Only Inductor, Only Capacitor and Only Resistor Circuit Questions in English

(B) Given: Initial frequency, $f_1 = 50 \, Hz$.
Initial reactance, $X_1 = 10 \, \Omega$.
Final frequency, $f_2 = 200 \, Hz$.
We know that the inductive reactance $X_L$ is given by $X_L = \omega L = 2 \pi f L$.
Since $L$ is constant, $X_L \propto f$.
Therefore, $\frac{X_2}{X_1} = \frac{f_2}{f_1}$.
Substituting the values: $\frac{X_2}{10} = \frac{200}{50}$.
$\frac{X_2}{10} = 4$.
$X_2 = 4 \times 10 = 40 \, \Omega$.

(A) Given,voltage $V = V_0 \sin \omega t$,where $V_0 = 10 \ V$ and inductance $L = 10 \ H$.
For a pure inductor,the induced $EMF$ is $E = -L \frac{di}{dt}$.
According to Kirchhoff's loop rule,$V + E = 0$,so $V = L \frac{di}{dt}$.
Rearranging for current $i$,we get $\frac{di}{dt} = \frac{V}{L} = \frac{V_0 \sin \omega t}{L}$.
Integrating with respect to time $t$:
$i = \int \frac{V_0}{L} \sin \omega t \ dt = \frac{V_0}{L} \left( -\frac{\cos \omega t}{\omega} \right) = -\frac{V_0}{\omega L} \cos \omega t$.
Using the trigonometric identity $-\cos \theta = \sin \left( \theta - \frac{\pi}{2} \right)$:
$i = \frac{V_0}{\omega L} \sin \left( \omega t - \frac{\pi}{2} \right)$.
Substituting the values $V_0 = 10$ and $L = 10$:
$i = \frac{10}{\omega \times 10} \sin \left( \omega t - \frac{\pi}{2} \right) = \frac{1}{\omega} \sin \left( \omega t - \frac{\pi}{2} \right)$.

(B) The capacitive reactance $X_C$ of a capacitor is given by the formula $X_C = \frac{1}{2 \pi f C}$,where $f$ is the frequency of the $AC$ source and $C$ is the capacitance.
From this relation,we can see that $X_C \propto \frac{1}{f}$.
Let the initial frequency be $f_1$ and the initial reactance be $X_{C1} = 6 \ k\Omega$.
Let the new frequency be $f_2 = 2f_1$.
The new reactance $X_{C2}$ is given by the ratio:
$\frac{X_{C2}}{X_{C1}} = \frac{f_1}{f_2} = \frac{f_1}{2f_1} = \frac{1}{2}$.
Therefore,$X_{C2} = \frac{X_{C1}}{2} = \frac{6 \ k\Omega}{2} = 3 \ k\Omega$.

(C) In an $AC$-circuit containing only capacitance,the current always leads the voltage by a phase angle of $90^{\circ}$.
This occurs because the charge $q$ on the capacitor is related to the voltage $V$ by $q = CV$.
The current $I$ is the rate of change of charge,$I = dq/dt = C(dV/dt)$.
If $V = V_0 \sin(\omega t)$,then $I = C \omega V_0 \cos(\omega t) = I_0 \sin(\omega t + 90^{\circ})$.
Thus,the current leads the voltage by $90^{\circ}$.

(B) In a pure inductive circuit,the alternating voltage is given by $V = V_0 \sin \omega t$.
The alternating current in the circuit is given by $I = I_0 \sin (\omega t - \frac{\pi}{2})$.
Comparing these two equations,we can see that the current lags behind the $EMF$ (voltage) by a phase angle of $\frac{\pi}{2}$.
Therefore,the current and $EMF$ differ in phase by $\frac{\pi}{2}$.

(D) In a purely resistive circuit,the alternating voltage $V = V_m \sin(\omega t)$ and the alternating current $I = I_m \sin(\omega t)$ are both in the same phase.
Since both reach their maximum and minimum values at the same time,the phase difference between the voltage and the current is $0^{\circ}$.

(D) Given: Capacitance $C = 50 \mu F = 50 \times 10^{-6} \text{ F}$ and voltage $V = 220 \sin 50 t \text{ V}$.
Comparing this with the standard equation $V = V_0 \sin \omega t$,we get peak voltage $V_0 = 220 \text{ V}$ and angular frequency $\omega = 50 \text{ rad/s}$.
The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C} = \frac{1}{50 \times 50 \times 10^{-6}} = \frac{1}{2500 \times 10^{-6}} = \frac{10^6}{2500} = 400 \Omega$.
The peak current $i_0$ is $i_0 = \frac{V_0}{X_C} = \frac{220}{400} = 0.55 \text{ A}$.
The $rms$ current $i_{rms}$ is given by $i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{0.55}{\sqrt{2}} \text{ A}$.

(C) For an $AC$ circuit containing an inductor only:
$I = I_0 \sin(\omega t)$
$V = V_0 \sin(\omega t + \frac{\pi}{2}) = V_0 \cos(\omega t)$
Instantaneous power $P = V \cdot I$
$P = (V_0 \cos(\omega t)) \cdot (I_0 \sin(\omega t))$
Using the trigonometric identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$:
$P = \frac{V_0 I_0}{2} \sin(2\omega t)$
The angular frequency of the power is $2\omega$. Since $\omega = 2\pi f$,the frequency of the power $f' = 2f$.
Given $f = 50 \ Hz$,the frequency of the instantaneous power is $f' = 2 \times 50 \ Hz = 100 \ Hz$.

(A) Given: Resistance $R = 100 \Omega$,$AC$ source $\varepsilon = 10 \sin (250 \pi t)$.
Comparing with $\varepsilon = \varepsilon_0 \sin (\omega t)$,we get $\varepsilon_0 = 10 \text{ V}$ and $\omega = 250 \pi \text{ rad/s}$.
The energy dissipated as heat $H$ is given by $H = \int_0^{t} \frac{\varepsilon^2}{R} dt$.
$H = \frac{1}{R} \int_0^{10^{-3}} (10 \sin (250 \pi t))^2 dt = \frac{100}{100} \int_0^{10^{-3}} \sin^2 (250 \pi t) dt$.
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $H = \int_0^{10^{-3}} \frac{1 - \cos (500 \pi t)}{2} dt$.
$H = \frac{1}{2} \left[ t - \frac{\sin (500 \pi t)}{500 \pi} \right]_0^{10^{-3}}$.
$H = \frac{1}{2} \left[ 10^{-3} - \frac{\sin (500 \pi \times 10^{-3})}{500 \pi} \right] = \frac{1}{2} \left[ 10^{-3} - \frac{\sin (0.5 \pi)}{500 \pi} \right]$.
Since $\sin (0.5 \pi) = 1$,$H = \frac{1}{2} \left[ \frac{1}{1000} - \frac{1}{500 \pi} \right] = \frac{1}{2} \left[ \frac{\pi - 2}{1000 \pi} \right] = \frac{\pi - 2}{2000 \pi} \text{ J}$.
Approximating $\pi \approx 3.14$,$\pi - 2 \approx 1.14$. So $H \approx \frac{1.14}{2000 \pi} \text{ J} = \frac{0.57}{\pi} \times 10^{-3} \text{ J} = \frac{0.57}{\pi} \text{ mJ}$.

(D) Given: Capacitance $C = 50 \mu F = 50 \times 10^{-6} \text{ F}$ and voltage $V = 220 \sin 50 t \text{ V}$.
Comparing with the standard equation $V = V_0 \sin \omega t$,we get peak voltage $V_0 = 220 \text{ V}$ and angular frequency $\omega = 50 \text{ rad/s}$.
The capacitive reactance $X_C$ is calculated as:
$X_C = \frac{1}{\omega C} = \frac{1}{50 \times 50 \times 10^{-6}} = \frac{1}{2500 \times 10^{-6}} = \frac{10^6}{2500} = 400 \Omega$.
The peak current $i_0$ is given by:
$i_0 = \frac{V_0}{X_C} = \frac{220}{400} = 0.55 \text{ A}$.
The rms current $i_{\text{rms}}$ is given by:
$i_{\text{rms}} = \frac{i_0}{\sqrt{2}} = \frac{0.55}{\sqrt{2}} \text{ A}$.

(C) The inductive reactance $X_L$ is given by the formula $X_L = \omega L$.
Since the angular frequency $\omega = 2 \pi \nu$, where $\nu$ is the frequency of the $AC$ source, we have $X_L = 2 \pi \nu L$.
Given: Inductance $L = 1 \ H$ and frequency $\nu = 50 \ Hz$.
Substituting these values into the formula:
$X_L = 2 \pi \times 50 \times 1 = 100 \pi \ \Omega$.

(C) In a pure inductive circuit, the current $I$ lags behind the voltage $V$ by a phase angle of $\phi = 90^{\circ}$ (or $\pi/2$ radians).
The average power consumed in an $A$.$C$. circuit is given by the formula $P = V_{rms} I_{rms} \cos \phi$.
Substituting the phase angle $\phi = 90^{\circ}$ into the formula, we get $\cos 90^{\circ} = 0$.
Therefore, $P = V_{rms} I_{rms} \times 0 = 0$.
Thus, a pure inductor consumes no average power.

(B) The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2\pi f C}$.
Given values are frequency $f = 50 \text{ Hz}$ and capacitance $C = 15.0 \mu\text{F} = 15.0 \times 10^{-6} \text{ F}$.
Substituting these values into the formula:
$X_C = \frac{1}{2 \times 3.14159 \times 50 \times 15.0 \times 10^{-6}}$
$X_C = \frac{1}{314.159 \times 15.0 \times 10^{-6}}$
$X_C = \frac{1}{4712.385 \times 10^{-6}}$
$X_C = \frac{10^6}{4712.385} \approx 212.2 \Omega$.
Rounding to the nearest integer, the capacitive reactance is $212 \Omega$.

(B) In a pure capacitive circuit, the current leads the voltage by a phase angle of $\pi/2$ radians.
The instantaneous voltage is given by $E = E_0 \sin(\omega t)$.
The capacitive reactance is $X_C = \frac{1}{\omega C}$.
The instantaneous current $I$ is given by $I = \frac{E}{X_C} = \frac{E_0 \sin(\omega t)}{1/(\omega C)} = E_0 \omega C \sin(\omega t)$.
Since the current leads the voltage by $\pi/2$, we add $\pi/2$ to the phase of the voltage.
Therefore, the instantaneous current is $I = E_0 \omega C \sin(\omega t + \pi/2)$.

(C) The given alternating voltage is $e = 150\sqrt{2} \sin 100t \text{ V}$.
Comparing this with the standard equation $e = E_0 \sin \omega t$, we get the peak voltage $E_0 = 150\sqrt{2} \text{ V}$ and angular frequency $\omega = 100 \text{ rad/s}$.
The root mean square $(RMS)$ voltage is $V_{rms} = E_0 / \sqrt{2} = 150 \text{ V}$.
The capacitive reactance $X_C$ is given by $X_C = 1 / (\omega C)$.
Given $C = 2 \mu F = 2 \times 10^{-6} \text{ F}$, we have $X_C = 1 / (100 \times 2 \times 10^{-6}) = 1 / (2 \times 10^{-4}) = 5000 \Omega$.
The $RMS$ current $I_{rms}$ is $I_{rms} = V_{rms} / X_C = 150 / 5000 = 0.03 \text{ A}$.
Converting to milliamperes, $I_{rms} = 0.03 \times 1000 \text{ mA} = 30 \text{ mA}$.

(C) कुंडली का प्रेरकत्व $L = 3 \ H$ है।
$a.c.$ स्रोत के लिए, प्रेरणिक प्रतिघात (inductive reactance) $X_L = \omega L = 2 \pi f L$ होता है।
$d.c.$ स्रोत के लिए, आवृत्ति $f = 0$ होती है, इसलिए प्रेरणिक प्रतिघात $X_{dc} = 2 \pi (0) L = 0$ होता है।
प्रश्न के अनुसार, $a.c.$ प्रतिघात और $d.c.$ प्रतिघात का अनुपात $\frac{X_L}{X_{dc}} = \frac{\omega L}{0} = \infty$ (अनंत) होगा।
अतः, सही विकल्प $C$ है।

(A) The capacitive reactance $X_C$ is given by the formula: $X_C = \frac{1}{2\pi f C}$.
Given values are $C = 10 \mu F = 10 \times 10^{-6} F = 10^{-5} F$ and $X_C = 2 \times 10^{-3} \Omega$.
Rearranging the formula to solve for frequency $f$: $f = \frac{1}{2\pi X_C C}$.
Substituting the values: $f = \frac{1}{2 \times \pi \times (2 \times 10^{-3}) \times 10^{-5}}$.
$f = \frac{1}{4 \times \pi \times 10^{-8}} = \frac{10^8}{4\pi} = \frac{100 \times 10^6}{4\pi} = \frac{25}{\pi} \times 10^6$ Hz.
Since $10^6$ Hz = $1$ MHz, the frequency is $\frac{25}{\pi}$ MHz.

(B) When an alternating current $(a.c.)$ source is connected across a pure capacitor, the voltage $(e_c)$ across the capacitor lags behind the current $(I_c)$ by a phase angle of $90^{\circ}$ (or $\pi/2$ radians).
Alternatively, we can say that the current leads the voltage by $90^{\circ}$.
Mathematically, if the voltage is $e_c = E_0 \sin(\omega t)$, then the current is $I_c = I_0 \sin(\omega t + \pi/2)$.
In a phasor diagram, if the voltage vector is along the positive $x$-axis, the current vector points along the positive $y$-axis, indicating a $90^{\circ}$ lead.
Therefore, option $B$ correctly represents this phase relationship.

(C) Step $1$: For a pure resistor, the instantaneous voltage is given by $V = V_m \sin(\omega t)$.
Step $2$: According to Ohm's law, the current is $I = \frac{V}{R} = \frac{V_m}{R} \sin(\omega t) = I_m \sin(\omega t)$.
Step $3$: Comparing the phase of voltage and current, both have the same phase angle $\omega t$.
Step $4$: Therefore, the voltage and current are in the same phase.