Alternating Current – JEE Main PhysicsPractice Questions & PYQs

Transformers operate on the principle of electromagnetic induction, where the secondary voltage is determined by the ratio of turns between the primary and secondary windings. With a primary supply of 230 V and a turns ratio of 10:1, the voltage across the secondary winding is reduced to $V_2 = \frac{230}{10} = 23 \text{ V}$. To determine the power consumed by the load resistance, we use the formula $P = \frac{V_2^2}{R}$. Substituting the secondary voltage and the provided resistance of 46 $\Omega$ into this expression yields $P = \frac{23^2}{46} = \frac{529}{46} = 11.5 \text{ W}$.

The average power dissipated in an AC circuit is determined by the product of the RMS voltage, the RMS current, and the cosine of the phase difference. With a peak voltage of 20, a peak current of 10, and a phase difference of 60 degrees, the average power is calculated using the following relationship:

$P_{avg} = \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^\circ$

This calculation simplifies directly to 50 W.

The condition where instantaneous current is zero while voltage is maximum indicates a phase difference of exactly $\frac{\pi}{2}$ between the two quantities. This phase shift is a hallmark of purely reactive components, such as a pure inductor or a pure capacitor, because these elements do not dissipate power and cause the current to either lag or lead the voltage by this angle. An LC circuit, being a combination of these reactive elements, also maintains a phase difference of $\frac{\pi}{2}$ when purely reactive, whereas a pure resistor causes voltage and current to be in phase. Therefore, the viable configurations that satisfy this criterion are a pure inductor, a pure capacitor, and a combination of an inductor and capacitor.

An AC generator functions based on the fundamental principle of electromagnetic induction, where a rotating coil within a magnetic field induces an electromotive force. A transformer operates through mutual inductance, a phenomenon where a varying current in the primary coil induces a voltage across a secondary coil. In alternating circuits, the resonance phenomenon occurs specifically when both an inductor $L$ and a capacitor $C$ are present to exchange energy. Finally, the sharpness of this resonance is quantitatively characterized by the quality factor $Q$, which determines how narrow or selective the resonance peak remains.

The instantaneous electromotive force (EMF) induced in a rotating coil is governed by the expression $E = N B A \omega \sin(\theta)$, where $N$ is the number of turns, $B$ is the magnetic field, $A$ is the area of the coil, and $\omega$ is the angular velocity. The angle $\theta$ in this formula represents the orientation between the area vector of the coil-which is perpendicular to its plane-and the magnetic field.

To calculate the EMF, we first convert the area to standard units, giving $A = 70 \times 10^{-4}$ m$^2$. The angular velocity is determined by the rotation frequency of 500 revolutions per minute, calculated as $\omega = 2\pi \times (500/60) = (100\pi / 6)$ rad/s. Since the plane of the coil is inclined at 60 degrees to the magnetic field, the angle $\theta$ between the area vector and the field is 90 degrees minus 60 degrees, which equals 30 degrees. This makes the sine component $\sin(30^\circ) = 1/2$. Substituting these values into the formula with $\pi = 22/7$ gives:

$E = 600 \times (70 \times 10^{-4}) \times 0.4 \times \left(\frac{100 \times 22}{6 \times 7}\right) \times \frac{1}{2}$

Solving this expression results in an instantaneous EMF of 44 V.

The current in an alternating current circuit is determined by the ratio of the rms voltage to the circuit's total impedance, expressed by the relationship $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$. In this series configuration, the total impedance is calculated using the formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$. By substituting the given resistance and reactances into this equation, the impedance becomes $Z = \sqrt{100^2 + (200 - 100)^2}$, which simplifies to $100\sqrt{2}$ ohm. Dividing the rms voltage of $200\sqrt{2}$ volt by this impedance value yields a final current of 2 ampere.

In a series LCR circuit, the average rate at which energy is supplied, known as power, reaches its maximum value during the condition of electrical resonance. This state occurs when the inductive reactance exactly offsets the capacitive reactance, satisfying the condition $X_L = X_C$. Given that the inductive reactance is $X_L = 79.6 \Omega$ at a frequency of $f = 50 \text{Hz}$, we use the relationship $X_C = \frac{1}{2 \pi f C}$ to find the required capacitance. Substituting the known values into the equation leads to $79.6 = \frac{1}{2 \pi \times 50 \times C}$. Rearranging to solve for the capacitance yields $C = \frac{1}{2 \pi \times 50 \times 79.6}$, which results in $C \approx 40 \mu\text{F}$.

The resonant frequency of an LC oscillator depends on the values of inductance and capacitance according to the formula $\omega_0 = \frac{1}{\sqrt{LC}}$, showing that the frequency is inversely proportional to the square root of their product. When the inductance is modified to $2L$ and the capacitance is modified to $8C$, the new frequency $\omega'$ is determined by substituting these updated values into the expression:
$\omega' = \frac{1}{\sqrt{(2L)(8C)}} = \frac{1}{\sqrt{16LC}}$
By factoring out the constant from the square root, this simplifies to $\omega' = \frac{1}{4\sqrt{LC}}$. Because the initial resonant frequency $\omega_0$ is defined as $\frac{1}{\sqrt{LC}}$, the new frequency is related to the original by $\omega' = \frac{1}{4} \omega_0$, which identifies the value of $x$ as $1/4$.

At steady state, an inductor acts as a short circuit because the current reaches a constant value, meaning the potential difference across it is determined solely by the coil's resistance. We calculate this constant current using the relationship $I = \frac{V}{R}$, which simplifies to $10 / 4$, resulting in 2.5 A. The energy stored in the magnetic field is then determined by the expression $E = \frac{1}{2} L I^2$. Substituting the given values into the equation $E = \frac{1}{2} \times 2 \times (2.5)^2$ yields 6.25 J, which is equivalent to $625 \times 10^{-2}$ J.

In an AC circuit containing an inductor and a capacitor in series, the condition for drawing maximum current is known as electrical resonance. This state occurs when the opposition offered by the inductor, called inductive reactance, is exactly equal in magnitude to the opposition offered by the capacitor, called capacitive reactance. Mathematically, this condition is expressed by setting $2 \pi f L$ equal to $\frac{1}{2 \pi f C}$, which leads to the resonance equation $C = \frac{1}{4 \pi^2 f^2 L}$.

Using the provided values where frequency $f = 7\,\text{kHz}$ is $7000\,\text{Hz}$, inductance $L = 2\,\mu\text{H}$ is $2 \times 10^{-6}\,\text{H}$, and $\pi = \frac{22}{7}$, we substitute these values into the resonance formula to calculate the capacitance. This gives $C = \frac{1}{4 \times (\frac{22}{7})^2 \times (7000)^2 \times 2 \times 10^{-6}}$. Simplifying the denominator, the square of the frequency term $7000^2$ combined with the square of the fraction $\frac{22}{7}$ results in $484 \times 10^6$. Multiplying this by the remaining factors of $4$, $2$, and $10^{-6}$ simplifies the denominator to $4 \times 484 \times 2$, which equals $3872$. Thus, $C = \frac{1}{3872}\,\text{F}$, confirming the value of $x$ is $3872$.

In an LCR series circuit, the total impedance is determined by the equation $Z = \sqrt{R^2 + (X_L - X_C)^2}$. When the source frequency reaches the point where the inductive reactance matches the capacitive reactance, given by $X_L = X_C$, these components effectively cancel each other out. This results in the minimum possible impedance for the circuit, $Z_{\min} = R$, which allows the current to reach its maximum magnitude, thereby maximizing the power dissipated in the circuit.

Similarly, in a circuit consisting only of a pure resistor, the voltage and current are always in phase, meaning there is no phase difference between them. This alignment ensures that the power factor is at its maximum value of unity, allowing the full amount of electrical energy to be converted into heat or work. Because there are no reactive components present to introduce a phase shift and diminish power transfer, this condition of zero phase difference inherently results in maximum power dissipation for the circuit.

When a series LCR circuit exhibits current in phase with the source electromotive force, the system is in resonance, where the inductive and capacitive reactances cancel each other out and reduce the total impedance to the resistance, such that $Z = 20 \Omega$. Dividing the source voltage of 220 V by this impedance yields an RMS current of $I_{rms} = \frac{220}{20} = 11 \text{ A}$. Because the amplitude of the current is defined by the relation $I_{max} = I_{rms} \sqrt{2}$, we calculate $I_{max} = 11 \sqrt{2} \text{ A}$, which simplifies to $\sqrt{121 \times 2} \text{ A} = \sqrt{242} \text{ A}$, confirming that the value of $x$ is 242.

Electrical resonance occurs in an AC circuit only when both an inductor and a capacitor are present, as their reactances must effectively cancel out for the circuit to reach a phase angle of zero, making the claim that resonance requires only one of these components false. Regarding power consumption, an ideal capacitor or inductor maintains a phase difference of $90^\circ$ between voltage and current, which results in a power factor of zero and zero average power dissipation, directly contradicting the assertion that such components consume high power. Because both initial claims misrepresent the fundamental physical behavior of these AC elements, both statements are incorrect.

The quality factor $Q$ of an RLC circuit is defined as the ratio of the resonant angular frequency to the bandwidth, while the bandwidth itself is expressed as the ratio of resistance to inductance. The ratio of the quality factor to the bandwidth simplifies to $\frac{Q}{\Delta \omega} = \frac{\omega_0}{(\Delta \omega)^2} = \omega_0 \cdot \frac{L^2}{R^2}$.

Using the resonance condition $\omega_0 = \frac{1}{\sqrt{LC}}$ with the parameters $L = 3$, $C = 27 \times 10^{-6}$, and $R = 10$ yields $\omega_0 = \frac{1}{\sqrt{3 \times 27 \times 10^{-6}}} = \frac{1}{9 \times 10^{-3}}$. Substituting these values into the expression provides $\frac{1}{9 \times 10^{-3}} \cdot \frac{3^2}{10^2}$, which simplifies to $\frac{1}{9 \times 10^{-3}} \cdot \frac{9}{100}$ and equals 10 s.

The quality factor of a series LCR circuit quantifies the sharpness of the resonance, reflecting how effectively the circuit stores energy relative to the energy dissipated per cycle. This parameter is governed by the values of resistance, inductance, and capacitance according to the formula:
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$

Substituting the given values $R = 100 \, \Omega$, $L = 1 \, \text{H}$, and $C = 6.25 \times 10^{-6} \, \text{F}$ into the equation, we first determine the square root of the ratio $\frac{L}{C}$, which simplifies to $\sqrt{160000}$ or 400. Multiplying this result by the inverse of the resistance, $\frac{1}{100}$, provides the final quality factor of 4.

In a series LCR circuit, the total impedance is defined by the formula $Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}$, which reaches its minimum value of $R$ exactly at the resonant frequency $\omega = \omega_R$. Because the current $I = V/Z$ is inversely related to impedance, it begins to increase as the frequency moves toward resonance, peaks at that frequency, and subsequently decreases as the frequency rises further. Finally, the power factor, calculated as the ratio $\cos \phi = R/Z$, effectively becomes unity at resonance since the impedance perfectly matches the resistance $R$.

In an AC circuit containing a capacitor, the current is determined by the rate of change of charge on the plates, defined by the relationship $I = \frac{dQ}{dt}$. Since the charge stored is related to the applied voltage by $Q = CE$, and the voltage varies sinusoidally as $E = E_0 \sin(\omega t)$, the current is derived by differentiating this expression with respect to time, yielding $I = \omega C E_0 \cos(\omega t)$. The maximum value of this current, $I_0$, is simply the product of the capacitance, the angular frequency, and the peak voltage, expressed as $I_0 = \omega C E_0$

Substituting the values from the problem, the angular frequency $\omega$ is $120 \pi \, \text{rad/s}$, the capacitance $C$ is $150 \times 10^{-6} \, \text{F}$, and the peak voltage $E_0$ is $36 \, \text{V}$. Calculating this product gives $I_0 = 120 \pi \times 150 \times 10^{-6} \times 36$, which simplifies to $0.648 \pi \, \text{A}$. Given that $\pi$ is approximately $3.14$, this calculation results in $I_0 \approx 2.03 \, \text{A}$. Comparing this result to the provided options, the value is closest to $2 \, \text{A}$.

In electrical circuits, resistance $R$, inductive reactance $X_L$, and capacitive reactance $X_C$ all measure opposition to current flow, meaning they fundamentally share the same physical dimension of ohms. When calculating the geometric mean of the reactances, represented by $\sqrt{X_L X_C}$, the resulting value maintains this same dimensional property of ohms. Because the numerator $R$ and the denominator $\sqrt{X_L X_C}$ both possess the same dimensions, their ratio cancels out these units entirely, making the expression $\frac{R}{\sqrt{X_L X_C}}$ dimensionless.

In an ideal LC circuit, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor, remaining conserved throughout the cycle. This energy conservation dictates that the maximum energy stored in the capacitor, which occurs when the current is zero, must equal the maximum energy stored in the inductor, which occurs when the charge on the capacitor is zero. We can express this balance as:

$\frac{q_{\max}^2}{2C} = \frac{1}{2} L I_{\max}^2$

Solving this expression for the maximum current provides the relationship:

$I_{\max} = \frac{q_{\max}}{\sqrt{LC}}$

Substituting the given values of 75 mH for the inductance, 1.2 µF for the capacitance, and 2.7 µC for the maximum charge, we calculate the denominator:

$\sqrt{LC} = \sqrt{75 \times 10^{-3} \times 1.2 \times 10^{-6}} = 3 \times 10^{-4} \text{s}$

Dividing the maximum charge by this time constant results in the maximum current:

$I_{\max} = \frac{2.7 \times 10^{-6} \text{C}}{3 \times 10^{-4} \text{s}} = 9 \times 10^{-3} \text{A} = 9 \text{mA}$

The power factor of an AC circuit represents the efficiency of power delivery and is defined as the ratio of resistance to total impedance, $\cos \phi = \frac{R}{Z}$. In the initial series LR configuration, the impedance is determined by $Z = \sqrt{R^2 + X_L^2}$. Given that the inductive reactance equals the resistance, $X_L = R$, the impedance simplifies to $Z = \sqrt{R^2 + R^2} = R\sqrt{2}$, which results in an initial power factor $P_1 = \frac{1}{\sqrt{2}}$.

When a capacitor is added such that its capacitive reactance equals the inductive reactance, $X_C = X_L$, the circuit reaches resonance. In this condition, the reactive components cancel each other out, and the net impedance reduces to $Z = R$. This yields a new power factor $P_2 = 1$, making the final ratio $P_1 : P_2 = 1 : \sqrt{2}$.