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

(C) The resonant frequency of an $LCR$ circuit is given by the formula $\omega = \frac{1}{\sqrt{LC}}$.
Since the resonant frequency remains the same,we have $\omega_1 = \omega_2$.
Therefore,$\frac{1}{\sqrt{L_1 C_1}} = \frac{1}{\sqrt{L_2 C_2}}$.
Squaring both sides,we get $L_1 C_1 = L_2 C_2$.
Given $L_1 = L$,$C_1 = C$,and $C_2 = 4C$,we substitute these values into the equation:
$L \cdot C = L_2 \cdot (4C)$.
Solving for $L_2$,we get $L_2 = \frac{L \cdot C}{4C} = \frac{L}{4}$.

(C) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given: $R = 100\, \Omega$,$L = 0.5\, H$,$C = 10 \times 10^{-6}\, F$,and $f = 50\, Hz$.
Angular frequency $\omega = 2\pi f = 2 \times \pi \times 50 = 100\pi\, rad/s$.
Inductive reactance $X_L = \omega L = 100\pi \times 0.5 = 50\pi \approx 157.08\, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 10 \times 10^{-6}} = \frac{1}{10^{-3}\pi} = \frac{1000}{\pi} \approx 318.31\, \Omega$.
Now,$Z = \sqrt{100^2 + (157.08 - 318.31)^2} = \sqrt{10000 + (-161.23)^2} = \sqrt{10000 + 25995.11} = \sqrt{35995.11} \approx 189.72\, \Omega$.

(C) The impedance $Z$ of a series $R-L-C$ circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
First,calculate the inductive reactance $X_L = \omega L = 1000 \times 0.9 = 900 \, \Omega$.
Next,calculate the capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 2.0 \times 10^{-6}} = \frac{10^6}{2000} = 500 \, \Omega$.
Now,substitute these values into the impedance formula:
$Z = \sqrt{300^2 + (900 - 500)^2}$
$Z = \sqrt{300^2 + 400^2}$
$Z = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega$.

(C) In a series $LCR$ circuit,the total applied voltage $V$ is given by the phasor sum of the voltages across the individual components:
$V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given:
$V_R = 5\, V$
$V_L = 10\, V$
$V_C = 10\, V$
Substituting these values into the formula:
$V = \sqrt{5^2 + (10 - 10)^2}$
$V = \sqrt{25 + 0^2}$
$V = \sqrt{25} = 5\, V$
Therefore,the applied $ac$ voltage is $5\, V$.

(C) 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.
$2$. The potential difference across the inductor $(V_L)$ leads the current by $\pi / 2$.
$3$. The potential difference across the capacitor $(V_C)$ lags behind the current by $\pi / 2$.
$4$. Therefore,the phase difference between $V_L$ and $V_C$ is $\pi / 2 - (-\pi / 2) = \pi$.
Thus,the statement that the potential difference across the capacitor and inductor have a phase difference of $\pi / 2$ is incorrect.

(D) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given: $R = 11\,\Omega$,$X_L = 25\,\Omega$,$X_C = 18\,\Omega$.
Substituting the values: $Z = \sqrt{11^2 + (25 - 18)^2} = \sqrt{121 + 7^2} = \sqrt{121 + 49} = \sqrt{170}$.
Wait,recalculating: $Z = \sqrt{121 + 49} = \sqrt{170} \approx 13.03\,\Omega$. Given the options,let's re-verify the values: $11^2 = 121$,$(25-18)^2 = 7^2 = 49$. $121+49 = 170$. If $R=12$,$Z = \sqrt{144+49} = \sqrt{193}$. If $R=12$ and $X_L-X_C=5$,$Z=13$. Assuming the intended calculation was $Z = \sqrt{12^2 + 5^2} = 13$,but based on the provided values $R=11, X_L=25, X_C=18$,$Z = \sqrt{170} \approx 13.03$. Using $Z \approx 13\,\Omega$,the current $I = V/Z = 260/13 = 20\,A$.

(A) In a series $LCR$ circuit,the supply voltage $V$ is given by the phasor sum of the individual potential differences across the components.
The formula is: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given:
$V_L = 60\,V$
$V_C = 30\,V$
$V_R = 40\,V$
Substituting the values:
$V = \sqrt{40^2 + (60 - 30)^2}$
$V = \sqrt{1600 + 30^2}$
$V = \sqrt{1600 + 900}$
$V = \sqrt{2500}$
$V = 50\,V$
Therefore,the supply voltage is $50\,V$.

(D) In an $LCR$ series $ac$ 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 $V_L = 50\,V$ and $V_C = 50\,V$.
The net voltage across the $LC$ combination is given by the phasor sum: $V_{LC} = |V_L - V_C|$.
Substituting the values: $V_{LC} = |50\,V - 50\,V| = 0\,V$.
Therefore,the voltage across the $LC$ combination is $0\,V$.

(C) The circuit is an $LCR$ series circuit connected to an $AC$ source of $200 \ V$.
From the diagram,the voltage across the inductor is $V_L = 100 \ V$ and the voltage across the capacitor is $V_C = 100 \ V$.
The total voltage $V_{total}$ in an $LCR$ series circuit is given by the formula:
$V_{total}^2 = V_R^2 + (V_L - V_C)^2$
Given $V_{total} = 200 \ V$,$V_L = 100 \ V$,and $V_C = 100 \ V$.
Substituting these values:
$(200)^2 = V_R^2 + (100 - 100)^2$
$40000 = V_R^2 + 0$
$V_R = 200 \ V$
The voltmeter $V$ is connected across the resistor,so its reading is equal to $V_R$.
Therefore,the reading of the voltmeter is $200 \ V$.

(D) In an $LCR$ series circuit, when the capacitor is removed, the circuit becomes an $LR$ circuit. The phase angle is given by $\tan \phi = \frac{X_L}{R}$. Given $\phi = 60^o$, we have $\tan 60^o = \frac{X_L}{R} \Rightarrow X_L = R \sqrt{3}$.
When the inductor is removed, the circuit becomes an $RC$ circuit. The phase angle is given by $\tan \phi = \frac{X_C}{R}$. Given $\phi = 60^o$, we have $\tan 60^o = \frac{X_C}{R} \Rightarrow X_C = R \sqrt{3}$.
Since $X_L = X_C$, the circuit is in resonance when both components are present.
At resonance, the impedance $Z = R = 100 \, \Omega$.
The average power dissipated is $P = \frac{V^2}{R} = \frac{200^2}{100} = \frac{40000}{100} = 400 \, W$.

(C) The circuit consists of an inductor with inductive reactance $X_L = 30 \, \Omega$ and a capacitor with capacitive reactance $X_C = 20 \, \Omega$ connected in parallel to an $AC$ source of $V = 90 \, V$.
The current through the inductor is $i_L = \frac{V}{X_L} = \frac{90}{30} = 3 \, A$.
The current through the capacitor is $i_C = \frac{V}{X_C} = \frac{90}{20} = 4.5 \, A$.
Since the current in the inductor lags the voltage by $90^\circ$ and the current in the capacitor leads the voltage by $90^\circ$,the currents are in opposite phases.
The net current through the circuit is $i = |i_C - i_L| = |4.5 - 3| = 1.5 \, A$.
The impedance of the circuit is $Z = \frac{V}{i} = \frac{90}{1.5} = 60 \, \Omega$.

(C) The impedance $(Z)$ of a series $LCR$ circuit is given by the formula:
$Z = \sqrt{R^2 + \left( 2\pi fL - \frac{1}{2\pi fC} \right)^2}$
$1$. At low frequencies $(f \to 0)$,the capacitive reactance $X_C = \frac{1}{2\pi fC}$ becomes very large,so $Z \to \infty$.
$2$. At the resonant frequency $f_r = \frac{1}{2\pi \sqrt{LC}}$,the inductive reactance $X_L = 2\pi fL$ equals the capacitive reactance $X_C$,making the term $(X_L - X_C) = 0$. Thus,the impedance reaches its minimum value,$Z = R$.
$3$. At high frequencies $(f > f_r)$,the inductive reactance $X_L = 2\pi fL$ becomes very large,causing $Z$ to increase again.
Therefore,the graph of $Z$ versus $f$ starts from a high value,decreases to a minimum at $f_r$,and then increases. This behavior is correctly represented by graph $C$.

(B) For the $RL$ circuit,the phase angle $\phi = \pi / 3$. The tangent of the phase angle is given by $\tan \phi = \frac{X_L}{R}$.
Thus,$\tan(\pi / 3) = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{100} \implies X_L = 100\sqrt{3} \ \Omega$.
For the $RC$ circuit,the phase angle $\phi = \pi / 3$. The tangent of the phase angle is given by $\tan \phi = \frac{X_C}{R}$.
Thus,$\tan(\pi / 3) = \frac{X_C}{R} \implies \sqrt{3} = \frac{X_C}{100} \implies X_C = 100\sqrt{3} \ \Omega$.
In the $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $X_L = X_C$,the term $(X_L - X_C) = 0$.
Therefore,$Z = \sqrt{R^2 + 0} = R = 100 \ \Omega$.

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

(C) The circuit consists of an inductor and a capacitor connected in parallel across an $AC$ source of $90 \; V$.
For the inductor branch,the current is $I_L = \frac{V}{X_L} = \frac{90}{30} = 3 \; A$.
For the capacitor branch,the current is $I_C = \frac{V}{X_C} = \frac{90}{20} = 4.5 \; A$.
Since the currents in the inductor and capacitor are $180^{\circ}$ out of phase,the total current $I$ drawn from the source is $I = |I_C - I_L| = |4.5 - 3| = 1.5 \; A$.
The impedance $Z$ of the parallel circuit is given by $Z = \frac{V}{I} = \frac{90}{1.5} = 60 \; \Omega$.

(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, meaning $X_C > X_L$. The phase angle is $\phi = 45^{\circ}$.
Substituting the values: $\tan 45^{\circ} = \frac{X_C - X_L}{R}$.
Since $\tan 45^{\circ} = 1$, we have $1 = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.
Rearranging the terms: $R = \frac{1}{2\pi fC} - 2\pi fL$.
$\frac{1}{2\pi fC} = 2\pi fL + R$.
Therefore, $C = \frac{1}{2\pi f(2\pi fL + R)}$.

(A) When $L$ is removed,the circuit becomes an $RC$ circuit. The phase difference $\phi_1$ is given by $\tan \phi_1 = \frac{X_C}{R}$.
Given $\phi_1 = \pi / 3$,so $\tan(\pi / 3) = \frac{X_C}{R} \implies X_C = R \sqrt{3}$.
When $C$ is removed,the circuit becomes an $RL$ circuit. The phase difference $\phi_2$ is given by $\tan \phi_2 = \frac{X_L}{R}$.
Given $\phi_2 = \pi / 3$,so $\tan(\pi / 3) = \frac{X_L}{R} \implies X_L = R \sqrt{3}$.
Since $X_L = X_C$,the circuit is in resonance.
In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting $X_L = X_C$,we get $Z = \sqrt{R^2 + 0} = R$.
The power factor is $\cos \phi = \frac{R}{Z} = \frac{R}{R} = 1$.

(D) Given: $L = 20 \, mH = 20 \times 10^{-3} \, H$,$C = 50 \, \mu F = 50 \times 10^{-6} \, F$,$R = 40 \, \Omega$,$V = 10 \, \sin \, 340t$.
Comparing with $V = V_0 \sin \omega t$,we get $V_0 = 10 \, V$ and $\omega = 340 \, rad/s$.
Inductive reactance $X_L = \omega L = 340 \times 20 \times 10^{-3} = 6.8 \, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{340 \times 50 \times 10^{-6}} = \frac{10^6}{17000} \approx 58.82 \, \Omega$.
Impedance $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{40^2 + (58.82 - 6.8)^2} = \sqrt{1600 + (52.02)^2} = \sqrt{1600 + 2706.08} = \sqrt{4306.08} \approx 65.62 \, \Omega$.
Peak current $I_0 = \frac{V_0}{Z} = \frac{10}{65.62} \, A$.
Power factor $\cos \phi = \frac{R}{Z} = \frac{40}{65.62}$.
Power loss $P = V_{rms} I_{rms} \cos \phi = \frac{1}{2} V_0 I_0 \cos \phi = \frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62} = \frac{2000}{4306} \approx 0.46 \, W$.

(B) The resonant frequency of an $LC$ circuit is given by $\nu = \frac{1}{2\pi \sqrt{LC}}$.
Since the same capacitor $C$ is used for both,we have $\nu \propto \frac{1}{\sqrt{L}}$.
Short waves have shorter wavelengths $(\lambda_s < \lambda_m)$,which implies they have higher frequencies $(\nu_s > \nu_m)$ because $\nu = \frac{c}{\lambda}$.
Since $\nu_s > \nu_m$,it follows that $\frac{1}{\sqrt{L_s}} > \frac{1}{\sqrt{L_m}}$.
Squaring both sides and rearranging,we get $L_s < L_m$.

(C) From the circuit diagram,we have resistance $R = 10 \ \Omega$,inductance $L = 0.1 \ H$,and angular frequency $\omega = 100 \ rad/s$.
The power factor is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}} = \frac{1}{\sqrt{2}}$.
This implies $Z = R\sqrt{2}$,so $R^2 + (X_L - X_C)^2 = 2R^2$,which simplifies to $(X_L - X_C)^2 = R^2$.
Thus,$X_L - X_C = \pm R$.
Given $X_L = \omega L = 100 \times 0.1 = 10 \ \Omega$.
Case $1$: $10 - X_C = 10 \implies X_C = 0$ (not possible for finite $C$).
Case $2$: $10 - X_C = -10 \implies X_C = 20 \ \Omega$.
Since $X_C = \frac{1}{\omega C}$,we have $20 = \frac{1}{100 \times C}$.
$C = \frac{1}{2000} \ F = 0.5 \times 10^{-3} \ F = 500 \ \mu F$.

(D) For the $L-R$ circuit: $\cos \phi_1 = \frac{R_1}{Z_1} = 0.6 = \frac{3}{5}$. Thus,$\sin \phi_1 = \frac{X_L}{Z_1} = 0.8 = \frac{4}{5}$. So,$X_L = \frac{4}{3} R_1$.
For the $C-R$ circuit: $\cos \phi_2 = \frac{R_2}{Z_2} = 0.5 = \frac{1}{2}$. Thus,$\sin \phi_2 = \frac{X_C}{Z_2} = \frac{\sqrt{3}}{2}$. So,$X_C = \sqrt{3} R_2$.
When joined in series,the circuit becomes an $L-C-R$ circuit. For the power factor to be $1$,the circuit must be in resonance,meaning $X_L = X_C$.
Equating the reactances: $\frac{4}{3} R_1 = \sqrt{3} R_2$.
Therefore,the ratio $\frac{R_1}{R_2} = \frac{3\sqrt{3}}{4}$.

(D) The circuit consists of a resistor $R$ in parallel with a series combination of an inductor $L$ and a capacitor $C$.
The peak voltage is $V_0 = 200 \, \text{V}$. The $RMS$ voltage is $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, \text{V}$.
The current through the resistor is $I_R = \frac{V_{\text{rms}}}{R} = \frac{100\sqrt{2}}{100} = \sqrt{2} \, \text{A}$.
The current through the $LC$ branch is $I_{LC} = \frac{V_{\text{rms}}}{|X_L - X_C|} = \frac{100\sqrt{2}}{|200 - 100|} = \frac{100\sqrt{2}}{100} = \sqrt{2} \, \text{A}$.
Since the current through the resistor is in phase with the voltage and the current through the $LC$ branch is $90^\circ$ out of phase with the voltage,the total $RMS$ current is given by $I_{\text{rms}} = \sqrt{I_R^2 + I_{LC}^2}$.
$I_{\text{rms}} = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \, \text{A}$.

(D) In the initial series $LR$ circuit,the impedance is $Z_1 = \sqrt{R^2 + X_L^2}$.
Given $X_L = 3R$,so $Z_1 = \sqrt{R^2 + (3R)^2} = \sqrt{10R^2} = R\sqrt{10}$.
The initial power factor is $\cos \phi_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}}$.
When a capacitor with $X_C = R$ is added in series,the circuit becomes an $LCR$ circuit.
The new impedance is $Z_2 = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values,$Z_2 = \sqrt{R^2 + (3R - R)^2} = \sqrt{R^2 + (2R)^2} = \sqrt{5R^2} = R\sqrt{5}$.
The new power factor is $\cos \phi_2 = \frac{R}{Z_2} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}}$.
The ratio of the new power factor to the old power factor is $\frac{\cos \phi_2}{\cos \phi_1} = \frac{1/\sqrt{5}}{1/\sqrt{10}} = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}$.

(A) In a series $LCR$ circuit,the current $I$ is the same through all elements.
The potential difference across each element is given by $V_R = I \cdot R$,$V_L = I \cdot X_L$,and $V_C = I \cdot X_C$.
Given: $R = 11 \, \Omega$,$X_L = 120 \, \Omega$,$X_C = 120 \, \Omega$.
The total impedance $Z$ is $\sqrt{R^2 + (X_L - X_C)^2} = \sqrt{11^2 + (120 - 120)^2} = 11 \, \Omega$.
The current $I = V / Z = 110 / 11 = 10 \, A$.
Now,calculating potential differences:
$V_R = I \cdot R = 10 \times 11 = 110 \, V$.
$V_L = I \cdot X_L = 10 \times 120 = 1200 \, V$.
$V_C = I \cdot X_C = 10 \times 120 = 1200 \, V$.
Comparing the values,$V_R = 110 \, V$ is the minimum potential difference.

(A) At resonance, the current $I_{rms}$ in the circuit is given by $I_{rms} = \frac{V_R}{R} = \frac{60 \, V}{120 \, \Omega} = 0.5 \, A.$
The voltage across the inductor is $V_L = I_{rms} \times X_L = I_{rms} \times (\omega L).$
Substituting the values: $40 = 0.5 \times (4 \times 10^3) \times L.$
$40 = 2000 \times L \implies L = \frac{40}{2000} = 0.02 \, H = 20 \, mH.$
At resonance, the voltage across the capacitor $V_C$ is equal to the voltage across the inductor $V_L$, so $V_C = 40 \, V.$
$V_C = I_{rms} \times X_C = I_{rms} \times \left(\frac{1}{\omega C}\right).$
$40 = 0.5 \times \left(\frac{1}{4 \times 10^3 \times C}\right).$
$40 = \frac{0.5}{4000 \times C} \implies C = \frac{0.5}{4000 \times 40} = \frac{0.5}{160000} = \frac{5}{1600000} = \frac{1}{320000} \, F.$
$C = \frac{1}{320000} \times 10^6 \, \mu F = \frac{100}{32} \, \mu F = 3.125 \, \mu F = \frac{25}{8} \, \mu F.$

(C) The circuit consists of an inductor with inductive reactance $X_L = 24 \ \Omega$ and a capacitor with capacitive reactance $X_C = 48 \ \Omega$ connected in parallel to an $AC$ source of $V = 240 \ V$.
The current through the inductor is $I_L = \frac{V}{X_L} = \frac{240}{24} = 10 \ A$.
The current through the capacitor is $I_C = \frac{V}{X_C} = \frac{240}{48} = 5 \ A$.
In a parallel $LC$ circuit,the currents $I_L$ and $I_C$ are $180^{\circ}$ out of phase. Therefore,the total current $I$ measured by the ammeter is the magnitude of the difference between the two currents:
$I = |I_L - I_C| = |10 \ A - 5 \ A| = 5 \ A$.

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

(D) The circuit consists of two parallel branches connected to an $AC$ source $V = V_0 \sin(\omega t)$.
Branch $1$ has $R$ and $X_L$ in series. The potential at point $A$ (relative to the bottom wire) is the voltage across the inductor: $V_A = V_0 \frac{X_L}{\sqrt{R^2 + X_L^2}} \sin(\omega t + \phi)$,where $\tan \phi = R/X_L$.
Branch $2$ has $R$ and $X_C$ in series. The potential at point $B$ (relative to the bottom wire) is the voltage across the capacitor: $V_B = V_0 \frac{X_C}{\sqrt{R^2 + X_C^2}} \sin(\omega t - \phi)$,where $\tan \phi = R/X_C$.
Given $R_1 = R_2 = R$ and $X_L = X_C = X$,let $Z = \sqrt{R^2 + X^2}$. Then $V_A = \frac{V_0 X}{Z} \sin(\omega t + \phi)$ and $V_B = \frac{V_0 X}{Z} \sin(\omega t - \phi)$.
The potential difference $V_A - V_B = \frac{V_0 X}{Z} [\sin(\omega t + \phi) - \sin(\omega t - \phi)] = \frac{V_0 X}{Z} [2 \cos(\omega t) \sin(\phi)]$.
Since $\sin \phi = R/Z$,we have $V_A - V_B = \frac{V_0 X}{Z} [2 \cos(\omega t) \frac{R}{Z}] = \frac{2 V_0 X R}{Z^2} \cos(\omega t) = \frac{2 V_0 X R}{R^2 + X^2} \cos(\omega t)$.
The peak value is $V_{peak} = \frac{2 V_0 X R}{R^2 + X^2}$. The $RMS$ value is $V_{RMS} = \frac{V_{peak}}{\sqrt{2}} = \frac{\sqrt{2} V_0 X R}{R^2 + X^2}$.

(A) The maximum current in an $RLC$ series circuit is given by $I_0 = \frac{V_0}{Z}$,where $Z$ is the impedance of the circuit.
The current in the circuit is $i(t) = I_0 \sin(\omega t - \phi)$.
The charge on the capacitor $q(t)$ is related to the current $i(t)$ by $i = \frac{dq}{dt}$.
Integrating the current,we get $q(t) = \int i(t) dt = \int I_0 \sin(\omega t - \phi) dt = -\frac{I_0}{\omega} \cos(\omega t - \phi)$.
The maximum charge $Q_0$ on the capacitor is the amplitude of this expression:
$Q_0 = \frac{I_0}{\omega}$.
Substituting $I_0 = \frac{V_0}{Z}$,we get:
$Q_0 = \frac{V_0}{\omega Z}$.

(B) Given: $R = 8.0\ \Omega$,$C = 5.0\ \mu F = 5.0 \times 10^{-6}\ F$,$L = 50.0\ mH = 50.0 \times 10^{-3}\ H$.
Resonance frequency $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{50 \times 10^{-3} \times 5 \times 10^{-6}}} = \frac{1}{\sqrt{250 \times 10^{-9}}} = \frac{1}{\sqrt{2.5 \times 10^{-7}}} = 2000\ rad/s$.
The operating frequency is $\omega = \frac{1}{2} \omega_0 = 1000\ rad/s$.
Inductive reactance $X_L = \omega L = 1000 \times 50 \times 10^{-3} = 50\ \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{1000 \times 5 \times 10^{-6}} = \frac{1}{0.005} = 200\ \Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{8^2 + (50 - 200)^2} = \sqrt{64 + (-150)^2} = \sqrt{64 + 22500} = \sqrt{22564} \approx 150.21\ \Omega$.
$I_{rms} = \frac{V_{rms}}{Z} = \frac{400}{150.21} \approx 2.663\ A$.
Power $P = I_{rms}^2 R = (2.663)^2 \times 8 \approx 7.09 \times 8 \approx 56.72\ W$.
Rounding to the nearest integer,$P \approx 57\ W$.

(D) In a series $LCR$ circuit,the total voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Given: $V = 220 \ V$,$V_L = 300 \ V$,$V_C = 300 \ V$.
Substituting these values into the formula: $220 = \sqrt{V_R^2 + (300 - 300)^2}$.
$220 = \sqrt{V_R^2 + 0^2} = V_R$.
Thus,the reading of the voltmeter across the resistor is $V_R = 220 \ V$.
The current $I$ in the circuit is given by $I = \frac{V_R}{R}$.
Given $R = 100 \ \Omega$,we have $I = \frac{220 \ V}{100 \ \Omega} = 2.2 \ A$.
Therefore,the readings are $220 \ V$ and $2.2 \ A$.

(A) From the given circuit,the voltage is $V = 200 \sin(100t) \, V$. Comparing this with $V = V_m \sin(\omega t)$,we get $V_m = 200 \, V$ and $\omega = 100 \, rad/s$.
The $RMS$ voltage is $V_{\text{rms}} = \frac{V_m}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, V$.
The circuit components are $R = 50 \, \Omega$,$L = 0.5 \, H$,and $C = 100 \, \mu F = 100 \times 10^{-6} \, F$.
Inductive reactance $X_L = \omega L = 100 \times 0.5 = 50 \, \Omega$.
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = \frac{1}{0.01} = 100 \, \Omega$.
The impedance $Z$ of the $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
$Z = \sqrt{50^2 + (50 - 100)^2} = \sqrt{50^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \, \Omega$.
The effective $(RMS)$ current is $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\sqrt{2}}{50\sqrt{2}} = 2 \, A$.

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

(A) The circuit is a series $LCR$ circuit with resistance $R = 18 \ \Omega$,inductive reactance $X_L = 40 \ \Omega$,and capacitive reactance $X_C = 16 \ \Omega$.
The formula for the impedance $Z$ of a series $LCR$ circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Substituting the given values:
$Z = \sqrt{18^2 + (40 - 16)^2}$
$Z = \sqrt{18^2 + 24^2}$
$Z = \sqrt{324 + 576}$
$Z = \sqrt{900}$
$Z = 30 \ \Omega$
Therefore,the impedance of the circuit is $30 \ \Omega$.

(B) From the given circuit,the voltage is $V = 200\sqrt{2} \sin(300t)$,so the $RMS$ voltage is $V_{rms} = 200 \, V$ and angular frequency $\omega = 300 \, rad/s$.
Given $R = 3000 \, \Omega$,$L = \frac{50}{3} \, H$,and $C = \frac{10}{3} \, \mu F = \frac{10}{3} \times 10^{-6} \, F$.
Inductive reactance $X_{L} = \omega L = 300 \times \frac{50}{3} = 5000 \, \Omega$.
Capacitive reactance $X_{C} = \frac{1}{\omega C} = \frac{1}{300 \times \frac{10}{3} \times 10^{-6}} = \frac{1}{10^{-3}} = 1000 \, \Omega$.
Impedance $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{3000^{2} + (5000 - 1000)^{2}} = \sqrt{3000^{2} + 4000^{2}} = 5000 \, \Omega$.
The average power consumed is $P = I_{rms}^{2} R = \left(\frac{V_{rms}}{Z}\right)^{2} R$.
$P = \left(\frac{200}{5000}\right)^{2} \times 3000 = \left(\frac{1}{25}\right)^{2} \times 3000 = \frac{3000}{625} = 4.8 \, W$.

(C) The total resistance of the circuit is $R_{total} = R_1 + R_2 = 44 \,\Omega + 36 \,\Omega = 80 \,\Omega$.
The total reactance of the circuit is $X_{total} = X_L - X_C = 90 \,\Omega - 30 \,\Omega = 60 \,\Omega$.
The impedance of the circuit is $Z = \sqrt{R_{total}^2 + X_{total}^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \,\Omega$.
The $RMS$ current in the circuit is $I_{rms} = \frac{V_{rms}}{Z} = \frac{200 \,V}{100 \,\Omega} = 2 \,A$.
The power dissipated in the coil is given by $P_{coil} = I_{rms}^2 \times R_2$.
Substituting the values,$P_{coil} = (2 \,A)^2 \times 36 \,\Omega = 4 \times 36 = 144 \,W$.

(B) The average power dissipated in an $a.c.$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
Here,$V_{rms} = \varepsilon$ and $I_{rms} = \frac{\varepsilon}{Z}$,where $Z$ is the impedance of the $LCR$ circuit.
The power factor is $\cos \phi = \frac{R}{Z}$.
Substituting these into the power formula:
$P = \varepsilon \cdot \left( \frac{\varepsilon}{Z} \right) \cdot \left( \frac{R}{Z} \right) = \frac{\varepsilon^2 R}{Z^2}$.
The impedance $Z$ of an $LCR$ series circuit is $Z = \sqrt{R^2 + (L\omega - \frac{1}{C\omega})^2}$.
Therefore,$Z^2 = R^2 + (L\omega - \frac{1}{C\omega})^2$.
Substituting $Z^2$ into the power equation,we get:
$P = \frac{\varepsilon^2 R}{R^2 + (L\omega - \frac{1}{C\omega})^2}$.

(C) In a series $LCR$ circuit,the current $I$ is given by $I = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}$.
From the circuit diagram,the voltage across the inductor $V_L = 10 \text{ V}$ and the voltage across the capacitor $V_C = 10 \text{ V}$.
Since $V_L = V_C$,the circuit is in resonance,meaning $X_L = X_C$.
The voltage across the $L-C$ combination is $V_{LC} = |V_L - V_C|$.
Since $V_L = V_C = 10 \text{ V}$,$V_{LC} = |10 - 10| = 0 \text{ V}$.
If the resistance $R$ is changed,the current $I$ in the circuit changes.
However,the condition for resonance $(X_L = X_C)$ depends only on the frequency of the source and the values of $L$ and $C$,not on $R$.
Therefore,$V_L = I X_L$ and $V_C = I X_C$ will change as $I$ changes,but their difference $V_{LC} = I(X_L - X_C)$ will remain $0 \text{ V}$ because $X_L - X_C = 0$.
Thus,the voltage across the $L-C$ combination remains the same (zero).

(A) Given: $R = 100\,\Omega$,$V = 200\,V$.
When capacitance is removed,the circuit becomes an $LR$ series circuit. The phase angle $\phi = 60^o$ is given by $\tan(60^o) = \frac{X_L}{R}$.
Thus,$X_L = R \tan(60^o) = 100 \times \sqrt{3} = 100\sqrt{3}\,\Omega$.
When inductance is removed,the circuit becomes an $RC$ series circuit. The phase angle $\phi = 60^o$ is given by $\tan(60^o) = \frac{X_C}{R}$.
Thus,$X_C = R \tan(60^o) = 100 \times \sqrt{3} = 100\sqrt{3}\,\Omega$.
In the original $LCR$ circuit,the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $X_L = X_C$,$Z = \sqrt{R^2 + 0} = R = 100\,\Omega$.
The current in the circuit is $I = \frac{V}{Z} = \frac{200}{100} = 2\,A$.

(A) The phase angle $\phi$ in an $LCR$ circuit is given by the formula: $\tan \phi = \frac{X_C - X_L}{R}$.
Substituting the given values: $\tan \phi = \frac{8\,\Omega - 5\,\Omega}{4\,\Omega} = \frac{3\,\Omega}{4\,\Omega} = \frac{3}{4}$.
Therefore,$\phi = \tan^{-1}(3/4)$.
Since $X_C > X_L$,the circuit is capacitive in nature,which means the current leads the voltage by the phase angle $\phi$.

(C) The given voltage is $V = 20 \cos(2000t)$,so the peak voltage $V_0 = 20 \, V$. The $RMS$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, V \approx 14.14 \, V$.
The angular frequency is $\omega = 2000 \, rad/s$. The inductive reactance is $X_L = \omega L = 2000 \times 5 \times 10^{-3} = 10 \, \Omega$. The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 50 \times 10^{-6}} = 10 \, \Omega$.
Since $X_L = X_C$,the series $LC$ branch is in resonance. The impedance of the $LC$ branch is $Z_{LC} = \sqrt{R_{LC}^2 + (X_L - X_C)^2} = \sqrt{4^2 + 0^2} = 4 \, \Omega$.
The total impedance of the circuit is $Z = R_{total} + Z_{LC} = 6 \, \Omega + 4 \, \Omega = 10 \, \Omega$.
The $RMS$ current in the circuit is $I_{rms} = \frac{V_{rms}}{Z} = \frac{10\sqrt{2}}{10} = \sqrt{2} \approx 1.414 \, A$. Thus,the ammeter reading is $1.4 \, A$.
The voltmeter is connected across the $LC$ branch. The voltage across this branch is $V_{LC} = I_{rms} \times Z_{LC} = 1.414 \times 4 = 5.656 \, V \approx 5.6 \, V$.

(C) In a series $LCR$ circuit,the total $rms$ voltage $V$ is given by the relation: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Given the ratio $V_L : V_C : V_R = 1 : 2 : 3$,we can express these voltages in terms of a constant $x$ as $V_L = x$,$V_C = 2x$,and $V_R = 3x$.
Substituting these values into the formula for $V$:
$100 = \sqrt{(3x)^2 + (x - 2x)^2}$
$100 = \sqrt{9x^2 + (-x)^2}$
$100 = \sqrt{9x^2 + x^2}$
$100 = \sqrt{10x^2}$
$100 = x\sqrt{10}$
$x = \frac{100}{\sqrt{10}} = 10\sqrt{10} \approx 10 \times 3.162 = 31.62 \, V$.
Now,calculate $V_R$:
$V_R = 3x = 3 \times 31.62 = 94.86 \, V$.
Rounding to the nearest given option,$V_R$ is close to $90 \, V$.

(D) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.
Given $X_L = 2R$,the old power factor is $\cos \phi_1 = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.
When a capacitor with $X_C = R$ is added in series,the new impedance $Z_{new} = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values,$Z_{new} = \sqrt{R^2 + (2R - R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.
The new power factor is $\cos \phi_2 = \frac{R}{Z_{new}} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The ratio of the new power factor to the old one is $\frac{\cos \phi_2}{\cos \phi_1} = \frac{1/\sqrt{2}}{1/\sqrt{5}} = \sqrt{\frac{5}{2}}$.

(D) 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 from the figure:
$V = 220\,V$
$V_L = 300\,V$
$V_C = 300\,V$
Substituting these values into the equation:
$220 = \sqrt{V_R^2 + (300 - 300)^2}$
$220 = \sqrt{V_R^2 + 0^2}$
$V_R = 220\,V$
Now,the current $I$ in the circuit is given by:
$I = \frac{V_R}{R}$
Given $R = 100\,\Omega$:
$I = \frac{220}{100} = 2.2\,A$
Thus,the reading of the voltmeter is $220\,V$ and the reading of the ammeter is $2.2\,A$.

(C) Given: $L = 20\,mH = 20 \times 10^{-3}\,H,$ $C = 120\,\mu F = 120 \times 10^{-6}\,F,$ $R = 60\,\Omega,$ $V_{rms} = 24\,V,$ $f = 50\,Hz,$ $t = 60\,s.$
Inductive reactance: $X_L = 2\pi fL = 2\pi \times 50 \times 20 \times 10^{-3} = 2\pi \approx 6.28\,\Omega.$
Capacitive reactance: $X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50 \times 120 \times 10^{-6}} = \frac{1}{0.012\pi} \approx 26.53\,\Omega.$
Impedance: $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{60^2 + (26.53 - 6.28)^2} = \sqrt{3600 + (20.25)^2} = \sqrt{3600 + 410.06} = \sqrt{4010.06} \approx 63.33\,\Omega.$
Power dissipated: $P = I_{rms}^2 R = \left(\frac{V_{rms}}{Z}\right)^2 R = \left(\frac{24}{63.33}\right)^2 \times 60 \approx (0.379)^2 \times 60 \approx 0.1436 \times 60 \approx 8.616\,W.$
Energy dissipated: $E = P \times t = 8.616 \times 60 \approx 516.96\,J \approx 5.17 \times 10^2\,J.$

(C) Given: $\omega = 100 \, rad/s$,$L = \frac{\sqrt{3}}{10} \, H$,$R_1 = 10 \, \Omega$,$C = \frac{\sqrt{3}}{2} \times 10^{-3} \, F$,$R_2 = 20 \, \Omega$.
For the $L-R_1$ branch:
Inductive reactance $X_L = \omega L = 100 \times \frac{\sqrt{3}}{10} = 10\sqrt{3} \, \Omega$.
The phase angle $\phi_1$ of current $I_1$ with respect to voltage $V$ is given by $\tan \phi_1 = -\frac{X_L}{R_1} = -\frac{10\sqrt{3}}{10} = -\sqrt{3}$.
Thus,$\phi_1 = -60^\circ$.
For the $C-R_2$ branch:
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \times \frac{\sqrt{3}}{2} \times 10^{-3}} = \frac{1}{\frac{\sqrt{3}}{20}} = \frac{20}{\sqrt{3}} \, \Omega$.
The phase angle $\phi_2$ of current $I_2$ with respect to voltage $V$ is given by $\tan \phi_2 = \frac{X_C}{R_2} = \frac{20/\sqrt{3}}{20} = \frac{1}{\sqrt{3}}$.
Thus,$\phi_2 = 30^\circ$.
The phase difference between $I_1$ and $I_2$ is $|\phi_2 - \phi_1| = |30^\circ - (-60^\circ)| = 90^\circ$.

(C) In an $LCR$ series circuit,the total voltage $V$ is given by the relation:
$V = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$
Given from the figure:
$V_{L} = 100 \, V$
$V_{C} = 100 \, V$
$V = 200 \, V$
Substituting these values into the formula:
$200 = \sqrt{V_{R}^{2} + (100 - 100)^{2}}$
$200 = \sqrt{V_{R}^{2} + 0^{2}}$
$200 = V_{R}$
Therefore,the reading of the voltmeter connected across the resistor $R$ is $200 \, V$.

(C) In an $LCR$ series circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
When $L$ is removed,the circuit becomes an $RC$ circuit. The current leads the voltage by $45^o$,so $\phi = -45^o$. Thus,$\tan(-45^o) = -\frac{X_C}{R}$,which implies $\frac{X_C}{R} = 1$,or $X_C = R$.
When $C$ is removed,the circuit becomes an $RL$ circuit. The current lags the voltage by $45^o$,so $\phi = 45^o$. Thus,$\tan(45^o) = \frac{X_L}{R}$,which implies $\frac{X_L}{R} = 1$,or $X_L = R$.
Since $X_L = R$ and $X_C = R$,it follows that $X_L = X_C$. This means the circuit is at resonance.
At resonance,the impedance of the circuit is $Z = R = 100 \, \Omega$.
The current in the original circuit is $I = \frac{V}{Z} = \frac{100 \, V}{100 \, \Omega} = 1 \, A$.

(C) The circuit consists of an inductor and a capacitor connected in parallel to an $AC$ voltage source of $V = 75 \, V$.
The current through the inductor is $i_L = 5 \, A$ and the current through the capacitor is $i_C = 2 \, A$.
In a parallel $LC$ circuit,the currents $i_L$ and $i_C$ are $180^{\circ}$ out of phase. Therefore,the net current $i_{LC}$ drawn from the source is the difference between the magnitudes of the two currents:
$i_{LC} = |i_L - i_C| = |5 \, A - 2 \, A| = 3 \, A$.
The equivalent impedance $Z$ of the parallel combination is given by Ohm's law:
$Z = \frac{V}{i_{LC}} = \frac{75 \, V}{3 \, A} = 25 \, \Omega$.
Thus,the impedance of the circuit is $25 \, \Omega$.

(C) The total resistance of the circuit is $R = R_1 + R_2 = 44\,\Omega + 36\,\Omega = 80\,\Omega$.
The total reactance of the circuit is $X = X_L - X_C = 90\,\Omega - 30\,\Omega = 60\,\Omega$.
The impedance of the circuit is $Z = \sqrt{R^2 + X^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$.
The $RMS$ current in the circuit is $I_{rms} = \frac{V_{rms}}{Z} = \frac{200\,V}{100\,\Omega} = 2\,A$.
The power dissipated in the coil is given by $P = I_{rms}^2 \times R_2$,where $R_2$ is the resistance of the coil.
$P = (2\,A)^2 \times 36\,\Omega = 4 \times 36 = 144\,W$.