RL, RC and LC AC Circuits Questions in English

(A) The given $\text{emf}$ is $E = E_{0} \cos(\omega t)$,where $E_{0} = 4 \text{ V}$ and $\omega = 1000 \text{ rad/s}$.
The impedance $Z$ of an $LR$ circuit is given by $Z = \sqrt{R^{2} + X_{L}^{2}}$,where $X_{L} = \omega L$.
Given $R = 4 \text{ } \Omega$,$L = 3 \text{ mH} = 3 \times 10^{-3} \text{ H}$,and $\omega = 1000 \text{ rad/s}$.
First,calculate the inductive reactance $X_{L} = \omega L = 1000 \times 3 \times 10^{-3} = 3 \text{ } \Omega$.
Now,calculate the impedance $Z = \sqrt{4^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ } \Omega$.
The amplitude of the current $i_{0}$ is given by $i_{0} = \frac{E_{0}}{Z} = \frac{4}{5} = 0.8 \text{ A}$.

(C) Given that the inductive reactance $X_L = \sqrt{3} R$.
The impedance $Z$ of the $L-R$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$.
Substituting $X_L = \sqrt{3} R$,we get $Z = \sqrt{R^2 + (\sqrt{3} R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.
The peak current in the circuit is $I_0 = \frac{E_0}{Z} = \frac{E_0}{2R}$.
The average power consumed in an $AC$ circuit is given by $P = I_{rms}^2 R = \left(\frac{I_0}{\sqrt{2}}\right)^2 R$.
Substituting $I_0 = \frac{E_0}{2R}$,we get $P = \frac{1}{2} \left(\frac{E_0}{2R}\right)^2 R = \frac{1}{2} \cdot \frac{E_0^2}{4R^2} \cdot R = \frac{E_0^2}{8R}$.

(B) Given: Resistance $R = 450 \Omega$,Inductance $L = 1.5 \text{ H}$,Frequency $f = \frac{150}{\pi} \text{ Hz}$.
The inductive reactance $X_L$ is given by $X_L = \omega L = 2\pi f L$.
Substituting the values: $X_L = 2 \pi \times \left(\frac{150}{\pi}\right) \times 1.5 = 2 \times 150 \times 1.5 = 300 \times 1.5 = 450 \Omega$.
The phase difference $\phi$ between voltage and current in an $RL$ circuit is given by $\tan \phi = \frac{X_L}{R}$.
Substituting the values: $\tan \phi = \frac{450}{450} = 1$.
Therefore,$\phi = \tan^{-1}(1)$.

(D) Given: Resistance $R = 200 \ \Omega$,Inductance $L = \frac{1}{2 \pi} \ H$,Frequency $f = 100 \ Hz$.
The inductive reactance is given by $X_L = 2 \pi f L$.
Substituting the values: $X_L = 2 \pi \times 100 \times \frac{1}{2 \pi} = 100 \ \Omega$.
The phase angle $\phi$ in an $RL$ series circuit is given by $\tan \phi = \frac{X_L}{R}$.
Substituting the values: $\tan \phi = \frac{100}{200} = \frac{1}{2} = 0.5$.
Therefore,the phase angle is $\phi = \tan^{-1}(0.5)$.

(B) The given values are: Inductance $L = \frac{1}{\pi} \ H$,Resistance $R = 300 \ \Omega$,Frequency $f = 200 \ Hz$.
First,calculate the inductive reactance $X_L = 2\pi f L$.
$X_L = 2 \times \pi \times 200 \times \frac{1}{\pi} = 400 \ \Omega$.
The phase difference $\phi$ in an $LR$ series circuit is given by $\tan \phi = \frac{X_L}{R}$.
Substituting the values,$\tan \phi = \frac{400}{300} = \frac{4}{3}$.
Therefore,$\phi = \tan^{-1}\left(\frac{4}{3}\right)$.

(A) In an $L-R$ 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 $\phi = 45^{\circ}$ and $\tan 45^{\circ} = 1$.
Substituting these values into the formula: $1 = \frac{X_L}{R}$.
Therefore,$X_L = R$.
Thus,the inductive reactance is equal to the resistance $R$.

(D) In an $LR$ series circuit, the phase angle $\phi$ between the current and the potential difference is given by the formula: $\tan \phi = \frac{X_L}{R}$.
Given: Resistance $R = 12 \ \Omega$ and inductive reactance $X_L = 5 \ \Omega$.
Substituting the values: $\tan \phi = \frac{5}{12}$.
This implies $\phi = \tan^{-1}\left(\frac{5}{12}\right)$.
To express this in terms of $\sin$ or $\cos$, we consider a right-angled triangle with opposite side $5$ and adjacent side $12$. The hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
Therefore, $\sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}$, which means $\phi = \sin^{-1}\left(\frac{5}{13}\right)$.
Also, $\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}$, which means $\phi = \cos^{-1}\left(\frac{12}{13}\right)$.
Comparing this with the given options, the correct option is $\cos^{-1}\left(\frac{12}{13}\right)$.

(B) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$.
Given $X_L = 2R$,we have $Z^2 = R^2 + (2R)^2 = R^2 + 4R^2 = 5R^2$.
The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$ is the power factor.
Since $V_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{V_{rms}}{Z} = \frac{E_0}{\sqrt{2}Z}$,the power is $P = \frac{E_0}{\sqrt{2}} \times \frac{E_0}{\sqrt{2}Z} \times \frac{R}{Z} = \frac{E_0^2 R}{2 Z^2}$.
Substituting $Z^2 = 5R^2$,we get $P = \frac{E_0^2 R}{2(5R^2)} = \frac{E_0^2 R}{10 R^2} = \frac{E_0^2}{10 R}$.

(C) 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:
$P = \left( \frac{E_0}{\sqrt{2}} \right) \left( \frac{E_0}{\sqrt{2} Z} \right) \left( \frac{R}{Z} \right) = \frac{E_0^2 R}{2 Z^2} \dots (i)$.
Given that the inductive reactance $X_L = R$,the impedance $Z$ of the $L-R$ circuit is:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = \sqrt{2} R$.
Substituting $Z^2 = 2 R^2$ into equation $(i)$:
$P = \frac{E_0^2 R}{2 (2 R^2)} = \frac{E_0^2 R}{4 R^2} = \frac{E_0^2}{4 R}$.

(A) The capacitive reactance is given by $X_{C} = \frac{1}{2 \pi f C}$.
If the capacitance $(C)$ is increased,the capacitive reactance $(X_{C})$ will decrease.
The total impedance $(Z)$ of the circuit is given by $Z = \sqrt{R^2 + X_{C}^2}$.
As $X_{C}$ decreases,the total impedance $(Z)$ of the circuit decreases.
According to Ohm's law for $A.C.$ circuits,the current $(I = V/Z)$ will increase.
Since the brightness of the lamp depends on the power dissipated $(P = I^2 R)$,an increase in current leads to an increase in the brightness of the lamp.

(A) The circuit is an $LCR$ series circuit with $X_L = 200 \, \Omega$, $X_C = 100 \, \Omega$, $R = 100 \, \Omega$, and $V_{rms} = 200 \sqrt{2} \, V$.
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 r.m.s. current $i_{rms}$ is given by:
$i_{rms} = \frac{V_{rms}}{Z}$
$i_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}} = 2 \, A$
Thus, the r.m.s. value of the current is $2 \, A$.

(C) Given data: $E = 15 \, V$, $f = 50 \, Hz$, $I = 0.5 \, A$, $\phi = \frac{\pi}{3} \, rad$.
The impedance of the circuit is given by $Z = \frac{E}{I} = \frac{15}{0.5} = 30 \, \Omega$.
The phase difference $\phi$ in an $LR$ circuit is given by $\tan \phi = \frac{X_L}{R}$.
Substituting the values: $\tan \frac{\pi}{3} = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{R} \implies X_L = \sqrt{3}R$.
The impedance formula is $Z = \sqrt{R^2 + X_L^2}$.
Substituting $X_L = \sqrt{3}R$: $Z = \sqrt{R^2 + (\sqrt{3}R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.
Since $Z = 30 \, \Omega$, we have $2R = 30 \, \Omega$.
Therefore, $R = \frac{30}{2} = 15 \, \Omega$.

(D) Given: $E_v = 15 \, V$, $I = 0.3 \, A$, $\phi = \frac{\pi}{3} \, rad$.
The impedance of the $LR$ circuit is given by $Z = \frac{E_v}{I} = \frac{15}{0.3} = 50 \, \Omega$.
The phase angle $\phi$ in an $LR$ circuit is given by $\cos \phi = \frac{R}{Z}$.
Substituting the values: $\cos(\frac{\pi}{3}) = \frac{R}{50}$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, we have $\frac{1}{2} = \frac{R}{50}$.
Therefore, $R = \frac{50}{2} = 25 \, \Omega$.

(D) When a d.c. voltage is applied, the coil acts as a pure resistor because the inductive reactance $X_L = 2 \pi f L = 0$ for $f = 0$.
$R = \frac{V}{I_{dc}} = \frac{200}{1} = 200 \, \Omega$
When an a.c. voltage is applied, the impedance $Z$ of the $LR$ circuit is given by:
$Z = \frac{V}{I_{ac}} = \frac{200}{0.5} = 400 \, \Omega$
The impedance is related to resistance and inductive reactance by $Z^2 = R^2 + X_L^2$.
$(400)^2 = (200)^2 + X_L^2$
$160000 = 40000 + X_L^2$
$X_L^2 = 120000$
$X_L = \sqrt{120000} = 200 \sqrt{3} \, \Omega$
Since $X_L = 2 \pi f L$, we have:
$2 \pi f \left( \frac{2 \sqrt{3}}{\pi} \right) = 200 \sqrt{3}$
$4 \sqrt{3} f = 200 \sqrt{3}$
$f = \frac{200}{4} = 50 \, Hz$

(C) Given: Resistance $R = 12 \ \Omega$ and inductive reactance $X_L = 5 \ \Omega$.
In an $L-R$ series circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2}$.
Substituting the values: $Z = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \ \Omega$.
The phase angle $\phi$ between the current and the potential difference is given by $\tan \phi = \frac{X_L}{R}$.
Therefore,$\tan \phi = \frac{5}{12}$,which implies $\phi = \tan^{-1}\left(\frac{5}{12}\right)$.
Alternatively,using $\cos \phi = \frac{R}{Z} = \frac{12}{13}$,we get $\phi = \cos^{-1}\left(\frac{12}{13}\right)$.
Since the provided options in the original question were inconsistent with standard calculations,the correct expression is $\tan^{-1}\left(\frac{5}{12}\right)$ or $\cos^{-1}\left(\frac{12}{13}\right)$. Based on the provided choices,option $C$ is the most accurate representation.

(C) The power factor of an $L-R$ series circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.
Given $\cos \phi = \frac{\sqrt{3}}{2}$,this implies $\phi = \frac{\pi}{6}$.
The phase angle is also given by $\tan \phi = \frac{X_L}{R}$.
Since $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$,we have $\frac{X_L}{R} = \frac{1}{\sqrt{3}}$.
Substituting $R = 50 \ \Omega$,we get $X_L = \frac{50}{\sqrt{3}} \ \Omega$.
We know $X_L = 2 \pi f L$. Given $f = 50 \ Hz$,we have $\frac{50}{\sqrt{3}} = 2 \pi (50) L$.
Solving for $L$: $L = \frac{50}{\sqrt{3} \times 100 \pi} = \frac{1}{2 \sqrt{3} \pi} \ H$.

(B) The power factor of a $CR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.
Given $\cos \phi_1 = \frac{1}{\sqrt{2}}$,this implies $\frac{R}{Z} = \frac{1}{\sqrt{2}}$,so $R = X_{C1} = \frac{1}{2\pi f_1 C}$.
When the frequency is halved,$f_2 = \frac{f_1}{2}$.
The new capacitive reactance is $X_{C2} = \frac{1}{2\pi f_2 C} = \frac{1}{2\pi (f_1/2) C} = 2 X_{C1} = 2R$.
The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + X_{C2}^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

(B) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.
Given $X_L = R$,we have $P_1 = \frac{R}{\sqrt{R^2 + R^2}} = \frac{R}{\sqrt{2R^2}} = \frac{1}{\sqrt{2}}$.
When a capacitor $C$ is added such that $X_C = X_L$,the circuit becomes a series $LCR$ circuit at resonance.
The impedance of the $LCR$ circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $X_L = X_C$,the impedance $Z = \sqrt{R^2 + 0} = R$.
The new power factor is $P_2 = \frac{R}{Z} = \frac{R}{R} = 1$.
Therefore,the ratio $P_1 : P_2 = \frac{1}{\sqrt{2}} : 1 = 1 : \sqrt{2}$.

(B) The power factor of an $R-L$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.
Given $\cos \phi = \frac{1}{\sqrt{2}}$,we have $\frac{R}{\sqrt{R^2 + X_L^2}} = \frac{1}{\sqrt{2}}$.
Squaring both sides,$\frac{R^2}{R^2 + X_L^2} = \frac{1}{2}$,which implies $2R^2 = R^2 + X_L^2$,so $R^2 = X_L^2$ or $X_L = R$.
Since $X_L = \omega L = 2\pi f L$,if the frequency $f$ is doubled,the new inductive reactance $X_L'$ becomes $2X_L = 2R$.
The new power factor is $\cos \phi' = \frac{R}{\sqrt{R^2 + (X_L')^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

(D) The inductive reactance $X_L$ is given by $X_L = L\omega = L(2\pi f)$.
Substituting the given values: $X_L = \frac{1}{\pi} \times 2\pi \times 200 = 400 \text{ } \Omega$.
The phase angle $\phi$ in an $LR$ series circuit is given by $\tan \phi = \frac{X_L}{R}$.
Substituting the values: $\tan \phi = \frac{400}{300} = \frac{4}{3}$.
Therefore,$\phi = \tan^{-1}\left(\frac{4}{3}\right)$.

(A) Given: Inductance $L = 25.48 \ mH = 25.48 \times 10^{-3} \ H$,Resistance $R = 8 \ \Omega$,Frequency $\nu = 50 \ Hz$.
First,calculate the inductive reactance $X_L$:
$X_L = \omega L = 2 \pi \nu L$
$X_L = 2 \times 3.14 \times 50 \times 25.48 \times 10^{-3}$
$X_L = 314 \times 25.48 \times 10^{-3} \approx 8 \ \Omega$.
The phase difference $\phi$ in an $LR$ series circuit is given by:
$\tan \phi = \frac{X_L}{R}$
$\tan \phi = \frac{8}{8} = 1$
$\phi = \tan^{-1}(1) = 45^{\circ}$.

(B) In an $LC$ series circuit,the net reactance is $X = X_{L} - X_{C}$.
Given that $X_{C} > X_{L}$,the net reactance $X$ is negative,meaning the circuit is capacitive in nature.
In a purely capacitive circuit,the voltage lags behind the current by a phase angle of $\frac{\pi}{2}$.
Therefore,the potential lags behind the current by $\frac{\pi}{2}$ in phase.

(B) The given $AC$ voltage is $V = V_m \cos(\omega t)$, where $V_m = 5 \text{ V}$ and $\omega = 1000 \text{ rad/s}$.
For an $L-R$ circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2}$, where $X_L = \omega L$.
Given $L = 3 \text{ mH} = 3 \times 10^{-3} \text{ H}$ and $R = 4 \text{ } \Omega$.
Calculating inductive reactance: $X_L = 1000 \times 3 \times 10^{-3} = 3 \text{ } \Omega$.
Calculating impedance: $Z = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ } \Omega$.
The maximum current $I_m$ is given by $I_m = \frac{V_m}{Z}$.
Substituting the values: $I_m = \frac{5}{5} = 1 \text{ A}$.

(D) For an $RL$ series circuit,the phase difference $\phi$ between voltage and current is given by the formula: $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.
Given: Frequency $\nu = \frac{25}{\pi} \text{ Hz}$,Resistance $R = 100 \ \Omega$,Inductance $L = 2 \text{ H}$.
Since $\omega = 2 \pi \nu$,we have:
$\tan \phi = \frac{2 \pi \nu L}{R}$
$\tan \phi = \frac{2 \pi \times (\frac{25}{\pi}) \times 2}{100}$
$\tan \phi = \frac{2 \times 25 \times 2}{100} = \frac{100}{100} = 1$.
Therefore,$\phi = \tan^{-1}(1) = 45^{\circ}$.

(B) The power consumed in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
Given: $V_{m} = 100 \ V$,$I_{m} = 1.1 \ A$,$X_{C} = 60 \ \Omega$,$R = 80 \ \Omega$.
The impedance $Z$ is calculated as $Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{80^{2} + 60^{2}} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \ \Omega$.
The power factor is $\cos \phi = \frac{R}{Z} = \frac{80}{100} = 0.8$.
The $rms$ values are $V_{rms} = \frac{V_{m}}{\sqrt{2}}$ and $I_{rms} = \frac{I_{m}}{\sqrt{2}}$.
Substituting these into the power formula:
$P = \left(\frac{100}{\sqrt{2}}\right) \times \left(\frac{1.1}{\sqrt{2}}\right) \times 0.8$
$P = \frac{100 \times 1.1}{2} \times 0.8$
$P = 50 \times 1.1 \times 0.8 = 55 \times 0.8 = 44.0 \ W$.

(C) The phase difference $\phi$ in an $RL$ series circuit is given by the formula: $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.
Given:
Resistance $R = 1 \Omega$
Inductance $L = 2.5 \text{ mH} = 2.5 \times 10^{-3} \text{ H}$
Frequency $\nu = 50 \text{ Hz}$
Angular frequency $\omega = 2 \pi \nu = 2 \pi \times 50 = 100 \pi \text{ rad/s}$.
Substituting the values:
$\tan \phi = \frac{100 \pi \times 2.5 \times 10^{-3}}{1}$
$\tan \phi = 100 \times 2.5 \times 10^{-3} \times \pi$
$\tan \phi = 0.25 \pi = \frac{\pi}{4}$.
Therefore,$\phi = \tan^{-1} \left( \frac{\pi}{4} \right)$.

(A) Given: Self-inductance $L = 0.04 \ H$,Resistance $R = 12 \ \Omega$,Voltage $V = 220 \ V$,Frequency $f = 50 \ Hz$.
First,calculate the inductive reactance $X_L$:
$X_L = \omega L = 2 \pi f L$
$X_L = 2 \times 3.14 \times 50 \times 0.04 = 12.56 \ \Omega$.
Next,calculate the impedance $Z$ of the $RL$ circuit:
$Z = \sqrt{R^2 + X_L^2}$
$Z = \sqrt{12^2 + 12.56^2} = \sqrt{144 + 157.75} = \sqrt{301.75} \approx 17.37 \ \Omega$.
Finally,calculate the current $I$ using Ohm's law for $AC$ circuits:
$I = \frac{V}{Z} = \frac{220}{17.37} \approx 12.66 \ A \approx 12.7 \ A$.

(C) In an $RL$ series circuit,the phase angle $\phi$ between the voltage and the current is given by the formula: $\tan \phi = \frac{X_L}{R}$,where $X_L = \omega L$ is the inductive reactance.
Given that the phase angle $\phi = 45^{\circ}$.
Substituting the value into the formula: $\tan 45^{\circ} = \frac{X_L}{R}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{X_L}{R}$.
Therefore,$X_L = R$.

(D) The impedance $Z$ of an $LR$ series circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L$.
Thus,$Z = \sqrt{R^2 + \omega^2 L^2}$.
The $RMS$ current $I$ in the circuit is $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + \omega^2 L^2}}$.
The power dissipated in the circuit is given by $P = I^2 R$.
Substituting the value of $I$,we get $P = \left( \frac{V}{\sqrt{R^2 + \omega^2 L^2}} \right)^2 R = \frac{V^2 R}{R^2 + \omega^2 L^2}$.

(D) Given, voltage drop across $L, V_L = 40 \,V$; voltage drop across $C, V_C = 120 \,V$; voltage drop across $R, V_R = 60 \,V$.
The source voltage $V$ in a series $LCR$ circuit is given by the formula:
$V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Substituting the given values:
$V = \sqrt{60^2 + (40 - 120)^2}$
$V = \sqrt{3600 + (-80)^2}$
$V = \sqrt{3600 + 6400}$
$V = \sqrt{10000}$
$V = 100 \,V$
Therefore, the source voltage is $100 \,V$.

(A) Resistance of the resistor: $R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.
Inductive reactance of the choke at $50 \, Hz$: $X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.
Since $X_L = 2 \pi f L$, we have $12.5 = 2 \pi \times 50 \times L$, which gives $L = \frac{12.5}{100 \pi} = \frac{1}{8 \pi} \, H$.
Now, for the new source of $40 \, Hz$, the new inductive reactance is $X_L' = 2 \pi f' L = 2 \pi \times 40 \times \frac{1}{8 \pi} = 10 \, \Omega$.
The impedance of the series $RL$ circuit is $Z = \sqrt{R^2 + X_L'^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.
The current in the circuit is $I = \frac{V'}{Z} = \frac{150}{10\sqrt{2}} = \frac{15}{\sqrt{2}} \, A$.

(A) In an $RC$ series circuit, the voltage across the resistor $(V_R)$ and the voltage across the capacitor $(V_C)$ are out of phase by $90^{\circ}$.
Given: $V_R = 12 \,V$ and $V_C = 5 \,V$.
The total applied voltage $V$ is given by the phasor sum:
$V = \sqrt{V_R^2 + V_C^2}$
Substituting the values:
$V = \sqrt{(12)^2 + (5)^2}$
$V = \sqrt{144 + 25}$
$V = \sqrt{169}$
$V = 13 \,V$
Thus, the applied voltage is $13 \,V$.

(B) Given voltage $V = 100 \sqrt{2} \sin(1000 t)$.
Comparing this with the standard equation $V = V_0 \sin(\omega t)$,we get angular frequency $\omega = 1000 \ rad/s$.
The inductive reactance is given by $X_L = \omega L = 1000 \times 0.5 = 500 \ \Omega$.
The resistance is $R = 500 \ \Omega$.
The impedance of the $RL$ series circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{500^2 + 500^2} = 500 \sqrt{2} \ \Omega$.
The power factor is defined as $\cos \phi = \frac{R}{Z}$.
Substituting the values,$\cos \phi = \frac{500}{500 \sqrt{2}} = \frac{1}{\sqrt{2}}$.

(B) Given, $X_{L} = 4 \, \Omega$ and $X_{C} = 4 \, \Omega$.
In a series $LC$ circuit, the voltage across the inductor $(V_{L})$ and the capacitor $(V_{C})$ are $180^{\circ}$ out of phase.
Therefore, the net voltage across the series combination of the inductor and capacitor is $V_{\text{net}} = |V_{L} - V_{C}|$.
Since $X_{L} = X_{C}$, the magnitudes of the voltages are equal, i.e., $V_{L} = I X_{L}$ and $V_{C} = I X_{C}$.
Thus, $V_{\text{net}} = I(X_{L} - X_{C}) = I(4 - 4) = 0 \, V$.
The voltmeter is connected across this series $LC$ combination, so its reading is $0 \, V$.
The total impedance of the circuit is $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.
Substituting the values, $Z = \sqrt{45^{2} + (4 - 4)^{2}} = \sqrt{45^{2}} = 45 \, \Omega$.
The current in the circuit is $I = \frac{V}{Z} = \frac{90 \, V}{45 \, \Omega} = 2 \, A$.
Therefore, the voltmeter reading is $0 \, V$ and the ammeter reading is $2 \, A$.

(C) The resistance of the bulb is given by $R = \frac{V^2}{P} = \frac{100^2}{50} = 200 \, \Omega$.
For the bulb to operate at its rated power, the current $I$ must be $I = \frac{P}{V} = \frac{50}{100} = 0.5 \, A$.
When connected to a $200 \, V$ $AC$ source, the total impedance $Z$ of the circuit is $Z = \frac{V_{source}}{I} = \frac{200}{0.5} = 400 \, \Omega$.
The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = 2\pi fL$.
Substituting the values: $400 = \sqrt{200^2 + X_L^2}$.
$160000 = 40000 + X_L^2 \implies X_L^2 = 120000$.
$X_L = \sqrt{120000} = 200\sqrt{3} \, \Omega$.
Since $X_L = 2\pi fL$, we have $200\sqrt{3} = 2 \times \pi \times 50 \times L$.
$L = \frac{200\sqrt{3}}{100\pi} = \frac{2\sqrt{3}}{\pi} \approx \frac{2 \times 1.732}{3.14} \approx 1.1 \, H$.

(B) When an iron rod is inserted into the coil,the permeability of the core increases,which significantly increases the self-inductance $(L)$ of the coil.
The inductive reactance of the circuit is given by $X_L = \omega L$.
As $L$ increases,the inductive reactance $X_L$ increases.
The total impedance of the 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.
According to Ohm's law for $A$.$C$. circuits,the current $I = V / Z$. As the impedance $Z$ increases,the current $I$ flowing through the circuit decreases.
The brightness of the bulb depends on the power dissipated,given by $P = I^2 R$. Since the current $I$ decreases,the power dissipated in the bulb decreases.
Therefore,the bulb glows less brightly.

(B) Given: Inductive reactance $ X_L = \omega L = \frac{1}{\sqrt{3}} \ \Omega $,Resistance $ R = 1 \ \Omega $,Frequency $ f = 50 \ Hz $.
In an $ RL $ circuit,the phase difference $ \phi $ between voltage and current is given by $ \tan \phi = \frac{X_L}{R} $.
Substituting the values,$ \tan \phi = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}} $.
Thus,$ \phi = 30^{\circ} = \frac{\pi}{6} \text{ radians} $.
The phase difference $ \phi $ is related to time lag $ t $ by the relation $ \phi = \omega t $.
Since $ \omega = 2\pi f = 2\pi \times 50 = 100\pi \ rad/s $.
Therefore,$ t = \frac{\phi}{\omega} = \frac{\pi/6}{100\pi} = \frac{1}{600} \ s $.
The time lag between maximum voltage and current is $ \frac{1}{600} \ s $.

(B) Given: Inductive reactance $X_L = \frac{1}{\sqrt{3}} \ \Omega$,Resistance $R = 1 \ \Omega$,Frequency $f = 50 \ Hz$.
In an $LR$ series circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_L}{R}$.
$\tan \phi = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$.
Therefore,$\phi = 30^{\circ} = \frac{\pi}{6} \ \text{radians}$.
The phase difference $\phi$ is related to the time lag $\Delta t$ by the formula $\phi = \omega \Delta t$,where $\omega = 2\pi f$.
$\omega = 2 \times \pi \times 50 = 100\pi \ \text{rad/s}$.
Substituting the values: $\frac{\pi}{6} = 100\pi \times \Delta t$.
$\Delta t = \frac{\pi}{6 \times 100\pi} = \frac{1}{600} \ s$.
Thus,the time lag is $\frac{1}{600} \ s$.

(D) Let the voltages across the capacitor,resistor,and inductor be $V_C = 2x$,$V_R = 3x$,and $V_L = 6x$ respectively.
In a series $LCR$ circuit,the total voltage $V$ is given by the relation $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Given $V = 240 \ V$,we substitute the values:
$240 = \sqrt{(3x)^2 + (6x - 2x)^2}$
$240 = \sqrt{9x^2 + (4x)^2}$
$240 = \sqrt{9x^2 + 16x^2}$
$240 = \sqrt{25x^2}$
$240 = 5x$
$x = \frac{240}{5} = 48 \ V$.
The voltage across the inductor is $V_L = 6x = 6 \times 48 = 288 \ V$.

(C) Given voltage $V = 144 \sin \left(100 \pi t + \frac{\pi}{2}\right) \text{ V}$ and current $I = 6 \sin \left(100 \pi t + \frac{\pi}{6}\right) \text{ A}$.
Comparing with standard forms $V = V_0 \sin(\omega t + \phi_V)$ and $I = I_0 \sin(\omega t + \phi_I)$,we get $V_0 = 144 \text{ V}$,$I_0 = 6 \text{ A}$,and phase difference $\phi = \phi_V - \phi_I = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} = 60^\circ$.
In an $LR$ series circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_L}{R}$.
$\tan(60^\circ) = \frac{X_L}{R} \Rightarrow \sqrt{3} = \frac{X_L}{R} \Rightarrow X_L = \sqrt{3} R$.
The impedance $Z$ is given by $Z = \frac{V_0}{I_0} = \frac{144}{6} = 24 \ \Omega$.
Also,$Z = \sqrt{R^2 + X_L^2}$.
Substituting $X_L = \sqrt{3} R$,we get $24 = \sqrt{R^2 + (\sqrt{3} R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.
Therefore,$R = \frac{24}{2} = 12 \ \Omega$.

(D) The impedance $Z$ of an $LR$ circuit is given by the formula: $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L = 2 \pi f L$.
Given: $R = 8 \Omega$,$L = \frac{60}{\pi} \text{ mH} = \frac{60}{\pi} \times 10^{-3} \text{ H}$,and $f = 50 \text{ Hz}$.
First,calculate the inductive reactance $X_L$:
$X_L = 2 \pi \times 50 \times \left( \frac{60}{\pi} \times 10^{-3} \right) = 100 \pi \times \frac{60}{\pi} \times 10^{-3} = 6000 \times 10^{-3} = 6 \Omega$.
Now,calculate the impedance $Z$:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \Omega$.

(B) In a series $L-C-R$ circuit,the total voltage $V$ is given by the phasor sum of the voltages across the resistor,inductor,and capacitor.
The phase angle $\phi$ between the voltage and current is given by $\phi = \tan^{-1} \left( \frac{X_L - X_C}{R} \right)$.
If $X_L > X_C$,then $\phi$ is positive,which means the voltage leads the current.
If $X_L < X_C$,then $\phi$ is negative,which means the current leads the voltage.
Therefore,the voltage leads the current when $X_L > X_C$.

(C) Given for the $LR$ series $AC$ circuit:
Voltage, $V_{rms} = 12 \, V$
Frequency, $f = 50 \, Hz$
Current, $I = 0.5 \, A$
Phase difference, $\phi = \frac{\pi}{3}$
First, calculate the total impedance $Z$ of the circuit using Ohm's law for $AC$ circuits:
$Z = \frac{V_{rms}}{I} = \frac{12}{0.5} = 24 \, \Omega$
The power factor of an $LR$ series circuit is given by:
$\cos \phi = \frac{R}{Z}$
Substituting the known values:
$\cos(\frac{\pi}{3}) = \frac{R}{24}$
Since $\cos(\frac{\pi}{3}) = 0.5$:
$0.5 = \frac{R}{24}$
$R = 24 \times 0.5 = 12 \, \Omega$
Therefore, the value of $R$ is $12 \, \Omega$.

(C) Given: $L = 10 \mu H = 10 \times 10^{-6} H$,$C = 1 \mu F = 10^{-6} F$,$R = 10 \Omega$,and angular frequency $\omega = 70 \times 10^3 \text{ rad s}^{-1}$.
First,calculate the inductive reactance $X_L = \omega L = (70 \times 10^3) \times (10 \times 10^{-6}) = 0.7 \Omega$.
Next,calculate the capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{(70 \times 10^3) \times (10^{-6})} = \frac{1}{0.07} \approx 14.29 \Omega$.
Comparing the two,we find that $X_C > X_L$.
Since the capacitive reactance is significantly greater than the inductive reactance,the circuit behaves like a series $RC$ circuit.

(D) For an $L-R$ circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (L\omega)^2}$. The current is $I = V/Z$, so $R^2 + L^2\omega^2 = (V/I)^2$.
Case $1$: $I_1 = 0.5 \,A$, $f_1 = 175/\pi \,Hz$, $\omega_1 = 2\pi f_1 = 350 \,rad/s$, $V = 8\sqrt{2} \,V$.
$R^2 + L^2(350)^2 = (8\sqrt{2} / 0.5)^2 = (16\sqrt{2})^2 = 256 \times 2 = 512$.
Case $2$: $I_2 = 0.4 \,A$, $f_2 = 225/\pi \,Hz$, $\omega_2 = 2\pi f_2 = 450 \,rad/s$, $V = 8\sqrt{2} \,V$.
$R^2 + L^2(450)^2 = (8\sqrt{2} / 0.4)^2 = (20\sqrt{2})^2 = 400 \times 2 = 800$.
Subtracting the two equations: $L^2(450^2 - 350^2) = 800 - 512 = 288$.
$L^2(202500 - 122500) = 288 \Rightarrow L^2(80000) = 288$.
$L^2 = 288 / 80000 = 0.0036 \Rightarrow L = 0.06 \,H = 60 \,mH$.
Substituting $L$ into the first equation: $R^2 + (0.06 \times 350)^2 = 512$.
$R^2 + (21)^2 = 512 \Rightarrow R^2 + 441 = 512$.
$R^2 = 71 \Rightarrow R = \sqrt{71} \,\Omega$.

(A) Given: $E = 6 \cos(6000t) \ V$,$L = 4 \ mH = 4 \times 10^{-3} \ H$,$R = 7 \ \Omega$.
Comparing $E = E_0 \cos(\omega t)$ with the given equation,we get $E_0 = 6 \ V$ and $\omega = 6000 \ rad/s$.
The inductive reactance is $X_L = \omega L = 6000 \times 4 \times 10^{-3} = 24 \ \Omega$.
The impedance of the $L-R$ circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \ \Omega$.
The amplitude of the current $I_0$ is given by $I_0 = \frac{E_0}{Z} = \frac{6}{25} = 0.24 \ A$.

(C) Given: Inductance $L = 0.2 \ H$,Resistance $R = 100 \ \Omega$,Voltage $V_{rms} = 180 \ V$,Frequency $f = 50 \ Hz$.
First,calculate the inductive reactance $X_L = 2 \pi f L$.
$X_L = 2 \times \pi \times 50 \times 0.2 = 20 \pi \ \Omega$.
Using $\pi^2 = 10$,we approximate $\pi \approx \sqrt{10} \approx 3.162$.
So,$X_L = 20 \times 3.162 = 63.24 \ \Omega$.
The impedance $Z$ of the $RL$ series circuit is $Z = \sqrt{R^2 + X_L^2}$.
$Z = \sqrt{100^2 + (63.24)^2} = \sqrt{10000 + 3999.3} = \sqrt{13999.3} \approx 118.32 \ \Omega$.
The $RMS$ current $I_{rms} = \frac{V_{rms}}{Z}$.
$I_{rms} = \frac{180}{118.32} \approx 1.5213 \ A$.
Rounding to three decimal places,we get $1.522 \ A$.

(A) For bulb $A$ connected in series with an inductor $L = 100 \ mH$,the current is given by $I_1 = \frac{V_0}{X_L} = \frac{V_0}{\omega L} = \frac{V_0}{\omega \times 100 \times 10^{-3}} = \frac{10 V_0}{\omega}$.
For bulb $B$ connected in series with a capacitor $C = 10 \ pF$,the current is given by $I_2 = \frac{V_0}{X_C} = V_0 \omega C = V_0 \omega \times 10 \times 10^{-12}$.
Since the inductive reactance $X_L$ is typically much smaller than the capacitive reactance $X_C$ at standard frequencies,the current $I_1$ through bulb $A$ is significantly larger than the current $I_2$ through bulb $B$.
Therefore,$I_1 > I_2$,which implies that bulb $A$ will glow brighter.

(D) The inductive reactance is given by $X_L = \omega L = 2 \pi f L$.
Substituting the given values: $X_L = 2 \times \pi \times 350 \times 0.1 = 70 \pi \approx 70 \times 3.14159 = 219.9 \Omega$.
In an $RL$ series circuit,the phase difference $\phi$ between the voltage and the current is given by $\tan \phi = \frac{X_L}{R}$.
Substituting the values: $\tan \phi = \frac{219.9}{110} \approx 1.999 \approx 2$.
Therefore,the phase difference is $\phi = \tan ^{-1}(2)$.

(C) Given: Inductance $L = \frac{1}{6 \pi} \text{ H}$,Resistance $R = 15 \text{ } \Omega$,Voltage $V = 100 \text{ V}$,Frequency $f = 60 \text{ Hz}$.
First,calculate the inductive reactance $X_L$:
$X_L = L \omega = L(2 \pi f) = \left(\frac{1}{6 \pi}\right) \times 2 \pi \times 60 = 20 \text{ } \Omega$.
Next,calculate the impedance $Z$ of the $LR$ series circuit:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \text{ } \Omega$.
The current $I$ in the circuit is:
$I = \frac{V}{Z} = \frac{100}{25} = 4 \text{ A}$.
The phase difference $\phi$ between voltage and current is given by:
$\tan \phi = \frac{X_L}{R} = \frac{20}{15} = \frac{4}{3}$.
Therefore,$\phi = \tan^{-1}\left(\frac{4}{3}\right)$.