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

(C) The total resistance of the circuit is $R = R_1 + R_2 = 44\,\Omega + 36\,\Omega = 80\,\Omega$.
The net reactance of the circuit is $X = X_L - X_C = 90\,\Omega - 30\,\Omega = 60\,\Omega$.
The impedance of the circuit is $Z = \sqrt{R^2 + X^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$.
The current in the series circuit is $I = \frac{V}{Z} = \frac{200\, V}{100\,\Omega} = 2\, A$.
The power dissipated in the coil is due to its internal resistance $R_2 = 36\,\Omega$. Therefore,$P = I^2 R_2 = (2)^2 \times 36 = 4 \times 36 = 144\, W$.

(C) In an $AC$ circuit with an inductor and a capacitor connected in parallel,the current through the inductor $(I_L)$ lags the voltage by $90^{\circ}$,and the current through the capacitor $(I_C)$ leads the voltage by $90^{\circ}$.
Thus,the currents $I_L$ and $I_C$ are in opposite phases,i.e.,they have a phase difference of $180^{\circ}$.
The total current $I$ drawn from the generator is the vector sum of these currents:
$I = |I_L - I_C|$
Given $I_L = 0.9\,A$ and $I_C = 0.4\,A$.
Therefore,$I = |0.9 - 0.4| = 0.5\,A$.

(A) To find the impedance of the circuit,we first calculate $X_{L}$ and $X_{C}$.
$X_{L} = 2 \pi \nu L = 2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \; \Omega = 8 \; \Omega$.
$X_{C} = \frac{1}{2 \pi \nu C} = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}} = 4 \; \Omega$.
Therefore,$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{3^{2} + (8 - 4)^{2}} = 5 \; \Omega$.
$(b)$ Phase difference,$\phi = \tan^{-1} \left( \frac{X_{L} - X_{C}}{R} \right) = \tan^{-1} \left( \frac{8 - 4}{3} \right) = \tan^{-1} \left( \frac{4}{3} \right) \approx 53.1^{\circ}$.
Since $\phi$ is positive,the current lags the voltage.
$(c)$ The power dissipated in the circuit is $P = I_{rms}^{2} R$.
$I_{rms} = \frac{V_{m}}{\sqrt{2} Z} = \frac{283}{\sqrt{2} \times 5} \approx 40 \; A$.
Therefore,$P = (40)^{2} \times 3 = 4800 \; W$.
$(d)$ Power factor = $\cos \phi = \frac{R}{Z} = \frac{3}{5} = 0.6$.

(N/A) Given: Inductance $L = 80 \; mH = 80 \times 10^{-3} \; H$,Capacitance $C = 60 \; \mu F = 60 \times 10^{-6} \; F$,Resistance $R = 15 \; \Omega$,Voltage $V = 230 \; V$,Frequency $f = 50 \; Hz$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \; rad/s \approx 314.16 \; rad/s$.
Inductive reactance $X_L = \omega L = 100 \pi \times 80 \times 10^{-3} \approx 25.13 \; \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times 60 \times 10^{-6}} \approx 53.05 \; \Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{15^2 + (25.13 - 53.05)^2} = \sqrt{225 + (-27.92)^2} = \sqrt{225 + 779.53} = \sqrt{1004.53} \approx 31.69 \; \Omega$.
Current $I = \frac{V}{Z} = \frac{230}{31.69} \approx 7.26 \; A$.
Average power transferred to the inductor $P_L = 0 \; W$ (since phase difference is $90^\circ$).
Average power transferred to the capacitor $P_C = 0 \; W$ (since phase difference is $90^\circ$).
Average power transferred to the resistor $P_R = I^2 R = (7.26)^2 \times 15 \approx 790.5 \; W$.
Total power absorbed by the circuit $P_{total} = P_R + P_L + P_C = 790.5 + 0 + 0 = 790.5 \; W$.

(N/A) Consider a series $LCR$ circuit connected to an $AC$ source with electromotive force $\varepsilon = V_m \sin \omega t$.
Let $q$ be the charge on the capacitor and $I$ be the current in the circuit at any time $t$. According to Kirchhoff's voltage law (loop rule),the sum of potential drops across the inductor $(V_L)$,resistor $(V_R)$,and capacitor $(V_C)$ must equal the applied source voltage $V$.
$V = V_L + V_R + V_C$
Substituting the expressions for potential differences:
$V = L \frac{dI}{dt} + IR + \frac{q}{C}$
Where:
$V_L = L \frac{dI}{dt}$ is the potential difference across the inductor.
$V_R = IR$ is the potential difference across the resistor.
$V_C = \frac{q}{C}$ is the potential difference across the capacitor.

(N/A) In a series $LCR$ circuit,the current $I$ is the same through all components at any instant.
Let the current be $I = I_m \sin(\omega t)$.
The voltage across the resistor $V_R$ is in phase with the current $I$.
The voltage across the inductor $V_L$ leads the current $I$ by $\pi/2$.
The voltage across the capacitor $V_C$ lags behind the current $I$ by $\pi/2$.
The total voltage $V$ is the phasor sum of $V_R$,$V_L$,and $V_C$. Since $V_L$ and $V_C$ are in opposite directions,their resultant is $(V_L - V_C)$.
Using the phasor diagram,the resultant voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
The phase angle $\phi$ between the source voltage and current is given by $\tan \phi = \frac{V_L - V_C}{V_R} = \frac{I_m X_L - I_m X_C}{I_m R} = \frac{X_L - X_C}{R}$.
Thus,the phase relation is $\phi = \tan^{-1} \left( \frac{X_L - X_C}{R} \right)$.

(N/A) If $X_C > X_L$,the phase angle $\phi$ is positive and the circuit becomes capacitive; consequently,the current in the circuit leads the source voltage.
If $X_C < X_L$,the phase angle $\phi$ is negative and the circuit becomes inductive; consequently,the source voltage in the circuit leads the current.
The phasor diagram and the variation of $V$ and $I$ with $\omega t$ for the case $X_C > X_L$ are shown in the figure.
Thus,we have obtained the amplitude and phase of the current for an $LCR$ series circuit using the technique of phasors,but this method has certain disadvantages:
$1$. The phasor diagram provides no information about the initial conditions.
$2$. One can choose any arbitrary value of $t$ and draw different phasors. The solution obtained is called the steady-state solution,which is not a general solution.
$3$. Furthermore,there exists a transient solution that persists even for $V = 0$.
The general solution is the sum of the transient solution and the steady-state solution. After a sufficiently long time,the effects of the transient solution die out,and the behavior of the circuit is described by the steady-state solution.

(N/A) The voltage equation for an $L-C-R$ series circuit is given by Kirchhoff's voltage law:
$L \frac{dI}{dt} + RI + \frac{q}{C} = V$
Substituting $V = V_m \sin \omega t$,we get:
$L \frac{dI}{dt} + RI + \frac{q}{C} = V_m \sin \omega t$ ... $(1)$
Since $I = \frac{dq}{dt}$,we have $\frac{dI}{dt} = \frac{d^2q}{dt^2}$. Substituting this into $(1)$:
$L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = V_m \sin \omega t$
Dividing by $L$:
$\frac{d^2q}{dt^2} + \frac{R}{L} \frac{dq}{dt} + \frac{q}{LC} = \frac{V_m}{L} \sin \omega t$ ... $(2)$
This is the equation for a forced,damped oscillator. Let the solution be $q = q_m \sin(\omega t + \theta)$ ... $(3)$
Then $\frac{dq}{dt} = q_m \omega \cos(\omega t + \theta)$ ... $(4)$ and $\frac{d^2q}{dt^2} = -q_m \omega^2 \sin(\omega t + \theta)$ ... $(5)$
Substituting $(3), (4),$ and $(5)$ into $(2)$:
$-q_m \omega^2 L \sin(\omega t + \theta) + R q_m \omega \cos(\omega t + \theta) + \frac{q_m}{C} \sin(\omega t + \theta) = V_m \sin \omega t$
$q_m \omega [R \cos(\omega t + \theta) + (\frac{1}{\omega C} - \omega L) \sin(\omega t + \theta)] = V_m \sin \omega t$
Using $X_C = \frac{1}{\omega C}$ and $X_L = \omega L$,and impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$,we define $\cos \phi = \frac{R}{Z}$ and $\sin \phi = \frac{X_L - X_C}{Z}$.
This leads to the phase relation where the current $I = I_m \sin(\omega t + \phi)$ lags or leads the voltage by phase angle $\phi = \tan^{-1}(\frac{X_L - X_C}{R})$.

(A) In an $L-C-R$ series $AC$ circuit,the voltage across the resistor $(V_R)$ is in phase with the current $(I)$.
The voltage across the inductor $(V_L)$ leads the current by $90^\circ$,and the voltage across the capacitor $(V_C)$ lags behind the current by $90^\circ$.
The net reactance of the circuit is $(X_L - X_C)$.
Using the phasor diagram,the total voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Since $V = IZ$,$V_R = IR$,$V_L = IX_L$,and $V_C = IX_C$,we substitute these into the equation:
$IZ = \sqrt{(IR)^2 + (IX_L - IX_C)^2}$
$IZ = I \sqrt{R^2 + (X_L - X_C)^2}$
Therefore,the impedance $Z$ is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

(N/A) In an $L-C-R$ series circuit,the sum of potential drops across the inductor $(L)$,resistor $(R)$,and capacitor $(C)$ is equal to the applied electromotive force $(E(t))$.
According to Kirchhoff's voltage law:
$L \frac{dI}{dt} + IR + \frac{q}{C} = E(t)$
Since the current $I = \frac{dq}{dt}$,the rate of change of current is $\frac{dI}{dt} = \frac{d^2q}{dt^2}$.
Substituting these into the equation,we get:
$L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = E(t)$
This is the second-order linear differential equation for the charge $q$ in an $L-C-R$ series circuit.

(N/A) Let the $rms$ current in circuit $(a)$ be $I_{a}$ and in circuit $(b)$ be $I_{b}$.
For circuit $(a)$,the impedance is $Z_{a} = R$. Thus,$I_{a} = \frac{V_{rms}}{R}$.
For circuit $(b)$,the impedance is $Z_{b} = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$. Thus,$I_{b} = \frac{V_{rms}}{\sqrt{R^{2} + (X_{L} - X_{C})^{2}}}$.
$(a)$ For $I_{a} = I_{b}$,we must have $Z_{a} = Z_{b}$.
$\therefore R = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.
Squaring both sides,$R^{2} = R^{2} + (X_{L} - X_{C})^{2}$,which implies $(X_{L} - X_{C})^{2} = 0$,or $X_{L} = X_{C}$.
This is the condition of electrical resonance.
$(b)$ Since $Z_{b} = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$,it is clear that $Z_{b} \geq R$ for all values of $X_{L}$ and $X_{C}$.
Since $I_{b} = \frac{V_{rms}}{Z_{b}}$ and $I_{a} = \frac{V_{rms}}{R}$,and $Z_{b} \geq R$,it follows that $I_{b} \leq I_{a}$.
Therefore,the $rms$ current in circuit $(b)$ can never be larger than that in circuit $(a)$.

(D) The phase angle $\phi$ by which the supply voltage leads the current in an $LCR$ series circuit is given by the relation: $\tan \phi = \frac{X_L - X_C}{R} = \frac{2 \pi \nu L - \frac{1}{2 \pi \nu C}}{R}$.
At very low frequencies $(\nu \to 0)$,the capacitive reactance $X_C = \frac{1}{2 \pi \nu C}$ is very large,making $X_L - X_C$ negative. Thus,$\tan \phi$ is negative,meaning the voltage lags behind the current (or the phase angle is negative).
As the frequency $\nu$ increases and reaches the resonant frequency $\nu_r = \frac{1}{2 \pi \sqrt{LC}}$,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,so $X_L - X_C = 0$ and $\phi = 0$.
At very high frequencies $(\nu \to \infty)$,the inductive reactance $X_L = 2 \pi \nu L$ becomes very large,making $X_L - X_C$ positive. Thus,$\tan \phi$ is positive,meaning the voltage leads the current (or the phase angle is positive).
Therefore,as the frequency increases from very low to very high values,the sign of the phase angle $\phi$ changes from negative to positive.

(N/A) The circuit consists of a resistor $R$ in parallel with a series combination of a capacitor $C$ and an inductor $L$. The voltage across both branches is $V = V_m \sin \omega t$.
$1$. Current through the resistor branch $(i_R)$:
$i_R = \frac{V}{R} = \frac{V_m}{R} \sin \omega t$.
$2$. Current through the $LC$ branch $(i_{LC})$:
The impedance of the $LC$ series branch is $Z_{LC} = j(\omega L - \frac{1}{\omega C})$.
The current is $i_{LC} = \frac{V}{Z_{LC}} = \frac{V_m \sin \omega t}{j(\omega L - \frac{1}{\omega C})} = \frac{V_m \sin(\omega t - \pi/2)}{\omega L - 1/(\omega C)}$ (if $\omega L > 1/\omega C$).
$3$. Total current $i$:
$i = i_R + i_{LC} = \frac{V_m}{R} \sin \omega t + \frac{V_m}{\omega L - 1/(\omega C)} \sin(\omega t - \pi/2)$.
Using phasor addition,the total current is $i = I_m \sin(\omega t + \phi)$,where $I_m = V_m \sqrt{(\frac{1}{R})^2 + (\frac{1}{\omega L - 1/(\omega C)})^2}$.
$4$. Impedance $Z$:
Since $i = V/Z$,the total impedance $Z$ is given by $\frac{1}{Z} = \sqrt{(\frac{1}{R})^2 + (\frac{1}{\omega L - 1/(\omega C)})^2}$.
Therefore,$Z = \frac{R |\omega L - 1/(\omega C)|}{\sqrt{R^2 + (\omega L - 1/(\omega C))^2}}$.

(C) Given: $f = 750\, Hz$,$V_{rms} = 20\, V$,$R = 100\, \Omega$,$L = 0.1803\, H$,$C = 10\, \mu F = 10 \times 10^{-6}\, F$,$S = 2\, J/^{\circ}C$,$\Delta T = 10^{\circ}C$.
First,calculate the inductive reactance $X_L$ and capacitive reactance $X_C$:
$X_L = 2\pi fL = 2 \times 3.14159 \times 750 \times 0.1803 \approx 849.6\, \Omega$.
$X_C = \frac{1}{2\pi fC} = \frac{1}{2 \times 3.14159 \times 750 \times 10 \times 10^{-6}} \approx 21.2\, \Omega$.
Calculate the impedance $Z$:
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (849.6 - 21.2)^2} = \sqrt{10000 + (828.4)^2} \approx \sqrt{10000 + 686246} \approx \sqrt{696246} \approx 834.4\, \Omega$.
The power dissipated in the resistor is $P = I_{rms}^2 R = \left(\frac{V_{rms}}{Z}\right)^2 R = \left(\frac{20}{834.4}\right)^2 \times 100 \approx (0.02397)^2 \times 100 \approx 0.05746\, W$ (or $J/s$).
The heat required is $Q = S \cdot \Delta T = 2\, J/^{\circ}C \times 10^{\circ}C = 20\, J$.
The time $t$ required is $t = \frac{Q}{P} = \frac{20}{0.05746} \approx 348.07\, s$.
Thus,the time is close to $348\, s$.

(D) Given: $P = 400 \ W$,$V_{rms} = 250 \ V$,$f = 50 \ Hz$,$\cos \phi = 0.8$.
$1$. Calculate impedance $Z$:
$P = V_{rms} I_{rms} \cos \phi = \frac{V_{rms}^2}{Z} \cos \phi$
$400 = \frac{250^2}{Z} \times 0.8$
$Z = \frac{62500 \times 0.8}{400} = \frac{50000}{400} = 125 \ \Omega$.
$2$. Calculate resistance $R$ and inductive reactance $X_L$:
$R = Z \cos \phi = 125 \times 0.8 = 100 \ \Omega$.
$X_L = Z \sin \phi = 125 \times \sqrt{1 - 0.8^2} = 125 \times 0.6 = 75 \ \Omega$.
$3$. For unity power factor,the circuit must be in resonance,so $X_C = X_L = 75 \ \Omega$.
$X_C = \frac{1}{2 \pi f C} = 75$
$C = \frac{1}{2 \pi \times 50 \times 75} = \frac{1}{7500 \pi} \ F = \frac{10^6}{7500 \pi} \ \mu F = \frac{10000}{75 \pi} \ \mu F = \frac{400}{3 \pi} \ \mu F$.
Comparing with $C = (\frac{n}{3 \pi}) \ \mu F$,we get $n = 400$.

(B) Given angular frequency $\omega = 100 \, rad/s$.
$1$. For the capacitor branch:
The current through the $60 \, \Omega$ resistor is $I_C = \frac{V_R}{R} = \frac{15 \, V}{60 \, \Omega} = 0.25 \, A = \frac{1}{4} \, A$.
Since the capacitor is in series with this resistor,the same current flows through the capacitor.
Using $V_C = I_C X_C = I_C \cdot \frac{1}{\omega C}$:
$10 = \frac{1}{4} \cdot \frac{1}{100 \cdot C}$
$10 = \frac{1}{400 C} \Rightarrow C = \frac{1}{4000} \, F = 0.25 \times 10^{-3} \, F = 250 \, \mu F$.
$2$. For the inductor branch:
The current through the $40 \, \Omega$ resistor is $I_{R'} = \frac{V_{R'}}{R'} = \frac{20 \, V}{40 \, \Omega} = 0.5 \, A = \frac{1}{2} \, A$.
Since the inductor is in parallel with this resistor branch,the total current in this parallel section is the sum of currents. However,looking at the circuit,the inductor is in parallel with the $R'$ branch. The voltage across the inductor is $20 \, V$.
$V_L = I_L X_L = I_L \cdot \omega L$
$20 = I_L \cdot 100 \cdot L$.
Assuming the current through the inductor is $I_L = 0.25 \, A$ (as derived from the node analysis where the total current splits):
$20 = 0.25 \cdot 100 \cdot L$
$20 = 25 \cdot L \Rightarrow L = \frac{20}{25} = 0.8 \, H$.

(D) Given: $R = 110 \, \Omega$,$V = 220 \, V$,$\omega = 300 \, rad/s$.
When capacitance is removed,the circuit is an $LR$ circuit. The phase angle is given by $\tan \phi = \frac{X_L}{R}$. Since current lags by $45^{\circ}$,$\tan 45^{\circ} = \frac{X_L}{R} = 1$,so $X_L = R$.
When the inductor is removed,the circuit is an $RC$ circuit. The phase angle is given by $\tan \phi = \frac{X_C}{R}$. Since current leads by $45^{\circ}$,$\tan 45^{\circ} = \frac{X_C}{R} = 1$,so $X_C = R$.
Since $X_L = R$ and $X_C = R$,it implies $X_L = X_C$.
In an $LCR$ circuit,when $X_L = X_C$,the circuit is in resonance.
At resonance,the impedance $Z = R$.
The rms current $I = \frac{V}{Z} = \frac{V}{R} = \frac{220}{110} = 2 \, A$.

(C) The impedance $Z$ of an $LCR$ series circuit is given by the formula: $Z = \sqrt{(X_{L} - X_{C})^{2} + R^{2}}$.
Given that $X_{L} = R$ and $X_{C} = R$,we have $X_{L} = X_{C}$.
Substituting these values into the impedance formula:
$Z = \sqrt{(R - R)^{2} + R^{2}}$
$Z = \sqrt{0^{2} + R^{2}}$
$Z = \sqrt{R^{2}}$
$Z = R$.
Therefore,the impedance of the circuit is $R$.

(D) Given:
Amplitude of current,$I_{0} = 10 \sqrt{2} \, A$.
Potential difference across inductor,$V_{L} = 40 \, V$.
Potential difference across capacitor,$V_{C} = 10 \, V$.
Potential difference across resistor,$V_{R} = 40 \, V$.
Step $1$: Calculate the $RMS$ current $(I_{RMS})$.
$I_{RMS} = \frac{I_{0}}{\sqrt{2}} = \frac{10 \sqrt{2}}{\sqrt{2}} = 10 \, A$.
Step $2$: Calculate the $RMS$ voltage of the source $(V_{RMS})$.
In an $LCR$ series circuit,the total voltage is given by:
$V_{RMS} = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$
$V_{RMS} = \sqrt{(40)^{2} + (40 - 10)^{2}}$
$V_{RMS} = \sqrt{40^{2} + 30^{2}} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, V$.
Step $3$: Calculate the impedance $(Z)$ of the circuit.
$Z = \frac{V_{RMS}}{I_{RMS}} = \frac{50 \, V}{10 \, A} = 5 \, \Omega$.
Therefore,the impedance of the circuit is $5 \, \Omega$.

(D) Given: $L = 100 \, mH = 0.1 \, H$,$C = 100 \, \mu F = 10^{-4} \, F$,$R = 10 \, \Omega$,$V = 220 \, V$,$f = 50 \, Hz$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \, rad/s$.
Inductive reactance $X_L = \omega L = 100 \pi \times 0.1 = 10 \pi \approx 31.4 \, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times 10^{-4}} = \frac{100}{\pi} \approx 31.8 \, \Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + (31.4 - 31.8)^2} = \sqrt{100 + (-0.4)^2} = \sqrt{100 + 0.16} \approx \sqrt{100} = 10 \, \Omega$.
Current $I = \frac{V}{Z} = \frac{220}{10} = 22 \, A$.

(A) The lamp is rated at $V_L = 25 \; V$ and $P_L = 5 \; W$.
The current $I$ flowing through the lamp at its rated power is given by $P_L = V_L \times I$.
$I = \frac{P_L}{V_L} = \frac{5 \; W}{25 \; V} = 0.2 \; A$.
Since the lamp and the resistance $R$ are in series,the same current $I = 0.2 \; A$ flows through the resistance $R$.
The voltage across the resistance $R$ is $V_R = V_{source} - V_L = 220 \; V - 25 \; V = 195 \; V$.
Using Ohm's law for the resistance $R$,we have $V_R = I \times R$.
$195 \; V = 0.2 \; A \times R$.
$R = \frac{195}{0.2} = 975 \; \Omega$.

(D) Given: $V_{L} = V_{C} = 2V_{R}$.
Since $V = IR$,we have $I X_{L} = I X_{C} = 2(IR)$.
Therefore,$X_{L} = X_{C} = 2R$.
Given $R = 5\,\Omega$,so $X_{L} = 2 \times 5 = 10\,\Omega$.
We know $X_{L} = 2\pi f L$.
Substituting the values: $10 = 2 \times \pi \times 50 \times L$.
$10 = 100 \pi L$.
$L = \frac{10}{100\pi} = \frac{1}{10\pi}\,H$.
To convert to $mH$,$L = \frac{1}{10\pi} \times 1000\,mH = \frac{100}{\pi}\,mH$.
Comparing this with $\frac{1}{K\pi}\,mH$,we have $\frac{1}{K\pi} = \frac{100}{\pi}$.
$K = \frac{1}{100} = 0.01$.

(B) The heat energy produced is given by $H = \frac{V^2}{R} t$. Since the heat produced $H$ and the voltage $V$ are the same for both cases,we have $H = \frac{V^2}{R_1} t_1 = \frac{V^2}{R_2} t_2$.
Given $t_1 = 20 \ min$ and $t_2 = 60 \ min$,we get $\frac{20}{R_1} = \frac{60}{R_2}$,which implies $R_2 = 3R_1$.
When connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_1} + \frac{1}{3R_1} = \frac{4}{3R_1}$.
Thus,$R_{eq} = \frac{3R_1}{4}$.
For the same heat $H$ to be produced by the parallel combination in time $t$,we have $H = \frac{V^2}{R_{eq}} t$.
Equating this to the heat produced by the first coil: $\frac{V^2}{R_1} \times 20 = \frac{V^2}{(3R_1/4)} \times t$.
$20 = \frac{4}{3} t \Rightarrow t = \frac{20 \times 3}{4} = 15 \ min$.

(A) Given: Peak voltage $V_0 = 220 \,V$,frequency $f = 50 \,Hz$,resistance $R = 400 \,\Omega$,capacitance $C = 200 \,\mu F = 200 \times 10^{-6} \,F$,and inductance $L = 6 \,H$.
Angular frequency $\omega = 2\pi f = 2 \times \pi \times 50 = 100\pi \,rad/s$.
Inductive reactance $X_L = \omega L = 100\pi \times 6 = 600\pi \,\Omega \approx 1884.96 \,\Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 200 \times 10^{-6}} = \frac{1}{0.02\pi} = \frac{50}{\pi} \,\Omega \approx 15.92 \,\Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (1884.96 - 15.92)^2} = \sqrt{160000 + (1869.04)^2} \approx \sqrt{160000 + 3493310} \approx \sqrt{3653310} \approx 1911.36 \,\Omega$.
Maximum current $I_0 = \frac{V_0}{Z} = \frac{220}{1911.36} \approx 0.115 \,A$.
Rounding to the nearest value,the maximum current is approximately $0.12 \,A$.

(C) Let the voltage of the $AC$ mains be $V$. The resistance of a bulb is given by $R_b = \frac{V_{rated}^2}{P_{rated}}$.
For a $50 \,W$ bulb,$R_1 = \frac{V_{rated}^2}{50}$.
For a $100 \,W$ bulb,$R_2 = \frac{V_{rated}^2}{100} = \frac{R_1}{2}$.
Let the resistance of the heater coil be $R_h$. The power output of the heater is $P = I^2 R_h = \left( \frac{V}{R_b + R_h} \right)^2 R_h$.
In the first case,$P_1 = \left( \frac{V}{R_1 + R_h} \right)^2 R_h$.
In the second case,$P_2 = \left( \frac{V}{R_2 + R_h} \right)^2 R_h = \left( \frac{V}{R_1/2 + R_h} \right)^2 R_h$.
Since $R_1/2 < R_1$,the denominator in the expression for $P_2$ is smaller than that for $P_1$,which implies $P_2 > P_1$.
Therefore,the heater output will increase.

(A) Given: Resistance $R = 10 \,\Omega$,Inductive reactance $X_L = 8 \,\Omega$,and Capacitive reactance $X_C = 6 \,\Omega$.
The formula for the total impedance $Z$ in an $LCR$ series circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Substituting the given values into the formula:
$Z = \sqrt{10^2 + (8 - 6)^2}$
$Z = \sqrt{100 + 2^2}$
$Z = \sqrt{100 + 4}$
$Z = \sqrt{104}$
$Z \approx 10.198 \,\Omega \approx 10.2 \,\Omega$.
Therefore,the total impedance is $10.2 \,\Omega$.

(D) Given the applied voltage equation: $\varepsilon = 500 \sqrt{2} \sin \omega t$.
Comparing this with $\varepsilon = \varepsilon_0 \sin \omega t$,the peak voltage $\varepsilon_0 = 500 \sqrt{2} \,V$.
The $r.m.s.$ voltage is $\varepsilon_{rms} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{500 \sqrt{2}}{\sqrt{2}} = 500 \,V$.
In a series $RLC$ circuit,the relationship between $r.m.s.$ voltages is given by $\varepsilon_{rms} = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Given $V_R = 400 \,V$ and $V_L = 700 \,V$,we substitute these values:
$(500)^2 = (400)^2 + (700 - V_C)^2$
$250000 = 160000 + (700 - V_C)^2$
$(700 - V_C)^2 = 90000$
$700 - V_C = \pm 300$.
This gives two possibilities: $V_C = 700 - 300 = 400 \,V$ or $V_C = 700 + 300 = 1000 \,V$.
Assuming the standard case where $V_C = 400 \,V$ is the $r.m.s.$ voltage,the peak voltage across the capacitor is $V_{C,peak} = V_C \sqrt{2} = 400 \sqrt{2} \,V$.

(D) Given: $E_0 = 200 \, V$,$R = 20 \, \Omega$,$L = 0.1 \, H$,$C = 10.6 \, F$.
$1$. At frequency $f = 0$ ($DC$ source):
The inductive reactance is $X_L = 2 \pi f L = 0 \, \Omega$.
The capacitive reactance is $X_C = \frac{1}{2 \pi f C} = \infty$.
Since the components are in series,the total impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \infty$.
Therefore,the current $I = \frac{E_0}{Z} = 0 \, A$.
$2$. At frequency $f = \infty$:
The inductive reactance is $X_L = 2 \pi f L = \infty$.
The capacitive reactance is $X_C = \frac{1}{2 \pi f C} = 0 \, \Omega$.
Since the components are in series,the total impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \infty$.
Therefore,the current $I = \frac{E_0}{Z} = 0 \, A$.
Thus,in both cases,the current is zero.

(D) The impedance $Z$ of an $LCR$ series circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Here,$X_L = 2 \pi f L$ is the inductive reactance and $X_C = \frac{1}{2 \pi f C}$ is the capacitive reactance.
As the frequency $f$ increases,$X_L$ increases linearly,while $X_C$ decreases.
At low frequencies,$X_C$ dominates,so $Z$ decreases as $f$ increases.
At the resonant frequency $f_0 = \frac{1}{2 \pi \sqrt{LC}}$,$X_L = X_C$,making the impedance $Z = R$,which is the minimum value.
As $f$ increases beyond $f_0$,$X_L$ dominates,causing $Z$ to increase again.
Therefore,the impedance decreases at first,reaches a minimum at resonance,and then increases.

(A) From the given circuit diagram, the capacitor $(X_C = 2 \, \Omega)$ and the inductor $(X_L = 2 \, \Omega)$ are connected in series with each other, and this combination is connected in parallel with the voltmeter. However, the ammeter is in series with the resistor $(R = 55 \, \Omega)$ and the $AC$ source $(110 \, V)$.
Since the capacitor and inductor are in series, their net reactance is $X = X_L - X_C = 2 \, \Omega - 2 \, \Omega = 0 \, \Omega$.
This means the branch containing the capacitor and inductor acts as a short circuit (zero impedance) for the $AC$ source.
Therefore, the entire voltage of the source $(110 \, V)$ appears across the resistor $(R = 55 \, \Omega)$.
The current $I$ measured by the ammeter is given by Ohm's law:
$I = \frac{V}{R} = \frac{110 \, V}{55 \, \Omega} = 2 \, A$.

(C) In the given circuit,the resistor $R$ is in series with a parallel combination of the inductor $L$ and the capacitor $C$.
The current flowing through the resistor $I_R$ is the total current supplied by the source.
In a parallel $LC$ circuit,the currents through the inductor $(i_L)$ and the capacitor $(i_C)$ are $180^{\circ}$ out of phase with each other.
Therefore,the net current through the parallel $LC$ combination is given by the difference between the magnitudes of the currents through the inductor and the capacitor:
$I_{LC} = |i_L - i_C|$
Given $i_L = 0.8 \,A$ and $i_C = 0.4 \,A$:
$I_{LC} = |0.8 \,A - 0.4 \,A| = 0.4 \,A$
Since the resistor is in series with this parallel combination,the current through the resistor $I_R$ must be equal to the net current flowing through the $LC$ branch.
Thus,$I_R = 0.4 \,A$.
Therefore,the correct option is $C$.

(C) The bulb $B_1$ is in series with a capacitor $C$,and the bulb $B_2$ is in series with an inductor $L$.
The capacitive reactance is given by $X_C = \frac{1}{2\pi f C}$,which decreases as frequency $f$ increases.
The inductive reactance is given by $X_L = 2\pi f L$,which increases as frequency $f$ increases.
As the frequency $f$ increases,$X_C$ decreases,allowing more current to flow through the branch containing $B_1$,thus increasing its brightness.
Conversely,as $f$ increases,$X_L$ increases,reducing the current through the branch containing $B_2$,thus decreasing its brightness.
Therefore,option $(c)$ is correct.

(B) At resonance,$\omega L = \frac{1}{\omega C}$,the impedance of the circuit is $Z = R$. Therefore,the current is $I_0 = \frac{\varepsilon_0}{R}$.
When the angular frequency is changed to $\omega^{\prime}$,the new current is $I^{\prime} = \frac{I_0}{2} = \frac{\varepsilon_0}{2R}$.
The impedance at frequency $\omega^{\prime}$ is given by $Z^{\prime} = \sqrt{R^2 + \left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2}$.
Since $I^{\prime} = \frac{\varepsilon_0}{Z^{\prime}}$,we have $\frac{\varepsilon_0}{2R} = \frac{\varepsilon_0}{\sqrt{R^2 + \left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2}}$.
Squaring both sides,we get $4R^2 = R^2 + \left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2$.
Rearranging the terms,$\left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2 = 3R^2$.
Taking the square root,we find $\left|\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right| = \sqrt{3} R$.

(A) In circuit $(a)$,the impedance is $Z_a = R = 40\,\Omega$. The rms current is $I_a = \frac{V}{Z_a} = \frac{220}{40} = 5.5\,A$.
In circuit $(b)$,the impedance is $Z_b = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $Z_b = \sqrt{R^2 + (X_L - X_C)^2} \geq R$,it follows that $Z_b \geq Z_a$.
Therefore,the current $I_b = \frac{V}{Z_b} \leq \frac{V}{Z_a} = I_a$.
This means the rms current in circuit $(b)$ can never be larger than the rms current in circuit $(a)$.

(D) Given: Inductance $L = \frac{50}{\pi} \text{ mH} = \frac{50}{\pi} \times 10^{-3} \text{ H}$,Capacitance $C = \frac{10^3}{\pi} \text{ }\mu\text{F} = \frac{10^3}{\pi} \times 10^{-6} \text{ F}$,Resistance $R = 10 \,\Omega$,Frequency $f = 50 \text{ Hz}$.
First,calculate the inductive reactance $X_L$:
$X_L = 2 \pi f L = 2 \pi \times 50 \times \left( \frac{50}{\pi} \times 10^{-3} \right) = 100 \times 50 \times 10^{-3} = 5 \,\Omega$.
Next,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 50 \times \left( \frac{10^3}{\pi} \times 10^{-6} \right)} = \frac{1}{100 \times 10^{-3}} = \frac{1}{0.1} = 10 \,\Omega$.
The net impedance $Z$ of the $LCR$ series circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
$Z = \sqrt{10^2 + (5 - 10)^2}$
$Z = \sqrt{100 + (-5)^2} = \sqrt{100 + 25} = \sqrt{125}$
$Z = 5 \sqrt{5} \,\Omega$.

(D) Given: $L = 1 \text{ H}$,$C = 20 \mu F = 20 \times 10^{-6} \text{ F}$,$R = 300 \Omega$,$V = 50 \sqrt{2} \sin 100 t$.
$1$. The angular frequency $\omega = 100 \text{ rad/s}$.
$2$. The $RMS$ voltage $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{50 \sqrt{2}}{\sqrt{2}} = 50 \text{ V}$.
$3$. Inductive reactance $X_L = \omega L = 100 \times 1 = 100 \Omega$.
$4$. Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 20 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega$.
$5$. Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{300^2 + (100 - 500)^2} = \sqrt{300^2 + (-400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \Omega$.
$6$. $RMS$ current $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{50}{500} = 0.1 \text{ A}$.
$7$. $RMS$ potential difference across the capacitor $V_C = I_{\text{rms}} \times X_C = 0.1 \times 500 = 50 \text{ V}$.

(A, C, D) Given: $V_{rms} = 20 \ V$,$\omega = 100 \ rad/s$,$C = 100 \ \mu F$,$L = 0.5 \ H$.
$1$. Impedance of the upper branch ($RC$ series):
$X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = 100 \ \Omega$.
$Z_1 = \sqrt{R_1^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \ \Omega$.
$I_{1,rms} = \frac{V_{rms}}{Z_1} = \frac{20}{100\sqrt{2}} = \frac{1}{5\sqrt{2}} \ A$.
Voltage across $100 \ \Omega$ resistor: $V_{R1} = I_{1,rms} \times R_1 = \frac{1}{5\sqrt{2}} \times 100 = \frac{20}{\sqrt{2}} = 10\sqrt{2} \ V$. (Option $C$ is correct).
$2$. Impedance of the lower branch ($RL$ series):
$X_L = \omega L = 100 \times 0.5 = 50 \ \Omega$.
$Z_2 = \sqrt{R_2^2 + X_L^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2} \ \Omega$.
$I_{2,rms} = \frac{V_{rms}}{Z_2} = \frac{20}{50\sqrt{2}} = \frac{2}{5\sqrt{2}} \ A$.
Voltage across $50 \ \Omega$ resistor: $V_{R2} = I_{2,rms} \times R_2 = \frac{2}{5\sqrt{2}} \times 50 = \frac{20}{\sqrt{2}} = 10\sqrt{2} \ V$. (Option $D$ is correct).
$3$. Total current $I_{rms}$:
The phase angle $\phi_1 = \tan^{-1}(\frac{-X_C}{R_1}) = -45^\circ$ and $\phi_2 = \tan^{-1}(\frac{X_L}{R_2}) = 45^\circ$.
The phase difference between $I_1$ and $I_2$ is $90^\circ$.
$I_{rms} = \sqrt{I_{1,rms}^2 + I_{2,rms}^2} = \sqrt{(\frac{1}{5\sqrt{2}})^2 + (\frac{2}{5\sqrt{2}})^2} = \sqrt{\frac{1}{50} + \frac{4}{50}} = \sqrt{\frac{5}{50}} = \frac{1}{\sqrt{10}} \approx 0.316 \ A \approx 0.3 \ A$. (Option $A$ is correct).

(B) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = 50 \ \Omega$, $X_L = 100 \ \Omega$, and $X_C = 50 \ \Omega$.
$Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \ \Omega$.
The average power dissipated in an $AC$ circuit is $P = V_{rms} I_{rms} \cos \phi$, where $\cos \phi = \frac{R}{Z}$ is the power factor.
$P = V_{rms} \times \left( \frac{V_{rms}}{Z} \right) \times \left( \frac{R}{Z} \right) = \frac{V_{rms}^2 R}{Z^2}$.
Given $V_{rms} = 10 \ V$.
$P = \frac{10^2 \times 50}{(50\sqrt{2})^2} = \frac{100 \times 50}{2500 \times 2} = \frac{5000}{5000} = 1 \ W$.

(B) Given: $V_{rms} = 220 \ V$,$R = 20 \ \Omega$,$X_C = 25 \ \Omega$,$X_L = 45 \ \Omega$.
The impedance $Z$ of the $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values: $Z = \sqrt{20^2 + (45 - 25)^2} = \sqrt{400 + 20^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \ \Omega$.
The $RMS$ current $I_{rms}$ is $I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.778 \ A \approx 7.8 \ A$.
The phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Substituting the values: $\tan \phi = \frac{45 - 25}{20} = \frac{20}{20} = 1$.
Therefore,$\phi = \tan^{-1}(1) = 45^{\circ}$.

(D) In a series $LCR$ circuit,the current $I$ is the same through all components.
$1$. The potential difference across the resistor $(V_R)$ is in phase with the current $I$.
$2$. The potential difference across the inductor $(V_L)$ leads the current $I$ by $\pi / 2$.
$3$. The potential difference across the capacitor $(V_C)$ lags behind the current $I$ by $\pi / 2$.
Since $V_L$ is at $+\pi / 2$ relative to $I$ and $V_C$ is at $-\pi / 2$ relative to $I$,the phase difference between $V_L$ and $V_C$ is $\pi / 2 - (-\pi / 2) = \pi$.
Option $A$ states that the applied $e.m.f.$ and potential difference across the resistance are in the same phase,which is incorrect as the applied $e.m.f.$ $V$ is generally at an angle $\theta$ with $I$ (and $V_R$).
Option $B$ is incorrect because the phase difference between applied $e.m.f.$ and $V_L$ depends on the circuit parameters.
Option $C$ is incorrect because the phase difference between $V_L$ and $V_C$ is $\pi$.
Option $D$ is correct because $V_R$ is in phase with $I$ and $V_C$ is at $\pi / 2$ phase difference with $I$,so the phase difference between $V_R$ and $V_C$ is $\pi / 2$.

(B) In an $LR$ circuit,the impedance is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L$ is the inductive reactance.
When a capacitor is connected in series,the circuit becomes an $LCR$ circuit.
The new impedance of the $LCR$ circuit is given by $Z' = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
If the circuit was initially inductive ($X_L > X_C$ or $X_L$ is present),adding a capacitor introduces $X_C$,which partially cancels the effect of $X_L$ if $X_L > X_C$,thereby reducing the total impedance $Z$.
Since the current $I = \frac{V}{Z}$,a decrease in impedance $Z$ leads to an increase in the current $I$ flowing through the circuit.

(D) In the given circuit,the two inductors $L$ and $2L$ are connected in series. Therefore,the equivalent inductance $L_{eq}$ is given by:
$L_{eq} = L + 2L = 3L$
The two capacitors $C$ and $2C$ are connected in parallel. Therefore,the equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = C + 2C = 3C$
The resonant frequency $f$ of an $L-C$ circuit is given by the formula:
$f = \frac{1}{2 \pi \sqrt{L_{eq} C_{eq}}}$
Substituting the values of $L_{eq}$ and $C_{eq}$ into the formula:
$f = \frac{1}{2 \pi \sqrt{(3L)(3C)}}$
$f = \frac{1}{2 \pi \sqrt{9LC}}$
$f = \frac{1}{2 \pi \cdot 3 \sqrt{LC}}$
$f = \frac{1}{6 \pi \sqrt{LC}}$

(C) In a series $LCR$ circuit,the applied alternating e.m.f. $(E)$ is given by the phasor sum of the individual potential drops across the inductor $(V_L)$,capacitor $(V_C)$,and resistor $(V_R)$.
The formula for the total e.m.f. is:
$E = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given values from the circuit diagram:
$V_L = 20 \ V$
$V_C = 50 \ V$
$V_R = 40 \ V$
Substituting these values into the formula:
$E = \sqrt{40^2 + (20 - 50)^2}$
$E = \sqrt{1600 + (-30)^2}$
$E = \sqrt{1600 + 900}$
$E = \sqrt{2500}$
$E = 50 \ V$
Therefore,the value of the alternating e.m.f. is $50 \ V$.

(A) In circuit $(A)$,the impedance is $Z_A = R = 40 \ \Omega$.
In circuit $(B)$,the impedance is $Z_B = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $Z_B = \sqrt{R^2 + (X_L - X_C)^2} \ge R$,it follows that $Z_B \ge Z_A$.
The r.m.s. current is given by $I = V/Z$. Since the voltage $V$ is the same $(220 \ V)$ for both circuits,$I_B = V/Z_B$ and $I_A = V/Z_A$.
Since $Z_B \ge Z_A$,it implies $I_B \le I_A$. Thus,the r.m.s. current in circuit $(B)$ can never be greater than that in circuit $(A)$.

(D) In a series $LCR$ circuit,the current $I$ is given by $I = \frac{E}{Z} = \frac{E}{\sqrt{R^2 + (X_L - X_C)^2}}$.
Here,$X_L = 2 \pi f L$ and $X_C = \frac{1}{2 \pi f C}$.
To keep the current $I$ constant for a given voltage $E$ and resistance $R$,the impedance $Z$ must remain constant.
This requires the reactances $X_L$ and $X_C$ to remain unchanged.
If the frequency $f$ is doubled $(f' = 2f)$,then for $X_L$ to remain constant: $2 \pi (2f) L' = 2 \pi f L \implies L' = \frac{L}{2}$.
Similarly,for $X_C$ to remain constant: $\frac{1}{2 \pi (2f) C'} = \frac{1}{2 \pi f C} \implies C' = \frac{C}{2}$.
Therefore,both $L$ and $C$ should be halved simultaneously.

(C) The phase angle $\phi$ in an $LCR$ series circuit is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Given: $L = \frac{300}{\pi} \text{ mH} = \frac{0.3}{\pi} \text{ H}$,$C = \frac{1}{\pi} \text{ mF} = \frac{1}{\pi} \times 10^{-3} \text{ F}$,$R = 20 \ \Omega$,$f = 50 \text{ Hz}$.
Angular frequency $\omega = 2 \pi f = 2 \times \pi \times 50 = 100 \pi \text{ rad/s}$.
Inductive reactance $X_L = \omega L = 100 \pi \times \frac{0.3}{\pi} = 30 \ \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times \frac{1}{\pi} \times 10^{-3}} = \frac{1}{0.1} = 10 \ \Omega$.
Substituting these values into the formula:
$\tan \phi = \frac{30 - 10}{20} = \frac{20}{20} = 1$.
Therefore,$\phi = \tan^{-1}(1)$.

(B) The impedance $Z$ of a series $LCR$ circuit is given by the formula: $Z = \sqrt{R^2 + (2 \pi v L - \frac{1}{2 \pi v C})^2}$.
At very low frequencies $(v \to 0)$,the capacitive reactance $X_C = \frac{1}{2 \pi v C}$ approaches infinity,so $Z \to \infty$.
At the resonant frequency $v_r = \frac{1}{2 \pi \sqrt{LC}}$,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,making the net reactance zero. Thus,$Z = R$,which is the minimum value of impedance.
For frequencies higher than the resonant frequency $(v > v_r)$,the inductive reactance $X_L = 2 \pi v L$ dominates,and $Z$ increases as $v$ increases.
Therefore,the graph of $Z$ versus $v$ starts from a high value,decreases to a minimum value $R$ at $v_r$,and then increases again. This behavior is correctly represented by graph $(B)$.

(D) In a parallel $LC$ circuit,the total impedance $Z$ is given by $\frac{1}{Z} = \sqrt{(\frac{1}{X_L} - \frac{1}{X_C})^2}$.
At resonance,$\omega^2 = \frac{1}{LC}$,which implies $X_L = X_C$.
At this frequency,the net current drawn from the source is minimum because the impedance $Z$ becomes infinite.
Since the circuit is connected in parallel,the voltage across the inductance $L$ and capacitance $C$ is the same as the source voltage.
However,in a parallel $LC$ circuit at resonance,the circulating current between $L$ and $C$ is maximum,leading to maximum voltage across the components.

(D) The impedance $Z$ of an $L-C-R$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Here,$X_L = L\omega = 2\pi fL$ and $X_C = \frac{1}{C\omega} = \frac{1}{2\pi fC}$.
As the frequency $f$ increases,the inductive reactance $X_L$ increases linearly,while the capacitive reactance $X_C$ decreases.
At low frequencies,$X_C$ dominates,so $Z$ decreases as $f$ increases.
At the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$,$X_L = X_C$,making the impedance $Z = R$,which is the minimum value.
As $f$ increases beyond $f_0$,$X_L$ dominates,causing $Z$ to increase.
Therefore,the impedance decreases at first,reaches a minimum value,and then increases.