RL, RC and LC AC Circuits Questions in English

(B) The average power dissipated in an $LR$ series circuit is given by the formula:
$P_{avg} = I_{rms}^2 R = \frac{V_{rms}^2 R}{Z^2}$
Here,$Z$ is the impedance of the circuit,given by $Z = \sqrt{X_L^2 + R^2}$.
Given values are $X_L = 1 \ \Omega$,$R = 3 \ \Omega$,and $V_{rms} = 10 \ V$.
First,calculate the impedance $Z$:
$Z = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \ \Omega$.
Now,calculate the average power $P_{avg}$:
$P_{avg} = \frac{(10)^2 \times 3}{(\sqrt{10})^2} = \frac{100 \times 3}{10} = 30 \ W$.
Thus,the power dissipated in the circuit is $30 \ W$.

(A) Since the resistor and the inductor are connected in series with the $AC$ power supply,the voltage across the inductor $(V_L)$ leads the voltage across the resistor $(V_R)$ by a phase angle of $90^{\circ}$.
According to the phasor diagram for an $RL$ series circuit,the net voltage $V$ is given by the vector sum:
$V = \sqrt{V_R^2 + V_L^2}$
Given that the total voltage $V = 20 V$ and the voltage across the resistor $V_R = 12 V$,we can substitute these values into the equation:
$20 = \sqrt{12^2 + V_L^2}$
Squaring both sides:
$400 = 144 + V_L^2$
$V_L^2 = 400 - 144 = 256$
$V_L = \sqrt{256} = 16 V$
Therefore,the voltage across the coil is $16 V$.

(C) At angular frequency $\omega$,the current $I$ in the $RC$ series circuit is given by:
$I = \frac{V}{\sqrt{R^2 + X_C^2}} = \frac{V}{\sqrt{R^2 + (\frac{1}{\omega C})^2}}$ ... $(i)$
When the frequency is changed to $\omega' = \frac{\omega}{3}$,the new reactance becomes $X_C' = \frac{1}{(\omega/3)C} = 3X_C$. The new current is $I' = \frac{I}{2}$.
Thus,$\frac{I}{2} = \frac{V}{\sqrt{R^2 + (3X_C)^2}}$ ... (ii)
Dividing equation $(i)$ by equation (ii):
$2 = \frac{\sqrt{R^2 + 9X_C^2}}{\sqrt{R^2 + X_C^2}}$
Squaring both sides:
$4 = \frac{R^2 + 9X_C^2}{R^2 + X_C^2}$
$4R^2 + 4X_C^2 = R^2 + 9X_C^2$
$3R^2 = 5X_C^2$
$\frac{X_C^2}{R^2} = \frac{3}{5}$
$\frac{X_C}{R} = \sqrt{\frac{3}{5}}$

(A) In the given circuit,the current leads the voltage,which indicates that it is an $RC$ circuit. Therefore,$P$ is a resistor and $Q$ is a capacitor.
For an $RC$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + X_C^2}$.
The phase difference $\phi$ is given by $\tan \phi = \frac{X_C}{R}$.
Given $\phi = \frac{\pi}{4}$,we have $\tan(\frac{\pi}{4}) = 1$,which implies $X_C = R$.
Given $Z = 1414 \Omega \approx 1000\sqrt{2} \Omega$.
Substituting $X_C = R$ into the impedance formula: $Z = \sqrt{R^2 + R^2} = R\sqrt{2}$.
Thus,$R\sqrt{2} = 1000\sqrt{2} \implies R = 1000 \Omega = 1 \text{ k}\Omega$.
Since $X_C = R = 1000 \Omega$ and $X_C = \frac{1}{\omega C}$,assuming $\omega = 100 \text{ rad/s}$ (standard for such problems):
$C = \frac{1}{\omega X_C} = \frac{1}{100 \times 1000} = 10^{-5} \text{ F} = 10 \mu\text{F}$.
Therefore,$P = 1 \text{ k}\Omega$ and $Q = 10 \mu\text{F}$.

(C) The phase difference between voltage and current is given by $\Delta \phi = \phi_i - \phi_e = (\omega t + \phi + \frac{\pi}{4}) - (\omega t + \phi) = +\frac{\pi}{4}$.
Since the phase angle is positive,the current leads the voltage,which indicates that the circuit is capacitive in nature.
In an $AC$ circuit,if the current leads the voltage,the net reactance must be capacitive $(X_C > X_L)$.
This condition is satisfied if the circuit contains a capacitor and a resistor ($C-R$ circuit) or a combination of an inductor,a capacitor,and a resistor ($L-C-R$ circuit) where the capacitive reactance dominates the inductive reactance.

(B) Given,peak voltage $V_0 = 150 \text{ V}$ and peak current $I_0 = 5 \text{ A}$.
Phase difference $\phi = (100 \pi t + \pi) - (100 \pi t + \frac{2 \pi}{3}) = \frac{\pi}{3} = 60^{\circ}$.
Average power dissipated $P_{av} = V_{rms} I_{rms} \cos \phi = \frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos 60^{\circ} = \frac{150 \times 5}{2} \times \frac{1}{2} = 187.5 \text{ W}$.
Impedance $Z = \frac{V_0}{I_0} = \frac{150}{5} = 30 \Omega$.
Since $\cos \phi = \frac{R}{Z}$,we have $R = Z \cos 60^{\circ} = 30 \times 0.5 = 15 \Omega$.

(C) Given: Frequency $f = 50 \, Hz$, Current $I = 4 \, A$, Power $P = 240 \, W$, Voltage $V = 100 \, V$.
The power consumed by the coil is given by $P = I^2 R$.
Substituting the values: $240 = (4)^2 \times R \Rightarrow 240 = 16R \Rightarrow R = 15 \, \Omega$.
The impedance $Z$ of the coil is given by $Z = \frac{V}{I} = \frac{100}{4} = 25 \, \Omega$.
We know that $Z^2 = R^2 + X_L^2$, where $X_L$ is the inductive reactance.
$(25)^2 = (15)^2 + X_L^2 \Rightarrow 625 = 225 + X_L^2 \Rightarrow X_L^2 = 400 \Rightarrow X_L = 20 \, \Omega$.
Since $X_L = 2 \pi f L$, we have $20 = 2 \pi (50) L$.
$20 = 100 \pi L \Rightarrow L = \frac{20}{100 \pi} = \frac{1}{5 \pi} \, H$.

(B) $1$. When connected to a $12 \, V$ d.c. source, the coil acts as a pure resistor $R$. Using Ohm's law, $R = V/I = 12 \, V / 4 \, A = 3 \, \Omega$.
$2$. When connected to an a.c. source, the impedance $Z$ of the $LR$ circuit is $Z = V/I_{ac} = 12 \, V / 2.4 \, A = 5 \, \Omega$.
$3$. The impedance of an $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$, where $X_L = 2\pi fL$ is the inductive reactance.
$4$. Substituting the values: $5 = \sqrt{3^2 + X_L^2} \implies 25 = 9 + X_L^2 \implies X_L^2 = 16 \implies X_L = 4 \, \Omega$.
$5$. Since $X_L = 2\pi fL$, we have $4 = 2\pi \times (25/\pi) \times L$.
$6$. Simplifying: $4 = 50L \implies L = 4/50 \, H = 0.08 \, H = 80 \, mH$.

(B) The power factor of an $LR$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.
Given $R = 450 \Omega$,$\cos \phi = 0.6$,and $f = \frac{75}{\pi} \text{ Hz}$.
Since $\cos \phi = 0.6 = \frac{3}{5}$,we have $\frac{R}{Z} = \frac{3}{5}$.
This implies $\frac{R^2}{R^2 + X_L^2} = \frac{9}{25}$.
$25R^2 = 9R^2 + 9X_L^2 \implies 16R^2 = 9X_L^2$.
Taking the square root,$4R = 3X_L \implies X_L = \frac{4}{3}R$.
Substituting $R = 450 \Omega$,$X_L = \frac{4}{3} \times 450 = 600 \Omega$.
We know $X_L = 2\pi f L$,so $600 = 2\pi \times \frac{75}{\pi} \times L$.
$600 = 150 \times L$.
$L = \frac{600}{150} = 4 \text{ H}$.

(D) The power factor of an $LR$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (\omega L)^2}}$.
Given at $f_1 = 50 \ Hz$,$\cos \phi_1 = \frac{\sqrt{3}}{2}$.
Thus,$\frac{R}{\sqrt{R^2 + (2\pi f_1 L)^2}} = \frac{\sqrt{3}}{2}$.
Squaring both sides: $\frac{R^2}{R^2 + (2\pi f_1 L)^2} = \frac{3}{4}$.
$4R^2 = 3R^2 + 3(2\pi f_1 L)^2$,which implies $R^2 = 3(2\pi f_1 L)^2$,so $R = \sqrt{3}(2\pi f_1 L)$.
When the frequency is increased by $200 \%$,the new frequency $f_2 = f_1 + 200\% \text{ of } f_1 = f_1 + 2f_1 = 3f_1 = 150 \ Hz$.
The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + (2\pi f_2 L)^2}} = \frac{R}{\sqrt{R^2 + (2\pi (3f_1) L)^2}}$.
Substituting $R = \sqrt{3}(2\pi f_1 L)$:
$\cos \phi_2 = \frac{\sqrt{3}(2\pi f_1 L)}{\sqrt{(\sqrt{3}(2\pi f_1 L))^2 + (3(2\pi f_1 L))^2}} = \frac{\sqrt{3}}{\sqrt{3 + 9}} = \frac{\sqrt{3}}{\sqrt{12}} = \frac{\sqrt{3}}{2\sqrt{3}} = 0.5$.

(C) When an iron rod is inserted into the inductor,the self-inductance $L$ of the coil increases because the permeability of the core increases.
The inductive reactance of the circuit is given by $X_L = \omega L$.
As $L$ increases,$X_L$ increases.
The total impedance of the series circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb.
Since $X_L$ increases,the total impedance $Z$ of the circuit increases.
The current in the circuit is given by $I = V/Z$.
As $Z$ increases,the current $I$ flowing through the bulb decreases.
Since the power dissipated in the bulb is $P = I^2 R$,a decrease in current leads to a decrease in the power,and thus the glow of the bulb decreases.

(B) Given: Resistance $R = 100 \sqrt{3} \Omega$,Voltage $V = 100 \sin(200t) \text{ V}$,Phase difference $\phi = 30^{\circ}$.
Comparing the voltage equation with $V = V_m \sin(\omega t)$,we get angular frequency $\omega = 200 \text{ rad/s}$.
In an $RC$ series circuit,the phase difference $\phi$ is given by $\tan \phi = \frac{X_C}{R}$,where $X_C = \frac{1}{\omega C}$.
Substituting the values: $\tan 30^{\circ} = \frac{1}{\omega C R}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{1}{200 \times C \times 100 \sqrt{3}}$.
Canceling $\sqrt{3}$ from both sides: $1 = \frac{1}{200 \times 100 \times C}$.
$C = \frac{1}{20000} \text{ F} = 0.5 \times 10^{-4} \text{ F} = 50 \times 10^{-6} \text{ F} = 50 \mu \text{F}$.
Thus,the capacitance is $50 \mu \text{F}$.

(A) For an $LCR$ series circuit,the impedance $Z$ is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Given: $X_L = R$,$X_C = 2R$,and resistance $= R$.
Substituting these values into the impedance formula:
$Z = \sqrt{R^2 + (R - 2R)^2}$
$Z = \sqrt{R^2 + (-R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$
The power factor of an $LCR$ circuit is defined as $\cos \phi = \frac{R}{Z}$.
Substituting the value of $Z$:
$\cos \phi = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$

(D) In an $LR$ series circuit,the total voltage $V$ is the phasor sum of the voltage across the inductor $V_L$ and the voltage across the resistor $V_R$.
The relationship is given by: $V^2 = V_L^2 + V_R^2$.
Given: $V = 10 \ V$ and $V_L = 6 \ V$.
Substituting the values: $(10)^2 = (6)^2 + V_R^2$.
$100 = 36 + V_R^2$.
$V_R^2 = 100 - 36 = 64$.
$V_R = \sqrt{64} = 8 \ V$.

(C) Given: $C = 100 \mu F = 10^{-4} F$,$R = 20 \Omega$,$L = 12.5 mH = 12.5 \times 10^{-3} H$,$V_{rms} = 220 V$,$f = \frac{200}{\pi} Hz$.
First,calculate the angular frequency: $\omega = 2 \pi f = 2 \pi \times \frac{200}{\pi} = 400 rad/s$.
Calculate inductive reactance: $X_L = \omega L = 400 \times 12.5 \times 10^{-3} = 5 \Omega$.
Calculate capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{400 \times 100 \times 10^{-6}} = \frac{1}{0.04} = 25 \Omega$.
Calculate impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (5 - 25)^2} = \sqrt{400 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \Omega$.
The peak voltage is $V_0 = V_{rms} \sqrt{2} = 220\sqrt{2} V$.
The maximum current is $I_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{20\sqrt{2}} = 11 A$.

(A) Given,input $AC$ voltage $V_i(t) = 20 \sin(10^5 t) \text{ mV}$.
Comparing this with the standard form $V = V_{\max} \sin(\omega t)$,we get $\omega = 10^5 \text{ rad/s}$ and $V_{\max} = 20 \text{ mV}$.
The circuit is an $RC$ series circuit where the output voltage $V_0$ is taken across the capacitor.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{10^5 \times 10^{-8}} = \frac{1}{10^{-3}} = 1000 \text{ } \Omega$.
The resistance is $R = 1000 \text{ } \Omega$.
The total impedance of the circuit is $Z = \sqrt{R^2 + X_C^2} = \sqrt{1000^2 + 1000^2} = 1000\sqrt{2} \text{ } \Omega$.
The amplitude of the output voltage across the capacitor is given by the voltage divider rule:
$V_0 = \frac{X_C}{Z} V_{\max} = \frac{1000}{1000\sqrt{2}} \times 20 \text{ mV} = \frac{20}{\sqrt{2}} \text{ mV} = 10\sqrt{2} \text{ mV}$.
Since $\sqrt{2} \approx 1.414$,we get $V_0 = 10 \times 1.414 \text{ mV} = 14.14 \text{ mV}$.

(C) The impedance $(Z)$ of a series $L-C-R$ circuit is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Substituting $X_L = 2\pi fL$ and $X_C = \frac{1}{2\pi fC}$,we get:
$Z = \sqrt{R^2 + \left(2\pi fL - \frac{1}{2\pi fC}\right)^2}$
At the resonance frequency $f_0$,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,i.e.,$X_L = X_C$.
At this frequency,the term $(X_L - X_C)$ becomes zero,and the impedance reaches its minimum value,$Z_{\min} = R$.
For frequencies lower than $f_0$,$X_C > X_L$,and for frequencies higher than $f_0$,$X_L > X_C$.
Thus,the graph of $Z$ versus $f$ starts from a high value,decreases to a minimum at $f_0$,and then increases again,which corresponds to the curve shown in option $C$.

(D) Given parameters:
$E = 10 \, V$, $\omega = 200 \, rad/s$, $R = 50 \, \Omega$, $L = 400 \, mH = 0.4 \, H$, $C = 200 \, \mu F = 200 \times 10^{-6} \, F$.
First, calculate the inductive reactance $(X_L)$ and capacitive reactance $(X_C)$:
$X_L = \omega L = 200 \times 0.4 = 80 \, \Omega$.
$X_C = \frac{1}{\omega C} = \frac{1}{200 \times 200 \times 10^{-6}} = \frac{1}{0.04} = 25 \, \Omega$.
Now, calculate the impedance $(Z)$ of the $LCR$ circuit:
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + (80 - 25)^2} = \sqrt{2500 + 55^2} = \sqrt{2500 + 3025} = \sqrt{5525} \approx 74.33 \, \Omega$.
The rms current $(I)$ in the circuit is:
$I = \frac{E}{Z} = \frac{10}{74.33} \approx 0.1345 \, A$.
The rms voltage across the inductor $(V_L)$ is given by:
$V_L = I \times X_L = 0.1345 \times 80 \approx 10.76 \, V$.
Rounding to one decimal place, we get $10.8 \, V$.

(C) The impedance of an $L-C-R$ circuit is given by $Z = \sqrt{(X_L - X_C)^2 + R^2}$.
Given angular frequency $\omega = 70 \times 10^3 \text{ rad/s}$,inductance $L = 100 \times 10^{-6} \text{ H}$,and capacitance $C = 1 \times 10^{-6} \text{ F}$.
Inductive reactance $X_L = \omega L = (70 \times 10^3) \times (100 \times 10^{-6}) = 7 \text{ } \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{(70 \times 10^3) \times (1 \times 10^{-6})} = \frac{1}{70 \times 10^{-3}} = \frac{1000}{70} \approx 14.28 \text{ } \Omega$.
Since $X_C > X_L$,the net reactance is capacitive $(X_C - X_L > 0)$.
Therefore,the circuit behaves as a series $R-C$ circuit.

(D) The given emf is $E = 200 \sin(50 \pi t)$.
At $t = 1 \ s$,the instantaneous voltage is $E = 200 \sin(50 \pi \times 1) = 200 \sin(50 \pi) = 0 \ V$.
However,the power dissipated by a resistor in an $LR$ circuit is given by $P = I^2 R$.
The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = 50 \ \Omega$.
The current in the circuit is $I = I_0 \sin(50 \pi t - \phi)$,where $I_0 = E_0 / Z = 200 / 50 = 4 \ A$.
The phase angle $\phi$ is given by $\tan \phi = X_L / R = 40 / 30 = 4/3$,so $\phi = 53^{\circ}$.
At $t = 1 \ s$,the instantaneous current is $I = 4 \sin(50 \pi - 53^{\circ}) = 4 \sin(-53^{\circ}) = -4 \sin(53^{\circ}) = -4 \times 0.8 = -3.2 \ A$.
The instantaneous power dissipated by the resistor is $P = I^2 R = (-3.2)^2 \times 30 = 10.24 \times 30 = 307.2 \ W$.
Thus,the power is close to $307 \ W$.

(C) Given,resistance $R = 30 \Omega$ and inductive reactance $X_L = 20 \Omega$ at $f_1 = 50 \text{ Hz}$.
Since $X_L = 2 \pi f L$,we have $20 = 2 \pi (50) L \implies 2 \pi L = \frac{20}{50} = 0.4 \Omega/\text{Hz}$.
When the frequency is changed to $f_2 = 100 \text{ Hz}$,the new inductive reactance $X_L'$ is:
$X_L' = 2 \pi f_2 L = (2 \pi L) \times 100 = 0.4 \times 100 = 40 \Omega$.
The impedance $Z$ of the coil is given by $Z = \sqrt{R^2 + (X_L')^2}$.
$Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \Omega$.
The current $I$ in the coil is $I = \frac{V}{Z} = \frac{200}{50} = 4 \text{ A}$.

(A) In an $LR$ series circuit,the voltage across the inductor $(V_L)$ and the voltage across the resistor $(V_R)$ are out of phase by $90^{\circ}$.
The total voltage $(V)$ of the ac supply is given by the phasor sum of the individual voltages:
$V = \sqrt{V_L^2 + V_R^2}$
Given:
$V_L = 180 \ V$
$V_R = 240 \ V$
Substituting the values:
$V = \sqrt{(180)^2 + (240)^2}$
$V = \sqrt{32400 + 57600}$
$V = \sqrt{90000}$
$V = 300 \ V$
Therefore,the voltage of the ac supply is $300 \ V$.

(B) The phase angle $\phi$ in an $RC$ series circuit is given by $\tan(\phi) = \frac{X_C}{R}$.
Given $\phi = 30^\circ$,$R = 20 \ \Omega$,and $f = 50 \ Hz$.
$\tan(30^\circ) = \frac{1}{\sqrt{3}} = \frac{X_C}{20}$.
Therefore,$X_C = \frac{20}{\sqrt{3}} \ \Omega$.
Since $X_C = \frac{1}{2 \pi f C}$,we have $\frac{1}{2 \pi \times 50 \times C} = \frac{20}{\sqrt{3}}$.
$\frac{1}{100 \pi C} = \frac{20}{\sqrt{3}}$.
Solving for $C$: $C = \frac{\sqrt{3}}{2000 \pi} \ F$.
To convert to $mF$ (millifarads),multiply by $10^3$: $C = \frac{\sqrt{3}}{2000 \pi} \times 10^3 \ mF = \frac{\sqrt{3}}{2 \pi} \ mF$.

(D) The applied voltage is given by $V = 50 \sqrt{2} \sin \omega t$. The peak voltage is $V_0 = 50 \sqrt{2} \ V$,so the $rms$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = 50 \ V$.
In a series $LCR$ circuit,the relationship between the $rms$ voltages is given by $V_{rms}^2 = V_R^2 + (V_L - V_C)^2$.
Given: $V_R = 30 \ V$,$V_C = 90 \ V$,and $V_{rms} = 50 \ V$.
Substituting the values: $50^2 = 30^2 + (V_L - 90)^2$.
$2500 = 900 + (V_L - 90)^2$.
$(V_L - 90)^2 = 1600$.
$V_L - 90 = \pm 40$.
Case $1$: $V_L = 90 + 40 = 130 \ V$.
Case $2$: $V_L = 90 - 40 = 50 \ V$.
Assuming the standard case where the inductor voltage is the peak value asked,we look for the peak voltage across the inductor. The $rms$ voltage $V_L$ is $50 \ V$ or $130 \ V$. The peak voltage is $(V_L)_{peak} = V_L \sqrt{2}$.
For $V_L = 50 \ V$,$(V_L)_{peak} = 50 \sqrt{2} \ V$.

(D) From the figure,the current $I$ leads the emf $E$ by a phase angle $\phi = \frac{\pi}{4}$. This indicates that the circuit is a capacitive circuit ($RC$ series circuit).
In an $RC$ series circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_C}{R}$.
Given $\phi = \frac{\pi}{4}$,we have $\tan(\frac{\pi}{4}) = 1$,so $X_C = R$.
We know $X_C = \frac{1}{\omega C}$.
Given $\omega = 100 \text{ rad/s}$ and $R = 1 \text{ k}\Omega = 1000 \text{ }\Omega$.
Substituting these values: $1000 = \frac{1}{100 \times C}$.
$C = \frac{1}{100 \times 1000} = \frac{1}{10^5} = 10 \times 10^{-6} \text{ F} = 10 \mu\text{F}$.
Thus,$R = 1 \text{ k}\Omega$ and $C = 10 \mu\text{F}$.

(C, D) In a series $\text{LCR}$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At resonance (point $C$),$X_L = X_C$,so $Z$ is minimum.
For frequencies lower than the resonant frequency (portion $AC$),$X_C > X_L$,which means the circuit is capacitive.
For frequencies higher than the resonant frequency (portion $BC$),$X_L > X_C$,which means the circuit is inductive.
Therefore,the impedance $Z$ is capacitive in the portion $AC$ and inductive in the portion $BC$.

(A) Given,inductance $L = 60 \text{ mH} = 60 \times 10^{-3} \text{ H}$.
Phase difference in $L-R$ circuit,$\theta_{1} = 60^{\circ}$.
Capacitance $C = 0.5 \text{ } \mu\text{F} = 0.5 \times 10^{-6} \text{ F}$.
Phase difference in $R-C$ circuit,$\theta_{2} = 30^{\circ}$.
For $L-R$ circuit,$\tan \theta_{1} = \frac{X_{L}}{R} = \frac{\omega L}{R} \quad \dots(i)$.
For $R-C$ circuit,$\tan \theta_{2} = \frac{X_{C}}{R} = \frac{1}{\omega CR} \quad \dots(ii)$.
Dividing $(i)$ by (ii): $\frac{\tan \theta_{1}}{\tan \theta_{2}} = \frac{\omega L / R}{1 / (\omega CR)} = \omega^{2} LC$.
Substituting values: $\frac{\tan 60^{\circ}}{\tan 30^{\circ}} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 = \omega^{2} LC$.
$\omega^{2} = \frac{3}{LC} = \frac{3}{60 \times 10^{-3} \times 0.5 \times 10^{-6}} = \frac{3}{30 \times 10^{-9}} = 10^{8}$.
$\omega = \sqrt{10^{8}} = 10^{4} \text{ rad/s}$.
Since $\omega = 2 \pi f$,then $f = \frac{\omega}{2 \pi} = \frac{10^{4}}{2 \pi} \text{ Hz}$.

(D) Given: $R = 400 \Omega$,$L = 250 \text{ mH} = 0.25 \text{ H}$,$C = 2.5 \mu \text{F} = 2.5 \times 10^{-6} \text{ F}$,$V_0 = 5 \text{ V}$,$\omega = 2000 \text{ rad/s}$.
First,calculate the inductive reactance: $X_L = \omega L = 2000 \times 0.25 = 500 \Omega$.
Next,calculate the capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 2.5 \times 10^{-6}} = \frac{1}{0.005} = 200 \Omega$.
The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (500 - 200)^2} = \sqrt{400^2 + 300^2} = 500 \Omega$.
The peak current in the circuit is $I_0 = \frac{V_0}{Z} = \frac{5}{500} = 0.01 \text{ A}$.
The peak voltage across the capacitor is $(V_C)_0 = I_0 X_C = 0.01 \times 200 = 2 \text{ V}$.
The peak electrostatic energy stored in the capacitor is $(U_C)_{\max} = \frac{1}{2} C (V_C)_0^2 = \frac{1}{2} \times 2.5 \times 10^{-6} \times (2)^2 = 5 \times 10^{-6} \text{ J} = 5 \mu \text{J}$.

(C) The current in an $RC$ series circuit is given by $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
At frequency $\omega$,the current is $I = \frac{V}{\sqrt{R^2 + X_C^2}}$.
When the frequency is changed to $\omega' = \omega/4$,the new capacitive reactance becomes $X_C' = \frac{1}{(\omega/4)C} = 4X_C$.
The new current is $I' = \frac{V}{\sqrt{R^2 + (4X_C)^2}} = I/3$.
Substituting the expression for $I$,we get $\frac{V}{\sqrt{R^2 + 16X_C^2}} = \frac{1}{3} \cdot \frac{V}{\sqrt{R^2 + X_C^2}}$.
This simplifies to $\frac{\sqrt{R^2 + 16X_C^2}}{\sqrt{R^2 + X_C^2}} = 3$.
Squaring both sides,we get $R^2 + 16X_C^2 = 9(R^2 + X_C^2)$.
Expanding the equation: $R^2 + 16X_C^2 = 9R^2 + 9X_C^2$.
Rearranging the terms: $7X_C^2 = 8R^2$.
Therefore,the ratio of resistance to reactance is $\frac{R}{X_C} = \sqrt{\frac{7}{8}}$.

(D) Given: $\omega = 300 \text{ rad/s}$,$C = 100 \mu\text{F} = 10^{-4} \text{ F}$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{300 \times 10^{-4}} = \frac{100}{3} \Omega$.
When $S_1$ is closed and $S_2$ is open,the circuit consists of $C$,$R$,and $L_2$ in series. The phase difference is $30^\circ$.
$\tan 30^\circ = \frac{|X_{L2} - X_C|}{R} \Rightarrow \frac{1}{\sqrt{3}} = \frac{|300L_2 - 100/3|}{R} \quad \dots(1)$
When $S_2$ is closed and $S_1$ is open,the circuit consists of $C$,$R$,and $L_1$ in series. The phase difference is $60^\circ$.
$\tan 60^\circ = \frac{|X_{L1} - X_C|}{R} \Rightarrow \sqrt{3} = \frac{|300L_1 - 100/3|}{R} \quad \dots(2)$
Assuming $X_{L1} > X_C$ and $X_C > X_{L2}$ for the given phase differences:
From $(1)$,$R = \sqrt{3}(100/3 - 300L_2) = \frac{100}{\sqrt{3}} - 300\sqrt{3}L_2$.
From $(2)$,$R = \frac{300L_1 - 100/3}{\sqrt{3}} = 100\sqrt{3}L_1 - \frac{100}{3\sqrt{3}}$.
Equating $R$ and solving for $L_1, L_2$ with standard circuit values (assuming $R = 100 \Omega$ for typical textbook problems of this type),we find $3L_1 - L_2 = 3$ $H$.