RL, RC and LC AC Circuits Questions in English

(C) The time constant of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
This implies that the dimensions of $\frac{L}{R}$ are the same as the dimensions of time,i.e.,$[T]$.
Therefore,the dimensions of $\frac{R}{L}$ are the reciprocal of the dimensions of time.
Dimensions of $\frac{R}{L} = \frac{1}{[T]} = [T^{-1}]$.

(B) In an $LR$ series circuit,the impedance $Z$ is given by the formula $Z = \sqrt{R^2 + X_L^2}$.
Here,$X_L$ is the inductive reactance,which is defined as $X_L = \omega L$.
Since the angular frequency $\omega = 2\pi f$,we can substitute this into the expression for $X_L$ to get $X_L = 2\pi fL$.
Substituting $X_L$ back into the impedance formula,we get $Z = \sqrt{R^2 + (2\pi fL)^2}$.
Therefore,$Z = \sqrt{R^2 + 4\pi^2 f^2 L^2}$.

(B) In an $RL$ series circuit,the voltage across the resistor $(V_R)$ and the voltage across the inductor $(V_L)$ are out of phase by $90^{\circ}$.
Therefore,the total applied voltage $V$ is the phasor sum of $V_R$ and $V_L$,given by the formula:
$V = \sqrt{V_R^2 + V_L^2}$
Given $V_R = 200 \, V$ and $V_L = 150 \, V$.
Substituting the values:
$V = \sqrt{(200)^2 + (150)^2}$
$V = \sqrt{40000 + 22500}$
$V = \sqrt{62500}$
$V = 250 \, V$.

(B) The impedance $Z$ of an $RL$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$, where $X_L = 2\pi f L$.
Given: $R = 10 \, \Omega$, $L = 20 \, H$, $V = 120 \, V$, and $f = 60 \, Hz$.
First, calculate the inductive reactance $X_L = 2 \times \pi \times 60 \times 20 = 2400\pi \, \Omega$.
$X_L \approx 2400 \times 3.1416 \approx 7540 \, \Omega$.
Now, calculate the impedance $Z = \sqrt{10^2 + (7540)^2} \approx 7540 \, \Omega$.
The current $I = \frac{V}{Z} = \frac{120}{7540} \approx 0.0159 \, A$.
Rounding to the nearest value, $I \approx 0.016 \, A$.

(A) In an $L-R$ series circuit,the total applied voltage $V$ is the phasor sum of the voltage across the resistor $V_R$ and the voltage across the inductor $V_L$.
The relationship is given by $V^2 = V_R^2 + V_L^2$.
Given: $V = 20\,V$ and $V_R = 12\,V$.
Substituting the values: $20^2 = 12^2 + V_L^2$.
$400 = 144 + V_L^2$.
$V_L^2 = 400 - 144 = 256$.
$V_L = \sqrt{256} = 16\,V$.
Therefore,the voltage across the coil is $16\,V$.

(A) In an $LR$ series circuit,the phase angle $\phi$ between the voltage and current is given by the formula: $\tan \phi = \frac{X_L}{R}$.
Here,$R = 300 \ \Omega$,$L = \frac{1}{\pi} \ H$,and frequency $f = 200 \ Hz$.
The inductive reactance $X_L$ is calculated as: $X_L = 2\pi f L = 2\pi \times 200 \times \frac{1}{\pi} = 400 \ \Omega$.
Substituting the values into the phase angle formula:
$\tan \phi = \frac{400}{300} = \frac{4}{3}$.
Therefore,the phase angle is $\phi = \tan^{-1} \left( \frac{4}{3} \right)$.

(C) In an $LCR$ series circuit,the potential difference across the inductor $(V_L)$,resistor $(V_R)$,and capacitor $(V_C)$ are given as $V_L = 46\, V$,$V_R = 8\, V$,and $V_C = 40\, V$.
The total $EMF$ $(V)$ of the source is given by the formula: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Substituting the given values: $V = \sqrt{8^2 + (46 - 40)^2}$.
$V = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\, V$.
Thus,the $EMF$ of the source is $10\, V$.

(D) The impedance $Z$ of an $L-R$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi fL$.
Given $R = 30 \, \Omega$,$f = 50 \, Hz$,and $L = \frac{0.4}{\pi} \, H$.
First,calculate the inductive reactance $X_L = 2\pi \times 50 \times \frac{0.4}{\pi} = 100 \times 0.4 = 40 \, \Omega$.
Now,calculate the impedance $Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega$.
The current $I$ in the circuit is given by $I = \frac{V}{Z} = \frac{200}{50} = 4 \, A$.
Thus,the impedance is $50 \, \Omega$ and the current is $4 \, A$.

(B) In a series $LCR$ circuit,the impedance $Z$ is the vector sum of resistance $R$ and the net reactance $X = X_L - X_C$.
The inductive reactance is given by $X_L = L\omega$.
The capacitive reactance is given by $X_C = \frac{1}{C\omega}$.
The total impedance $Z$ is calculated using the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Substituting the values of $X_L$ and $X_C$ into the formula:
$Z = \sqrt{R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2}$
This can be written as:
$Z = \left[R^2 + \left(L\omega - \frac{1}{C\omega}\right)^2\right]^{1/2}$

(D) The given e.m.f. is $E = E_0 \cos(\omega t)$,where $E_0 = 4 \, V$ and $\omega = 1000 \, rad/s$.
Given inductance $L = 3 \, mH = 3 \times 10^{-3} \, H$ and resistance $R = 4 \, \Omega$.
The inductive reactance is $X_L = \omega L = 1000 \times 3 \times 10^{-3} = 3 \, \Omega$.
The impedance of the $LR$-circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \Omega$.
The amplitude of the current $I_0$ is given by $I_0 = E_0 / Z = 4 / 5 = 0.8 \, A$.

(C) In an $RL$ series circuit,the phase angle $\phi$ between voltage and current is given by the formula: $\tan \phi = \frac{X_L}{R}$.
Given that the phase angle $\phi = 45^o$,we substitute this into the equation:
$\tan 45^o = \frac{X_L}{R}$.
Since $\tan 45^o = 1$,we have $1 = \frac{X_L}{R}$.
Therefore,$X_L = R$.

(B) When $DC$ is supplied,the coil acts as a pure resistor because the frequency is zero. Therefore,the resistance $R$ is given by $R = \frac{V}{I_{DC}} = \frac{100 \, V}{1 \, A} = 100 \, \Omega$.
When $AC$ is supplied,the coil offers both resistance and inductive reactance. The total opposition to the current is called impedance $(Z)$.
Using Ohm's law for $AC$ circuits,$Z = \frac{V}{I_{AC}} = \frac{100 \, V}{0.5 \, A} = 200 \, \Omega$.
Thus,the impedance of the coil is $200 \, \Omega$.

(C) Given: Resistance $R = 12\,\Omega$,Inductance $L = 0.21\,H$,Frequency $f = 50\,Hz$.
The angular frequency is given by $\omega = 2\pi f = 2 \times \pi \times 50 = 100\pi\,rad/s$.
The inductive reactance is $X_L = \omega L = 100\pi \times 0.21 = 21\pi\,\Omega$.
Using $\pi \approx 3.14$,we get $X_L \approx 100 \times 3.14 \times 0.21 = 65.94\,\Omega$.
However,using the standard approximation $\pi \approx 22/7$,$X_L = 100 \times (22/7) \times 0.21 = 100 \times 22 \times 0.03 = 66\,\Omega$.
The phase angle $\phi$ in an $RL$ series circuit is given by $\tan \phi = \frac{X_L}{R}$.
$\tan \phi = \frac{66}{12} = 5.5$.
$\phi = \tan^{-1}(5.5) \approx 79.7^o \approx 80^o$.

(B) In a series $LR$ circuit,the potential difference across the resistance $(V_R)$ and the inductance $(V_L)$ are in quadrature,meaning they differ in phase by $90^{\circ}$.
Given: $V_L = 16\, V$ and $V_R = 20\, V$.
The total potential difference $(V)$ across the circuit is given by the phasor sum:
$V = \sqrt{V_R^2 + V_L^2}$
Substituting the given values:
$V = \sqrt{(20)^2 + (16)^2}$
$V = \sqrt{400 + 256}$
$V = \sqrt{656}$
$V \approx 25.6\, V$.

(D) Given: Voltage $V = 220\, V$,Frequency $f = 50\, Hz$,Inductance $L = 0.2\, H$,Resistance $R = 20\, \Omega$.
First,calculate the inductive reactance $X_L = 2 \pi f L = 2 \times 3.14 \times 50 \times 0.2 = 62.8\, \Omega$.
The impedance of the $LR$ series circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 62.8^2} = \sqrt{400 + 3943.84} = \sqrt{4343.84} \approx 65.9\, \Omega$.
The current in the circuit is $I = \frac{V}{Z} = \frac{220}{65.9} \approx 3.33\, A$.

(A) Given: $V = 20\cos(2000t) \,V$,so $V_0 = 20 \,V$ and $\omega = 2000 \,rad/s$.
The circuit consists of a resistor $R_1 = 6 \,\Omega$ in series with an inductor of $L = 5 \,mH$ having internal resistance $R_2 = 4 \,\Omega$,and a capacitor $C = 50 \,\mu F$.
Total resistance $R = R_1 + R_2 = 6 + 4 = 10 \,\Omega$.
Inductive reactance $X_L = \omega L = 2000 \times 5 \times 10^{-3} = 10 \,\Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 50 \times 10^{-6}} = \frac{1}{0.1} = 10 \,\Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + (10 - 10)^2} = 10 \,\Omega$.
Amplitude of current $i_0 = \frac{V_0}{Z} = \frac{20}{10} = 2 \,A$.

(A) In an $ac$ circuit containing a coil (inductor with resistance),the phase difference $\phi$ between the voltage and the current is given by the formula $\tan \phi = \frac{X_L}{R}$.
Given that the reactance $X_L = \sqrt{3} R$.
Substituting this into the formula,we get $\tan \phi = \frac{\sqrt{3} R}{R} = \sqrt{3}$.
Since $\tan \phi = \sqrt{3}$,the phase angle $\phi = 60^{\circ}$.
Converting degrees to radians,$\phi = 60^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{3}$.

(A) In an $ac$ circuit,if the current lags behind the voltage by a phase angle $\phi$,the circuit must be an inductive circuit containing a resistor $(R)$ and an inductor $(L)$.
In an $RL$ series circuit,the voltage across the resistor $(V_R)$ is in phase with the current $(I)$,while the voltage across the inductor $(V_L)$ leads the current by $90^\circ$.
The resultant voltage $V$ leads the current by a phase angle $\phi$,which is given by:
$\tan \phi = \frac{V_L}{V_R} = \frac{I_0 X_L}{I_0 R} = \frac{X_L}{R}$
Since the current lags behind the voltage,the circuit must contain $R$ and $L$.

(C) The impedance $Z$ of an $RL$ series circuit is given by the formula: $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi \nu L$.
Given: $R = 40\, \Omega$,$L = 95.5\, \text{mH} = 95.5 \times 10^{-3}\, \text{H}$,and $\nu = 50\, \text{Hz}$.
First,calculate the inductive reactance $X_L$:
$X_L = 2 \times 3.1416 \times 50 \times 95.5 \times 10^{-3} \approx 30\, \Omega$.
Now,calculate the impedance $Z$:
$Z = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\, \Omega$.

(C) Given: Resistance $R = 5\,\Omega$,Inductance $L = 0.1\,H$,and angular frequency $\omega = 50\,rad/s$.
In an $LR$ series circuit,the phase difference $\phi$ between the current and the $EMF$ is given by $\tan \phi = \frac{X_L}{R}$.
First,calculate the inductive reactance $X_L = \omega L = 50 \times 0.1 = 5\,\Omega$.
Now,substitute the values into the formula: $\tan \phi = \frac{5}{5} = 1$.
Therefore,$\phi = \tan^{-1}(1) = \frac{\pi}{4}$.

(B) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (\omega L)^2} = \sqrt{R^2 + \omega^2 L^2}$.
Power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
Since $I_{rms} = \frac{V_{rms}}{Z}$ and the power factor $\cos \phi = \frac{R}{Z}$, we substitute these into the power formula:
$P = V_{rms} \left( \frac{V_{rms}}{Z} \right) \left( \frac{R}{Z} \right) = \frac{V_{rms}^2 R}{Z^2}$.
Substituting $Z^2 = R^2 + \omega^2 L^2$, we get:
$P = \frac{V^2 R}{R^2 + \omega^2 L^2}$.

(C) The phase angle $\phi$ in an $RL$ circuit is given by the formula $\tan \phi = \frac{X_L}{R}$.
Here, the inductive reactance $X_L = 2\pi \nu L$.
Given: Resistance $R = 200 \, \Omega$, Inductance $L = 1.0 \, H$, and Frequency $\nu = \frac{200}{2\pi} \, Hz$.
Substituting the values: $X_L = 2\pi \times \left( \frac{200}{2\pi} \right) \times 1 = 200 \, \Omega$.
Now, $\tan \phi = \frac{200}{200} = 1$.
Therefore, $\phi = \tan^{-1}(1) = 45^o$.

(D) The impedance $Z$ of an $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi \nu L$.
Given $R = 12 \, \Omega$,$L = 0.04 \, H$,$\nu = 50 \, Hz$,and $V = 220 \, V$.
First,calculate the inductive reactance: $X_L = 2 \times 3.14 \times 50 \times 0.04 = 12.56 \, \Omega$.
Now,calculate the impedance: $Z = \sqrt{12^2 + 12.56^2} = \sqrt{144 + 157.75} = \sqrt{301.75} \approx 17.37 \, \Omega$.
The current $I$ is given by $I = \frac{V}{Z} = \frac{220}{17.37} \approx 12.66 \, A \approx 12.7 \, A$.

(C) In the given circuit,the inductor $L$ and capacitor $C$ are connected in parallel with an $AC$ power supply.
At resonance,the inductive reactance $X_L = \omega L$ and capacitive reactance $X_C = \frac{1}{\omega C}$ are equal,i.e.,$X_L = X_C$.
In a parallel $LC$ circuit at resonance,the current flowing through the inductor branch $(I_L)$ and the current flowing through the capacitor branch $(I_C)$ are equal in magnitude but opposite in phase (phase difference of $180^{\circ}$).
Therefore,the total current $I$ supplied by the source is $I = I_L + I_C = 0$.
Ammeter $A_3$ is placed in the main line to measure the total current supplied by the source.
Since the total current is zero at resonance,ammeter $A_3$ will read zero ampere.

(A) The phase angle $\phi$ in an $LCR$ series 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 $\phi = -45^o$ (or we use the magnitude of the lead as $\tan(45^o) = \frac{X_C - X_L}{R}$).
Substituting the values: $\tan 45^o = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.
Since $\tan 45^o = 1$,we have $1 = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.
$R = \frac{1}{2\pi fC} - 2\pi fL$.
$\frac{1}{2\pi fC} = 2\pi fL + R$.
Therefore,$C = \frac{1}{2\pi f(2\pi fL + R)}$.

(A) For $DC$,the solenoid acts as a pure resistor $R$. Given $V = 100 \, V$ and $I = 1.0 \, A$,we have $R = \frac{V}{I} = \frac{100}{1} = 100 \, \Omega$.
For $AC$,the solenoid acts as an $LR$ circuit with impedance $Z$. Given $V = 100 \, V$ and $I = 0.5 \, A$,we have $Z = \frac{V}{I} = \frac{100}{0.5} = 200 \, \Omega$.
The impedance is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi f L$.
Substituting the values: $200 = \sqrt{100^2 + (2 \cdot \pi \cdot 50 \cdot L)^2}$.
Squaring both sides: $40000 = 10000 + (100\pi L)^2$.
$30000 = (100\pi L)^2$.
Taking the square root: $100\sqrt{3} = 100\pi L$.
$L = \frac{\sqrt{3}}{\pi} \approx \frac{1.732}{3.1416} \approx 0.55 \, H$.

(C) The average power consumed in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.
We know that $E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{E_0}{Z\sqrt{2}}$.
The power factor is $\cos \phi = \frac{R}{Z}$.
Substituting these values,we get $P = \left(\frac{E_0}{\sqrt{2}}\right) \left(\frac{E_0}{Z\sqrt{2}}\right) \left(\frac{R}{Z}\right) = \frac{E_0^2 R}{2Z^2}$.
Given that the inductive reactance $X_L = R$,the impedance $Z$ is calculated as $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2}R$.
Substituting $Z^2 = 2R^2$ into the power formula: $P = \frac{E_0^2 R}{2(2R^2)} = \frac{E_0^2 R}{4R^2} = \frac{E_0^2}{4R}$.

(A) The rated current of the bulb is $i = \frac{P}{V} = \frac{60}{10} = 6\, A$.
In an $RL$ series circuit,the supply voltage $V$ is related to the voltage across the bulb $V_R$ and the voltage across the inductor $V_L$ by $V = \sqrt{V_R^2 + V_L^2}$.
Given $V = 100\, V$ and $V_R = 10\, V$,we have $100^2 = 10^2 + V_L^2$.
$V_L^2 = 10000 - 100 = 9900$.
$V_L = \sqrt{9900} \approx 99.5\, V$.
The inductive reactance is $X_L = \frac{V_L}{i} = \frac{99.5}{6} \approx 16.58\, \Omega$.
Since $X_L = 2\pi f L$,we have $L = \frac{X_L}{2\pi f} = \frac{16.58}{2 \times 3.14 \times 50} = \frac{16.58}{314} \approx 0.0528\, H$.
Thus,the required self-inductance is approximately $0.052\, H$.

(C) In an $RC$ series circuit,the current $I$ flows through both the resistor $R$ and the capacitor $C$.
$1$. The voltage across the resistor ($V_R$,measured by $V_2$) is always in phase with the current ($I$,measured by $A$). Thus,statement $I$ is correct.
$2$. The voltage across the capacitor ($V_C$,measured by $V_1$) lags behind the current by $\frac{\pi}{2}$.
$3$. Since $V_R$ is in phase with $I$,and $V_C$ lags behind $I$ by $\frac{\pi}{2}$,it follows that $V_C$ lags behind $V_R$ by $\frac{\pi}{2}$. This means $V_R$ is ahead of $V_C$ by $\frac{\pi}{2}$. Therefore,statement $II$ is correct.
$4$. Statement $III$ is incorrect because $V_1$ (voltage across capacitor) lags behind the current $A$ by $\frac{\pi}{2}$.
Thus,both statements $I$ and $II$ are correct.

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

(B) Given: $I_{rms} = 4 \, A$,$f = 50 \, Hz$,$P = 240 \, W$,$V_{rms} = 100 \, V$.
Power consumed in the coil is given by $P = I_{rms}^2 R$.
$R = \frac{P}{I_{rms}^2} = \frac{240}{4^2} = \frac{240}{16} = 15 \, \Omega$.
Impedance of the coil is $Z = \frac{V_{rms}}{I_{rms}} = \frac{100}{4} = 25 \, \Omega$.
We know that $Z^2 = R^2 + X_L^2$,where $X_L = 2\pi f L$.
$X_L = \sqrt{Z^2 - R^2} = \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20 \, \Omega$.
Since $X_L = 2\pi f L$,we have $20 = 2 \times \pi \times 50 \times L$.
$L = \frac{20}{100\pi} = \frac{1}{5\pi} \, H$.

(B) Given: $R = X_L$ and $X_L = 2X_C$,which implies $X_C = \frac{R}{2}$.
The impedance $Z$ of a series $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values: $Z = \sqrt{R^2 + (R - \frac{R}{2})^2} = \sqrt{R^2 + (\frac{R}{2})^2} = \sqrt{R^2 + \frac{R^2}{4}} = \sqrt{\frac{5R^2}{4}} = \frac{\sqrt{5}R}{2}$.
The phase difference $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Substituting the values: $\tan \phi = \frac{R - R/2}{R} = \frac{R/2}{R} = \frac{1}{2}$.
Therefore,$\phi = \tan^{-1}\left(\frac{1}{2}\right)$.
Thus,the impedance is $\frac{\sqrt{5}R}{2}$ and the phase difference is $\tan^{-1}\left(\frac{1}{2}\right)$.

(C) In a series $LCR$ circuit,the circuit is inductive when the inductive reactance is greater than the capacitive reactance,i.e.,$X_L > X_C$.
At point $A$: $X_C > X_L$,so the circuit is capacitive.
At point $B$: $X_C = X_L$,which is the resonance condition,so the circuit is purely resistive.
At point $C$: $X_L > X_C$,so the circuit is inductive.
Therefore,the correct option is $C$.

(C) In a parallel $RLC$ circuit,the same alternating emf $E$ is applied across each component.
$1$. The current $I_R$ through the resistor is in phase with the applied emf $E$.
$2$. The current $I_L$ through the inductor lags behind the applied emf $E$ by a phase angle of $\frac{\pi}{2}$.
$3$. The current $I_C$ through the capacitor leads the applied emf $E$ by a phase angle of $\frac{\pi}{2}$.
Comparing these relationships with the given diagrams,in option $C$,$I_R$ is in phase with $E$,$I_C$ is directed upwards (leading by $\frac{\pi}{2}$),and $I_L$ is directed downwards (lagging by $\frac{\pi}{2}$). Thus,the diagram in option $C$ correctly represents the phase relationships.

(A) From the given diagram,the current $i$ leads the voltage $e$ by a phase angle $\phi = \pi/4$. This indicates that the circuit is an $RC$ series circuit.
For an $RC$ circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_C}{R} = \frac{1}{\omega CR}$.
Given $\phi = \pi/4$,we have $\tan(\pi/4) = 1$,so $1 = \frac{1}{\omega CR}$,which implies $\omega CR = 1$.
Given $\omega = 100 \text{ rad/s}$,we have $100 \times C \times R = 1$,or $CR = 1/100 = 0.01 \text{ s}$.
Checking option $(A)$: $R = 1000 \, \Omega$ and $C = 10 \times 10^{-6} \text{ F} = 10^{-5} \text{ F}$.
Then $CR = 1000 \times 10^{-5} = 10^{-2} = 0.01 \text{ s}$.
This matches the condition,so option $(A)$ is correct.

(A) The net reactance $X$ of a series $LC$ circuit is given by the difference between the inductive reactance $X_L$ and the capacitive reactance $X_C$.
$X = X_L - X_C = 2\pi fL - \frac{1}{2\pi fC}$.
As the frequency $f$ increases,$X_L = 2\pi fL$ increases linearly,while $X_C = \frac{1}{2\pi fC}$ decreases hyperbolically.
At low frequencies,$X_C$ dominates,making the net reactance negative.
At high frequencies,$X_L$ dominates,making the net reactance positive.
At the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$,the net reactance is zero.
The curve that starts from negative values,passes through zero,and increases towards positive values as frequency increases represents this behavior. Looking at the provided graph,none of the curves $a, b, c, d$ perfectly represent the standard $X$ vs $f$ plot for a series $LC$ circuit,but in many textbook contexts for this specific problem,the curve that shows a transition from negative to positive values is sought. However,based on the provided image,curve $a$ represents $X_L$,curve $d$ represents $X_C$,and curve $b$ is a linear decrease. None of the curves correctly represent the $LC$ series reactance curve,which should be a hyperbola-like curve crossing the frequency axis. Given the options provided in standard sources for this specific question,the intended answer is often associated with the curve that crosses the axis.

(C) Given: $C = \frac{2.5}{\pi} \times 10^{-6} \, F$,$R = 3000 \, \Omega$,$V_{rms} = 200 \, V$,$\nu = 50 \, Hz$.
First,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{2\pi \nu C} = \frac{1}{2\pi \times 50 \times (\frac{2.5}{\pi} \times 10^{-6})} = \frac{1}{250 \times 10^{-6}} = 4000 \, \Omega$.
Calculate the impedance $Z$:
$Z = \sqrt{R^2 + X_C^2} = \sqrt{3000^2 + 4000^2} = \sqrt{9 \times 10^6 + 16 \times 10^6} = \sqrt{25 \times 10^6} = 5000 \, \Omega$.
Calculate the power factor $\cos \phi$:
$\cos \phi = \frac{R}{Z} = \frac{3000}{5000} = 0.6$.
Calculate the power dissipated $P$:
$P = \frac{V_{rms}^2 \cos \phi}{Z} = \frac{200^2 \times 0.6}{5000} = \frac{40000 \times 0.6}{5000} = 8 \times 0.6 = 4.8 \, W$.

(C) For $dc$ source: $P = \frac{V^2}{R}$. Given $V = 10 \, V$ and $P = 20 \, W$,we have $R = \frac{V^2}{P} = \frac{10^2}{20} = \frac{100}{20} = 5 \, \Omega$.
For $ac$ source: $P = I_{rms}^2 R = \left( \frac{V}{Z} \right)^2 R = \frac{V^2 R}{Z^2}$. Given $V = 10 \, V$ and $P = 10 \, W$,we have $Z^2 = \frac{V^2 R}{P} = \frac{10^2 \times 5}{10} = 50 \, \Omega^2$.
The impedance of an $RL$ circuit is $Z^2 = R^2 + X_L^2 = R^2 + (2 \pi \nu L)^2$.
Substituting the values: $50 = 5^2 + (2 \pi \nu \times 10 \times 10^{-3})^2$.
$50 = 25 + (2 \pi \nu \times 10^{-2})^2$.
$25 = (2 \pi \nu \times 10^{-2})^2$.
Taking the square root: $5 = 2 \pi \nu \times 10^{-2}$.
$\nu = \frac{5}{2 \pi \times 10^{-2}} = \frac{500}{2 \pi} \approx \frac{500}{6.28} \approx 79.6 \, Hz \approx 80 \, Hz$.

(A) The power factor of an $RL$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{1}{2}$.
Since $\cos \phi = 1/2$,the phase angle $\phi = 60^\circ$.
We know that $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.
Substituting the values: $\tan 60^\circ = \frac{2 \pi f L}{R}$.
$\sqrt{3} = \frac{2 \pi \times 50 \times L}{100}$.
$\sqrt{3} = \frac{100 \pi L}{100}$.
$\sqrt{3} = \pi L$.
Therefore,$L = \frac{\sqrt{3}}{\pi} \, H$.

(D) The phase difference $\varphi$ in an $RL$ circuit is given by $\tan \varphi = \frac{X_L}{R}$.
Given: $R = \pi \sqrt{3} \,\Omega$,$\varphi = 30^\circ$,and $\nu = 50 \,Hz$.
We know that $X_L = 2\pi \nu L$.
Substituting the values into the formula: $\tan 30^\circ = \frac{2\pi \nu L}{R}$.
$\frac{1}{\sqrt{3}} = \frac{2\pi \times 50 \times L}{\pi \sqrt{3}}$.
Canceling $\pi \sqrt{3}$ from both sides: $\frac{1}{\sqrt{3}} = \frac{100L}{\sqrt{3}}$.
$1 = 100L$.
$L = \frac{1}{100} = 0.01 \,H$.

(B) Given: Current $i = 100 \sin(314t) \ A$ and Voltage $e = 200 \sin(314t + \pi/3) \ V$.
From the peak values,the peak current $I_0 = 100 \ A$ and peak voltage $V_0 = 200 \ V$.
The impedance $Z$ is given by $Z = V_0 / I_0 = 200 / 100 = 2 \ \Omega$.
In an $RL$ circuit,the impedance is related to resistance $R$ and reactance $X_L$ by the formula $Z^2 = R^2 + X_L^2$.
Given $R = 1 \ \Omega$,we have $2^2 = 1^2 + X_L^2$.
$4 = 1 + X_L^2 \implies X_L^2 = 3$.
Therefore,$X_L = \sqrt{3} \ \Omega$.

(A) In an $RL$ series circuit,the phase difference $\varphi$ between the voltage and the current is given by the formula: $\tan \varphi = \frac{X_L}{R}$.
Given that the reactance $X_L = \sqrt{3} R$.
Substituting this into the formula: $\tan \varphi = \frac{\sqrt{3} R}{R} = \sqrt{3}$.
Since $\tan \varphi = \sqrt{3}$,we have $\varphi = 60^{\circ}$.
Converting degrees to radians: $\varphi = 60^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{3}$ radians.

(A) The current in an $RC$ series circuit is given by $I = \frac{V}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C}$.
At initial frequency $\omega$,$I = \frac{V}{\sqrt{R^2 + (1/\omega C)^2}}$ ... $(i)$
At frequency $\omega' = \omega/3$,the new reactance is $X_C' = \frac{1}{(\omega/3)C} = \frac{3}{\omega C} = 3X_C$.
The new current is $I' = I/2 = \frac{V}{\sqrt{R^2 + (3X_C)^2}}$ ... (ii)
Dividing $(i)$ by (ii),we get $2 = \frac{\sqrt{R^2 + 9X_C^2}}{\sqrt{R^2 + X_C^2}}$.
Squaring both sides: $4 = \frac{R^2 + 9X_C^2}{R^2 + X_C^2}$.
$4R^2 + 4X_C^2 = R^2 + 9X_C^2$.
$3R^2 = 5X_C^2$.
$\frac{X_C^2}{R^2} = \frac{3}{5}$.
Therefore,the ratio $\frac{X_C}{R} = \sqrt{\frac{3}{5}}$.

(A) The phase difference $\phi$ in an $RL$ series circuit is given by $\tan \phi = \frac{X_L}{R}$.
Here,$R = 300 \Omega$,$L = \frac{1}{\pi} \text{ H}$,and $f = 200 \text{ Hz}$.
The inductive reactance is $X_L = \omega L = 2\pi f L$.
Substituting the values: $X_L = 2\pi \times 200 \times \frac{1}{\pi} = 400 \Omega$.
Now,$\tan \phi = \frac{400}{300} = \frac{4}{3}$.
Therefore,the phase difference is $\phi = \tan^{-1} \frac{4}{3}$.

(C) The average power in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$.
Here,$E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{E_0}{Z\sqrt{2}}$.
The power factor is $\cos \phi = \frac{R}{Z}$.
Substituting these,$P = \frac{E_0}{\sqrt{2}} \times \frac{E_0}{Z\sqrt{2}} \times \frac{R}{Z} = \frac{E_0^2 R}{2Z^2}$.
Given that the inductive reactance $X_L = R$,the impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2}R$.
Substituting $Z^2 = 2R^2$ into the power formula:
$P = \frac{E_0^2 R}{2(2R^2)} = \frac{E_0^2 R}{4R^2} = \frac{E_0^2}{4R}$.

(B) The power consumed in an $AC$ circuit is given by $P = I_{rms}^2 R$.
Given $P = 240 W$ and $I_{rms} = 4 A$,we have $R = \frac{P}{I_{rms}^2} = \frac{240}{16} = 15 \Omega$.
The impedance $Z$ of the circuit is given by $Z = \frac{V_{rms}}{I_{rms}} = \frac{100}{4} = 25 \Omega$.
We know that $Z^2 = R^2 + X_L^2$,where $X_L$ is the inductive reactance.
$X_L = \sqrt{Z^2 - R^2} = \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20 \Omega$.
Since $X_L = 2\pi f L$,we have $20 = 2\pi(50)L$.
$L = \frac{20}{100\pi} = \frac{1}{5\pi} H$.

(B) The dimension of resistance $R$ is $[M L^2 T^{-3} I^{-2}]$.
The dimension of inductance $L$ is $[M L^2 T^{-2} I^{-2}]$.
The dimension of capacitance $C$ is $[M^{-1} L^{-2} T^4 I^2]$.
Checking option $B$:
$\frac{L}{R} = \frac{[M L^2 T^{-2} I^{-2}]}{[M L^2 T^{-3} I^{-2}]} = [T^1] = T$.
Checking option $C$:
$\sqrt{LC} = \sqrt{[M L^2 T^{-2} I^{-2}] \cdot [M^{-1} L^{-2} T^4 I^2]} = \sqrt{[T^2]} = [T^1] = T$.
Since both $B$ and $C$ result in the dimension of time,the question implies identifying expressions with time dimensions. Given the standard format,both $\frac{L}{R}$ and $\sqrt{LC}$ are correct.

(B) Given: Resistance $R = 3\,\Omega$ and Inductive reactance $X_L = 3\,\Omega$.
In an $LR$ series circuit,the phase difference $\phi$ between the applied voltage and the current is given by the formula:
$\tan \phi = \frac{X_L}{R}$
Substituting the given values:
$\tan \phi = \frac{3\,\Omega}{3\,\Omega} = 1$
Therefore,the phase difference is:
$\phi = \tan^{-1}(1) = \frac{\pi}{4}$ radians.

(B) Given: Resistance $R = 30\,\Omega$.
Inductive reactance $X_L = 20\,\Omega$ at frequency $f = 50\,Hz$.
Since $X_L = 2\pi f L$,the inductive reactance is directly proportional to the frequency $(X_L \propto f)$.
At a new frequency $f' = 100\,Hz$,the new inductive reactance $X_L'$ is:
$X_L' = X_L \times \left(\frac{f'}{f}\right) = 20\,\Omega \times \left(\frac{100\,Hz}{50\,Hz}\right) = 40\,\Omega$.
The impedance $Z$ of the $RL$ circuit 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 $I = \frac{V}{Z} = \frac{200\,V}{50\,\Omega} = 4\,A$.