2}{100} = 900 \,\Omega$.
When connected to a $V = 500 \,V$ supply,the total impedance $Z$ is given by $Z = \frac{V}{I} = \frac{500}{1/3} = 1500 \,\Omega$.
In an $LR$ circuit,$Z^2 = R^2 + X_L^2$,so $X_L^2 = Z^2 - R^2$.
$X_L^2 = 1500^2 - 900^2 = (1500 - 900)(1500 + 900) = 600 \times 2400 = 1440000$.
$X_L = \sqrt{1440000} = 1200 \,\Omega$.
Since $X_L = 2 \pi f L$,we have $1200 = 2 \pi \left(\frac{150}{\pi}\right) L$.
$1200 = 300 L$,which gives $L = 4 \,H$.

(A) Given: $R = 3\,\Omega$,$L = 25.48\,mH = 25.48 \times 10^{-3}\,H$,$C = 796\,\mu F = 796 \times 10^{-6}\,F$,and $f = 50\,Hz$.
First,calculate the inductive reactance $X_L$:
$X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \approx 8\,\Omega$.
Next,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}} \approx 4\,\Omega$.
The impedance $Z$ of the series $LCR$ circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values:
$Z = \sqrt{3^2 + (8 - 4)^2} = \sqrt{9 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,\Omega$.

(B) Given: $R = 400\,\Omega$,$L = (3/\pi)\,H$,$e = 400 \sin(100\pi t)$.
Comparing with $e = E_0 \sin(\omega t)$,we get $E_0 = 400\,V$ and $\omega = 100\pi\,rad/s$.
Inductive reactance $X_L = \omega L = 100\pi \times (3/\pi) = 300\,\Omega$.
Impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{400^2 + 300^2} = 500\,\Omega$.
Peak current $I_0 = E_0 / Z = 400 / 500 = 0.8\,A = (4/5)\,A$. Thus,option $(a)$ is incorrect.
In a series $RL$ circuit,the potential difference across the resistor is always in phase with the current. Thus,option $(b)$ is correct.
The potential difference across an inductor always leads the current by $90^{\circ}$. Thus,options $(c)$ and $(d)$ are incorrect.