RL, RC and LC AC Circuits Questions in English

(A) The circuit is an $LCR$ series circuit with $R = 100\,\Omega$,$X_L = 200\,\Omega$,and $X_C = 100\,\Omega$.
The impedance $Z$ of the circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
$Z = \sqrt{100^2 + (200 - 100)^2}$
$Z = \sqrt{100^2 + 100^2} = \sqrt{2 \times 100^2} = 100 \sqrt{2}\,\Omega$
The $rms$ current $I_{rms}$ is given by:
$I_{rms} = \frac{V_{rms}}{Z}$
Given $V_{rms} = 200 \sqrt{2}\,V$,we have:
$I_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}} = 2\,A$

(C) The given parameters are $R = 80\,\Omega$,$X_{L} = 100\,\Omega$,and $X_{C} = 40\,\Omega$.
The peak voltage (amplitude) is $V_{0} = 2500\,V$.
The impedance $Z$ of the series $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_{L} - X_{C})^2}$.
Substituting the values: $Z = \sqrt{80^2 + (100 - 40)^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$.
The amplitude of the current $I_{0}$ is given by $I_{0} = \frac{V_{0}}{Z}$.
$I_{0} = \frac{2500}{100} = 25\,A$.

(C) For a series $L, R$ circuit,the power factor is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$,where $X_L = \omega L$.
Given the initial angular frequency $\omega_1 = 1000 \ rad/s$ and power factor $\cos \phi_1 = \frac{1}{\sqrt{2}}$.
Since $\cos \phi_1 = \frac{1}{\sqrt{2}}$,the phase angle $\phi_1 = 45^{\circ}$.
Thus,$\tan \phi_1 = \frac{X_{L1}}{R} = \frac{\omega_1 L}{R} = \tan 45^{\circ} = 1$.
This implies $R = \omega_1 L = 1000 L$.
When the source is changed to $E = (20 \sin 2000 t) \ V$,the new angular frequency is $\omega_2 = 2000 \ rad/s$.
The new inductive reactance is $X_{L2} = \omega_2 L = 2000 L = 2(\omega_1 L) = 2R$.
The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + X_{L2}^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

(D) For $DC$ voltage, the inductor acts as a pure resistor because $X_L = 0$ for $DC$.
$R = \frac{V}{I} = \frac{100 \, V}{5 \, A} = 20 \, \Omega$.
For $AC$ voltage, the circuit is an $LR$ series circuit.
Given $X_L = 20\sqrt{3} \, \Omega$ and $R = 20 \, \Omega$.
The impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = \sqrt{400 + 1200} = \sqrt{1600} = 40 \, \Omega$.
The $RMS$ voltage is $V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{200}{\sqrt{2}} \, V$.
The $RMS$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{200 / \sqrt{2}}{40} = \frac{5}{\sqrt{2}} \, A$.
The power dissipated in the circuit is $P = I_{rms}^2 R$.
$P = \left( \frac{5}{\sqrt{2}} \right)^2 \times 20 = \frac{25}{2} \times 20 = 250 \, W$.

(A) Given: Resistance $R = 90 \Omega$, Supply voltage $V = 120 \text{ V}$, Frequency $f = 60 \text{ Hz}$, Voltage across resistor $V_R = 36 \text{ V}$.
The current in the series circuit is $I = \frac{V_R}{R} = \frac{36}{90} = 0.4 \text{ A}$.
The impedance of the circuit is $Z = \frac{V}{I} = \frac{120}{0.4} = 300 \Omega$.
We know $Z = \sqrt{R^2 + X_L^2}$, so $300 = \sqrt{90^2 + X_L^2}$.
Squaring both sides: $90000 = 8100 + X_L^2$.
$X_L^2 = 90000 - 8100 = 81900$.
$X_L = \sqrt{81900} \approx 286.18 \Omega$.
Since $X_L = 2 \pi f L$, we have $L = \frac{X_L}{2 \pi f} = \frac{286.18}{2 \times 3.14 \times 60} = \frac{286.18}{376.8} \approx 0.76 \text{ H}$.

(A) The impedance of the $RC$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
When a dielectric is placed between the plates of the capacitor,its capacitance $C$ increases $(C' = KC)$.
Since $X_C = \frac{1}{\omega C}$,an increase in $C$ leads to a decrease in capacitive reactance $(X_C \downarrow)$.
As $X_C$ decreases,the total impedance $Z = \sqrt{R^2 + X_C^2}$ of the circuit decreases $(Z \downarrow)$.
According to Ohm's law for $AC$ circuits,the current $I = \frac{V}{Z}$. Since $Z$ decreases,the current $I$ in the circuit increases.
Consequently,the power dissipated in the bulb $(P = I^2 R)$ increases,and the glow of the bulb increases.

(D) Case-$I$: $DC$ supply
For a $DC$ circuit,the inductor acts as a short circuit (resistance only).
$R = \frac{V}{I} = \frac{20 \ V}{5 \ A} = 4 \ \Omega$
Case-$II$: $AC$ supply
For an $AC$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2}$.
$Z = \frac{V}{I} = \frac{20 \ V}{4 \ A} = 5 \ \Omega$
Since $Z^2 = R^2 + X_L^2$,we have $5^2 = 4^2 + X_L^2$.
$25 = 16 + X_L^2 \Rightarrow X_L^2 = 9 \Rightarrow X_L = 3 \ \Omega$
We know $X_L = 2 \pi f L$,where $f = 50 \ Hz$ and $\pi = 3$.
$3 = 2 \times 3 \times 50 \times L$
$3 = 300 \times L$
$L = \frac{3}{300} \ H = 0.01 \ H$
Converting to millihenry $(mH)$:
$L = 0.01 \times 1000 \ mH = 10 \ mH$

(C) Given:
Reactance of capacitor $X_C = 4 \sqrt{3} \Omega$
Resistance $R = 4 \Omega$
Peak voltage $V_0 = 8 \sqrt{2} \text{ V}$
Step $1$: Calculate the impedance $Z$ of the $RC$ circuit.
$Z = \sqrt{R^2 + X_C^2}$
$Z = \sqrt{4^2 + (4 \sqrt{3})^2} = \sqrt{16 + 16 \times 3} = \sqrt{16 + 48} = \sqrt{64} = 8 \Omega$
Step $2$: Calculate the $RMS$ voltage $V_{\text{rms}}$.
$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8 \text{ V}$
Step $3$: Calculate the $RMS$ current $I_{\text{rms}}$.
$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{8}{8} = 1 \text{ A}$
Step $4$: Calculate the power dissipation $P$.
Power is dissipated only in the resistor.
$P = I_{\text{rms}}^2 \times R = (1)^2 \times 4 = 4 \text{ W}$

(B) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
Substituting $X_C$,we get $Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$.
As the angular frequency $\omega$ increases,the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.
Since $Z = \sqrt{R^2 + X_C^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$ of the circuit.
The current in the circuit is given by $I = \frac{V_0}{Z}$. Since $V_0$ is constant and $Z$ decreases,the current $I$ increases.
The brightness of the bulb is proportional to the power dissipated,$P = I^2 R$. As the current $I$ increases,the power dissipated increases,and the bulb glows brighter.

(A) In a series $R-C$ circuit,the impedance is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$.
When a dielectric of constant $K = 4$ is inserted,the new capacitance becomes $C' = KC = 4C$.
Consequently,the new capacitive reactance becomes $X_C' = \frac{1}{\omega (4C)} = \frac{X_C}{4}$.
Since $X_C' < X_C$,the total impedance $Z' = \sqrt{R^2 + (X_C')^2}$ is less than $Z = \sqrt{R^2 + X_C^2}$.
The current in the circuit is $I = \frac{V}{Z}$. Since $Z' < Z$,the current in case $(B)$ is greater than in case $(A)$,so $I_R^B > I_R^A$ (Option $B$ is true).
The voltage across the capacitor is $V_C = I X_C = \frac{V}{\sqrt{R^2 + X_C^2}} X_C = \frac{V}{\sqrt{(R/X_C)^2 + 1}}$.
As $X_C$ decreases,the term $(R/X_C)^2$ increases,which makes the denominator $\sqrt{(R/X_C)^2 + 1}$ larger.
Therefore,$V_C$ decreases when $X_C$ decreases. Thus,$V_C^A > V_C^B$ (Option $C$ is true).
Hence,the correct statements are $(B)$ and $(C)$.

(C) Given: $\omega = 500 \ rad/s$.
The impedance $Z$ of a series $R-C$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$.
Given $Z = R\sqrt{1.25}$,we have $Z^2 = 1.25R^2$.
Substituting into the impedance formula: $R^2 + X_C^2 = 1.25R^2$.
$X_C^2 = 0.25R^2$.
$X_C = 0.5R$.
Since $X_C = \frac{1}{\omega C}$,we have $\frac{1}{\omega C} = 0.5R$.
Rearranging for the time constant $\tau = RC$: $RC = \frac{1}{0.5\omega}$.
Substituting $\omega = 500 \ rad/s$: $\tau = \frac{1}{0.5 \times 500} = \frac{1}{250} = 0.004 \ s$.
Converting to milliseconds: $\tau = 0.004 \times 1000 \ ms = 4 \ ms$.

(A) Given: Supply voltage $V = 200 V$,Frequency $f = 50 Hz$,Power $P = 500 W$,Voltage across lamp $V_R = 100 V$.
$1$. Since the circuit is an $RC$ series circuit,the supply voltage is $V = \sqrt{V_R^2 + V_C^2}$.
$200^2 = 100^2 + V_C^2 \Rightarrow V_C^2 = 40000 - 10000 = 30000$.
$V_C = 100\sqrt{3} V$.
$2$. The phase angle $\varphi$ is given by $\tan \varphi = \frac{V_C}{V_R} = \frac{100\sqrt{3}}{100} = \sqrt{3}$.
$\varphi = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.
$3$. Power $P = V_R \cdot I \Rightarrow 500 = 100 \cdot I \Rightarrow I = 5 A$.
Also,$V_C = I \cdot X_C \Rightarrow 100\sqrt{3} = 5 \cdot X_C \Rightarrow X_C = 20\sqrt{3} \Omega$.
$4$. $X_C = \frac{1}{2\pi f C} \Rightarrow 20\sqrt{3} = \frac{1}{2 \cdot \pi \cdot 50 \cdot C \cdot 10^{-6}}$.
$C = \frac{1}{20\sqrt{3} \cdot 100 \cdot \pi} \cdot 10^6 = \frac{10^6}{2000 \cdot \pi \sqrt{3}} = \frac{1000}{2 \cdot 5} = 100 \mu F$.
Thus,$C = 100 \mu F$ and $\varphi = 60^{\circ}$.

(A) Given: Self-inductance $L = 1 \ H$,Resistance $R = 100 \pi \ \Omega$,Voltage $V_{rms} = 100 \pi \ V$,Frequency $f = 50 \ Hz$.
First,calculate the inductive reactance $X_L = \omega L = 2 \pi f L = 2 \pi \times 50 \times 1 = 100 \pi \ \Omega$.
The impedance of the $LR$ series circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{(100 \pi)^2 + (100 \pi)^2} = \sqrt{2 \times (100 \pi)^2} = 100 \pi \sqrt{2} \ \Omega$.
The $RMS$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{100 \pi}{100 \pi \sqrt{2}} = \frac{1}{\sqrt{2}} \ A$.
The maximum current $I_{max}$ is given by $I_{max} = I_{rms} \sqrt{2} = \frac{1}{\sqrt{2}} \times \sqrt{2} = 1 \ A$.

(B) Let the input voltage be $V_{\text{in}}$. The circuit consists of two potential dividers in parallel.
For the left branch,the voltage at point $A$ with respect to the bottom node is $V_A = V_{\text{in}} \cdot \frac{-jX_C}{R - jX_C}$.
For the right branch,the voltage at point $B$ with respect to the bottom node is $V_B = V_{\text{in}} \cdot \frac{R}{R - jX_C}$.
Thus,the potential difference $(V_B - V_A)$ is given by:
$V_B - V_A = V_{\text{in}} \cdot \frac{R + jX_C}{R - jX_C}$.
Let $Z = R - jX_C$. Then $V_B - V_A = V_{\text{in}} \cdot \frac{R + jX_C}{R - jX_C}$.
The phase of $(V_B - V_A)$ relative to $V_{\text{in}}$ is the phase of $\frac{R + jX_C}{R - jX_C}$.
Let $\tan \theta = \frac{X_C}{R}$. Then the phase of $(R + jX_C)$ is $\theta$ and the phase of $(R - jX_C)$ is $-\theta$.
The phase of the ratio is $\theta - (-\theta) = 2\theta$.
Given the phase difference is $90^{\circ}$,we have $2\theta = 90^{\circ}$,so $\theta = 45^{\circ}$.
Therefore,$\tan 45^{\circ} = \frac{X_C}{R} \implies 1 = \frac{1}{\omega RC} \implies \omega = \frac{1}{RC}$.
Substituting the values: $R = 10^5 \ \Omega$,$C = 100 \times 10^{-12} \ F = 10^{-10} \ F$.
$\omega = \frac{1}{10^5 \times 10^{-10}} = \frac{1}{10^{-5}} = 10^5 \ rad/sec$.
Comparing with $10^x$,we get $x = 5$.

(A) The circuit consists of a bulb (resistor $R$) and a capacitor $(C)$ connected in series,forming an $R-C$ circuit.
The impedance of an $R-C$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{2\pi fC}$ is the capacitive reactance.
When the frequency $f$ of the $AC$ source is increased,the capacitive reactance $X_C = \frac{1}{2\pi fC}$ decreases.
Since $Z = \sqrt{R^2 + X_C^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$ of the circuit.
According to Ohm's law for $AC$ circuits,the current $i$ is given by $i = \frac{V}{Z}$. Since the source voltage $V$ is constant and $Z$ decreases,the current $i$ flowing through the circuit increases.
The brightness of the bulb is proportional to the power dissipated,$P = i^2R$. As the current $i$ increases,the power dissipated by the bulb increases,and thus the bulb will give more intense light.

(B) For $DC$ supply,the coil acts as a pure resistor $R$. Given $V = 100 \ V$ and $I = 25 \ A$,the resistance is $R = \frac{V}{I} = \frac{100}{25} = 4 \ \Omega$.
For $AC$ supply,the coil acts as an $LR$ circuit with impedance $Z$. Given $V = 100 \ V$ and $I = 20 \ A$,the impedance is $Z = \frac{V}{I} = \frac{100}{20} = 5 \ \Omega$.
The impedance $Z$ is related to resistance $R$ and inductive reactance $X_L$ by the formula $Z^2 = R^2 + X_L^2$.
Substituting the values,$5^2 = 4^2 + X_L^2$.
$25 = 16 + X_L^2 \Rightarrow X_L^2 = 25 - 16 = 9$.
Therefore,the reactance $X_L = \sqrt{9} = 3 \ \Omega$.

(C) The impedance of the series $L-C-R$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $|X_L - X_C| = R$,we substitute this into the impedance formula:
$Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.
The peak current in the circuit is $I_0 = \frac{V_0}{Z} = \frac{V_0}{R\sqrt{2}}$.
The potential difference across the capacitor is $V_C = I_0 X_C$.
The r.m.s. value of the potential difference across the capacitor is $V_{C,rms} = I_{rms} X_C$.
Since $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{V_0}{R\sqrt{2} \cdot \sqrt{2}} = \frac{V_0}{2R}$,we have:
$V_{C,rms} = \frac{V_0}{2R} \cdot \frac{1}{\omega C} = \frac{V_0}{2R\omega C}$.

(D) In an $LR$ series circuit,the impedance $Z$ is given by the formula $Z = \sqrt{R^2 + X_L^2}$.
Here,$R$ is the resistance and $X_L$ is the inductive reactance.
The inductive reactance $X_L$ is defined as $X_L = \omega L$,where $\omega = 2 \pi f$.
Substituting the value of $\omega$,we get $X_L = 2 \pi f L$.
Now,substituting $X_L$ into the impedance formula:
$Z = \sqrt{R^2 + (2 \pi f L)^2}$
$Z = \sqrt{R^2 + 4 \pi^2 f^2 L^2}$.
Thus,the correct option is $D$.

(D) Given: Inductance $L = \frac{0.3}{\pi} \ H$,Resistance $R = 40 \ \Omega$,Voltage $V = 230 \ V$,Frequency $f = 50 \ Hz$.
First,calculate the inductive reactance $X_L = 2\pi f L$.
$X_L = 2 \times \pi \times 50 \times \frac{0.3}{\pi} = 100 \times 0.3 = 30 \ \Omega$.
The impedance $Z$ of the $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$.
$Z = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \ \Omega$.
The current $I$ in the circuit is given by $I = \frac{V}{Z}$.
$I = \frac{230}{50} = 4.6 \ A$.
Thus,the impedance is $50 \ \Omega$ and the current is $4.6 \ A$.

(D) Given: Inductance $L = \frac{100}{\pi} \text{ mH} = \frac{0.1}{\pi} \text{ H}$,Capacitance $C = \frac{10^{-3}}{2\pi} \text{ F}$,Resistance $R = 10 \Omega$,Frequency $f = 50 \text{ Hz}$.
First,calculate the inductive reactance $X_L = 2\pi f L = 2\pi \times 50 \times \frac{0.1}{\pi} = 100 \times 0.1 = 10 \Omega$.
Next,calculate the capacitive reactance $X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{10^{-3}}{2\pi}} = \frac{1}{50 \times 10^{-3}} = \frac{1000}{50} = 20 \Omega$.
The tangent of the phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Substituting the values: $\tan \phi = \frac{10 - 20}{10} = \frac{-10}{10} = -1$.
The magnitude of the tangent of the phase angle is $|\tan \phi| = 1$.

(D) The phase angle $\phi$ between voltage and current in an $RL$ circuit is given by $\tan \phi = \frac{X_L}{R}$.
Given $\phi = 45^{\circ}$,$R = 80 \Omega$,and $f = 480 \text{ Hz}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{X_L}{R}$,which implies $X_L = R$.
Substituting $X_L = 2 \pi f L$,we get $2 \pi f L = R$.
Solving for $L$: $L = \frac{R}{2 \pi f}$.
Substituting the values: $L = \frac{80}{2 \pi \times 480} = \frac{80}{960 \pi} = \frac{1}{12 \pi} \text{ H}$.

(A) When $80 \ V$ $d.c.$ is applied,the solenoid acts as a pure resistor because the frequency of $d.c.$ is zero,so inductive reactance $X_L = 2 \pi fL = 0$.
Resistance $R = \frac{V}{I_{dc}} = \frac{80 \ V}{0.8 \ A} = 100 \ \Omega$.
When $80 \ V$ $a.c.$ is applied,the impedance $Z$ is given by $Z = \frac{V}{I_{ac}} = \frac{80 \ V}{0.4 \ A} = 200 \ \Omega$.
The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$.
Substituting the values: $200 = \sqrt{100^2 + X_L^2}$.
$40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000$.
$X_L = \sqrt{30000} \approx 173.2 \ \Omega$.
Since $X_L = 2 \pi fL$,we have $L = \frac{X_L}{2 \pi f} = \frac{173.2}{2 \times 3.14 \times 50} = \frac{173.2}{314} \approx 0.55 \ H$.
Thus,the impedance is $200 \ \Omega$ and the inductance is $0.55 \ H$.

(B) The impedance $Z$ of a series $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = \frac{X_L}{2}$,we have $X_L = 2R$.
Given $R = 2X_C$,we have $X_C = \frac{R}{2}$.
Substituting these values into the impedance formula:
$Z = \sqrt{R^2 + (2R - \frac{R}{2})^2} = \sqrt{R^2 + (\frac{3R}{2})^2} = \sqrt{R^2 + \frac{9R^2}{4}} = \sqrt{\frac{13R^2}{4}} = \frac{\sqrt{13}}{2} R$.
The phase difference $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Substituting the values: $\tan \phi = \frac{2R - R/2}{R} = \frac{3R/2}{R} = \frac{3}{2}$.
Therefore,$\phi = \tan^{-1}(\frac{3}{2})$.

(B) Given: $R = X_C = \frac{1}{\omega C}$.
Initial impedance $Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + R^2} = \sqrt{2} R$.
Initial peak current $I_0 = \frac{V_0}{Z} = \frac{V_0}{\sqrt{2} R}$,which implies $\frac{V_0}{R} = \sqrt{2} I_0$.
When angular frequency $\omega' = \frac{\omega}{\sqrt{3}}$,the new capacitive reactance $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega / \sqrt{3}) C} = \sqrt{3} \left(\frac{1}{\omega C}\right) = \sqrt{3} R$.
New impedance $Z' = \sqrt{R^2 + (X_C')^2} = \sqrt{R^2 + (\sqrt{3} R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.
New peak current $I_0' = \frac{V_0}{Z'} = \frac{V_0}{2R} = \frac{1}{2} \left(\frac{V_0}{R}\right) = \frac{1}{2} (\sqrt{2} I_0) = \frac{I_0}{\sqrt{2}}$.

(A) The phase difference $\phi$ between the voltage and current in an $LCR$ series circuit is given by: $\tan \phi = \frac{X_L - X_C}{R}$.
Given $\phi = 45^{\circ}$,so $\tan 45^{\circ} = 1$.
Thus,$\frac{\omega L - \frac{1}{\omega C}}{R} = 1$.
$\omega L - \frac{1}{\omega C} = R$.
$\omega L = R + \frac{1}{\omega C} = \frac{R \omega C + 1}{\omega C}$.
Since $\omega = 2 \pi f$,we have $L = \frac{R \omega C + 1}{\omega^2 C} = \frac{R(2 \pi f)C + 1}{(2 \pi f)^2 C}$.
$L = \frac{1 + 2 \pi fCR}{4 \pi^2 f^2 C}$.

(A) For $d.c.$ supply,the inductor acts as a pure resistor:
$R = \frac{V}{I} = \frac{100 \ V}{1 \ A} = 100 \ \Omega$.
For $a.c.$ supply,the impedance $Z$ is given by:
$Z = \frac{V}{I} = \frac{100 \ V}{0.5 \ A} = 200 \ \Omega$.
The impedance of an $LR$ circuit is $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2 \pi f L$.
Substituting the values: $200 = \sqrt{100^2 + X_L^2}$.
Squaring both sides: $40000 = 10000 + X_L^2 \Rightarrow X_L^2 = 30000$.
$X_L = \sqrt{30000} = 100\sqrt{3} \ \Omega$.
Since $X_L = 2 \pi f L$,we have $100\sqrt{3} = 2 \pi (50) L$.
$100\sqrt{3} = 100 \pi L \Rightarrow L = \frac{\sqrt{3}}{\pi} \ H$.

(C) For a $d.c.$ source,the inductor acts as a simple resistor because the frequency is zero. Thus,the resistance $R$ is given by $R = \frac{V}{I} = \frac{12 \ V}{4 \ A} = 3 \ \Omega$.
For an $a.c.$ source,the impedance $Z$ is given by $Z = \frac{V}{I} = \frac{12 \ V}{2.4 \ A} = 5 \ \Omega$.
The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$,where $X_L$ is the inductive reactance.
Squaring both sides,$Z^2 = R^2 + X_L^2$,so $X_L^2 = Z^2 - R^2 = 5^2 - 3^2 = 25 - 9 = 16$.
Thus,$X_L = 4 \ \Omega$.
Since $X_L = 2 \pi f L$,we have $L = \frac{X_L}{2 \pi f} = \frac{4}{2 \pi \times 50} = \frac{4}{100 \pi} = \frac{4}{\pi} \times 10^{-2} \ H$.

(B) The capacitive reactance of a capacitor is given by the formula $X_{C} = \frac{1}{2 \pi f C}$.
Here,$f$ is the frequency of the $A.C.$ source and $C$ is the capacitance.
When the capacitance $C$ is reduced,the capacitive reactance $X_{C}$ increases.
Similarly,when the frequency $f$ is reduced,the capacitive reactance $X_{C}$ increases.
Since the bulb is in series with the capacitor,the total impedance $Z$ of the circuit is given by $Z = \sqrt{R^2 + X_{C}^2}$.
As $X_{C}$ increases in both cases,the total impedance $Z$ of the circuit increases.
According to Ohm's law for $A.C.$ circuits,the current $I = \frac{V}{Z}$.
Since the impedance $Z$ increases,the current $I$ flowing through the bulb decreases.
The brightness of the bulb depends on the power dissipated,$P = I^2 R$.
As the current $I$ decreases,the power dissipated and the brightness of the lamp decrease in both cases.

(C) In an $LCR$ series circuit,the current $I$ is the same through all components. The voltage across the resistor $V_R$ is in phase with the current. 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 $90^\circ$.
Thus,$V_L$ and $V_C$ are in opposite directions. The net reactive voltage is $(V_L - V_C)$.
Using the phasor diagram,the resultant voltage $V$ is the vector sum of $V_R$ and $(V_L - V_C)$,which are perpendicular to each other.
Applying the Pythagorean theorem: $V^2 = V_R^2 + (V_L - V_C)^2$.

(D) For an $LR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (\omega L)^2}$.
Comparing the given equation $E = 4 \cos(1000 t)$ with the standard equation $E = E_0 \cos(\omega t)$,we get $E_0 = 4 \ V$ and $\omega = 1000 \ rad/s$.
Given $L = 3 \ mH = 3 \times 10^{-3} \ H$ and $R = 4 \ \Omega$.
The inductive reactance is $X_L = \omega L = 1000 \times 3 \times 10^{-3} = 3 \ \Omega$.
Now,calculate the impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \ \Omega$.
The maximum current $I_0$ is given by $I_0 = \frac{E_0}{Z} = \frac{4}{5} = 0.8 \ A$.

(D) In a parallel $LC$ circuit,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 $180^{\circ}$ out of phase.
The total current $I$ drawn from the source is the magnitude of the difference between the two currents:
$I = |I_C - I_L|$
Given $I_L = 0.6 \,A$ and $I_C = 0.9 \,A$.
$I = |0.9 \,A - 0.6 \,A| = 0.3 \,A$.

(D) In a series $LCR$ circuit,the voltage across the inductor $(V_L)$ leads the current by $90^{\circ}$,and the voltage across the capacitor $(V_C)$ lags the current by $90^{\circ}$.
Since the current is the same for all components in a series circuit,the phase difference between $V_L$ and $V_C$ is $180^{\circ}$.
However,the individual voltages $V_L$ and $V_C$ are not necessarily in phase with the source voltage $(V_S)$ unless the circuit is at resonance.
Therefore,the statement that they are 'in phase with the source voltage' is incorrect,making $D$ the correct choice.

(B) In a series $LCR$ circuit, the total voltage (e.m.f.) '$e$' is given by the phasor sum of the voltages across the individual components:
$e = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given:
$V_R = 40 \,V$
$V_L = 80 \,V$
$V_C = 40 \,V$
Substituting these values into the formula:
$e = \sqrt{(40)^2 + (80 - 40)^2}$
$e = \sqrt{1600 + (40)^2}$
$e = \sqrt{1600 + 1600}$
$e = \sqrt{3200}$
$e = \sqrt{1600 \times 2}$
$e = 40 \sqrt{2} \,V$

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

(C) In a series $LCR$ circuit,the phase angle $\phi$ between the voltage and the current is given by the formula: $\tan \phi = \frac{X_L - X_C}{R}$.
Since the current leads the voltage by $45^{\circ}$,the phase angle $\phi$ is $-45^{\circ}$ (because current is ahead of voltage).
Substituting the values: $\tan(-45^{\circ}) = \frac{X_L - X_C}{R}$.
$-1 = \frac{X_L - X_C}{R}$.
$-R = X_L - X_C$.
Rearranging the terms,we get: $X_C = X_L + R$.

(D) In an $LR$ circuit,the phase angle $\phi$ between voltage and current is given by $\tan \phi = \frac{X_L}{R}$.
Given $\phi = 45^{\circ}$,we have $\tan 45^{\circ} = 1$.
Therefore,$\frac{X_L}{R} = 1$,which implies $X_L = R$.
Given $R = 100 \ \Omega$,so $X_L = 100 \ \Omega$.
The inductive reactance is $X_L = 2 \pi f L$.
Substituting the values,$100 = 2 \pi \times 100 \times L$.
Solving for $L$,we get $L = \frac{100}{2 \pi \times 100} = \frac{1}{2 \pi} \ H$.

(B) For an $LR$ series circuit,the phase angle $\phi$ between the voltage and the current is given by the formula: $\tan \phi = \frac{X_L}{R}$.
Here,$X_L = \omega L = 2 \pi f L$.
Given: $L = \frac{1}{\pi} \text{ H}$,$R = 300 \text{ } \Omega$,and $f = 200 \text{ Hz}$.
Substituting the values: $X_L = 2 \pi \times 200 \times \frac{1}{\pi} = 400 \text{ } \Omega$.
Now,$\tan \phi = \frac{400}{300} = \frac{4}{3}$.
Therefore,$\phi = \tan^{-1}(\frac{4}{3})$.

(B) The phase difference $\phi$ between voltage and current in an $RL$ circuit is given by $\tan \phi = \frac{X_L}{R}$.
Given $\phi = 45^{\circ}$,$R = 100 \ \Omega$,and $f = 1000 \ Hz$.
Since $\tan 45^{\circ} = 1$,we have $X_L = R$.
Substituting $X_L = 2 \pi f L$,we get $2 \pi f L = R$.
Therefore,$L = \frac{R}{2 \pi f}$.
Substituting the values: $L = \frac{100}{2 \pi \times 1000} = \frac{100}{2000 \pi} = \frac{1}{20 \pi} = \frac{0.05}{\pi} \ H$.

(A) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$.
The initial current is $I = \frac{V}{Z}$.
When the frequency changes to $\omega' = \frac{\omega}{3}$,the new capacitive reactance is $X_c' = \frac{1}{\omega' C} = \frac{1}{(\omega/3) C} = 3X_c$.
The new impedance is $Z' = \sqrt{R^2 + (X_c')^2} = \sqrt{R^2 + (3X_c)^2}$.
Given that the new current $I' = \frac{I}{2}$,we have $\frac{V}{Z'} = \frac{1}{2} \frac{V}{Z}$,which implies $Z' = 2Z$.
Squaring both sides,$(Z')^2 = 4Z^2$,so $R^2 + 9X_c^2 = 4(R^2 + X_c^2)$.
$R^2 + 9X_c^2 = 4R^2 + 4X_c^2$.
$5X_c^2 = 3R^2$.
$\frac{X_c^2}{R^2} = \frac{3}{5} = 0.6$.
Therefore,the ratio $\frac{X_c}{R} = \sqrt{0.6}$.

(D) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$. The initial current is $I = \frac{V}{Z}$.
When the frequency changes to $\omega' = \frac{\omega}{3}$,the new reactance is $X_c' = \frac{1}{\omega' C} = \frac{1}{(\omega/3)C} = 3X_c$.
The new impedance is $Z' = \sqrt{R^2 + (3X_c)^2} = \sqrt{R^2 + 9X_c^2}$.
The new current is $I' = \frac{V}{Z'} = \frac{I}{2}$,which implies $Z' = 2Z$.
Squaring both sides,$Z'^2 = 4Z^2$,so $R^2 + 9X_c^2 = 4(R^2 + X_c^2)$.
$R^2 + 9X_c^2 = 4R^2 + 4X_c^2$.
$5X_c^2 = 3R^2$.
$\frac{X_c^2}{R^2} = \frac{3}{5}$.
Therefore,the ratio $\frac{X_c}{R} = \sqrt{\frac{3}{5}}$.

(B) In a series $RL$ 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 $\phi = 45^{\circ}$ and the resistance $R = 40 \ \Omega$.
Substituting these values into the formula:
$\tan 45^{\circ} = \frac{X_L}{40}$.
Since $\tan 45^{\circ} = 1$,we have:
$1 = \frac{X_L}{40}$.
Therefore,the inductive reactance $X_L = 40 \ \Omega$.

(C) In an $LCR$ series circuit,the voltage across the inductor $V_L$ and the voltage across the capacitor $V_C$ are $180^{\circ}$ out of phase with each other.
Given that $V_L = 60 \,V$ and $V_C = 60 \,V$.
The resultant voltage across the $LC$ combination is given by $V_{LC} = |V_L - V_C|$.
Substituting the values,we get $V_{LC} = |60 \,V - 60 \,V| = 0 \,V$.
Therefore,the voltage across the $LC$ combination is $0 \,V$.

(A) 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 capacitor $(V_C)$ lags behind the current $I$ by $\frac{\pi}{2}$.
$3$. The potential difference across the inductor $(V_L)$ leads the current $I$ by $\frac{\pi}{2}$.
Evaluating the options:
$(A)$ The potential difference across $R$ $(V_R)$ is in phase with $I$,and the potential difference across the capacitor $(V_C)$ lags behind $I$ by $\frac{\pi}{2}$. Therefore,the phase difference between $V_R$ and $V_C$ is $\frac{\pi}{2}$. This is $CORRECT$.
$(B)$ The applied e.m.f. $(V)$ leads or lags the current $I$ by a phase angle $\phi$,while $V_R$ is in phase with $I$. Thus,$V$ and $V_R$ are not in the same phase. This is $INCORRECT$.
$(C)$ The phase difference between the applied e.m.f. $(V)$ and $V_L$ is $(\frac{\pi}{2} - \phi)$. This is $INCORRECT$.
$(D)$ $V_L$ leads $I$ by $\frac{\pi}{2}$ and $V_C$ lags $I$ by $\frac{\pi}{2}$. Thus,the phase difference between $V_L$ and $V_C$ is $\pi$. This is $INCORRECT$.

(A) The instantaneous e.m.f. is given by $E = E_{0} \sin \omega t$.
The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi$ is the power factor.
Here,$V_{rms} = \frac{E_{0}}{\sqrt{2}}$ and $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_{0}}{\sqrt{2} Z}$.
The power factor is $\cos \phi = \frac{R}{Z}$.
Thus,$P = \left( \frac{E_{0}}{\sqrt{2}} \right) \left( \frac{E_{0}}{\sqrt{2} Z} \right) \left( \frac{R}{Z} \right) = \frac{E_{0}^{2} R}{2 Z^{2}}$.
For an $LR$ series circuit,the impedance $Z$ is given by $Z = \sqrt{R^{2} + X_{L}^{2}}$.
Given $X_{L} = R$,we have $Z = \sqrt{R^{2} + R^{2}} = \sqrt{2 R^{2}} = R \sqrt{2}$.
Therefore,$Z^{2} = 2 R^{2}$.
Substituting $Z^{2}$ into the power equation: $P = \frac{E_{0}^{2} R}{2 (2 R^{2})} = \frac{E_{0}^{2} R}{4 R^{2}} = \frac{E_{0}^{2}}{4 R}$.

(A) The impedance $Z$ of a series $LCR$ circuit is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.
Step $1$: Calculate inductive reactance $X_L$.
$X_L = \omega L = 1000 \times 0.9 = 900 \Omega$.
Step $2$: Calculate capacitive reactance $X_C$.
$X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega$.
Step $3$: Calculate the impedance $Z$.
$Z = \sqrt{300^2 + (900 - 500)^2}$
$Z = \sqrt{300^2 + 400^2}$
$Z = \sqrt{90000 + 160000} = \sqrt{250000}$
$Z = 500 \Omega$.

(B) The brightness of the lamp depends on the current flowing through the circuit.
In an $RC$ series circuit,the capacitive reactance is given by $X_{C} = \frac{1}{\omega C}$.
When the capacitance $C$ is reduced,the capacitive reactance $X_{C}$ increases.
The total impedance of the circuit is $Z = \sqrt{R^2 + X_{C}^2}$.
As $X_{C}$ increases,the total impedance $Z$ of the circuit increases.
According to Ohm's law for $AC$ circuits,the current $I = \frac{V}{Z}$.
Since $Z$ increases,the current $I$ flowing through the lamp decreases.
Therefore,the lamp shines less brightly.