Alternating Current – JEE Main PhysicsPractice Questions & PYQs

The behavior of an inductor is defined by its capacity to oppose any instantaneous change in the current flowing through it, which ensures that the current magnitude remains identical at the exact moment the switch position changes. With $T_1$ connected to $T_2$ at steady state, the current in the circuit is calculated as $I = 6 / 6 = 1$ A. Immediately after the connection is shifted from $T_2$ to $T_3$, the inductor maintains this same flow of $1$ A through the new segment containing the $3\ \Omega$ resistor. Applying Ohm's law, the potential drop across this resistor is $V = I \times r = 1 \times 3 = 3$ V.

In a series L-C-R circuit operating at resonance, the inductive and capacitive reactances cancel each other out, meaning the total impedance equals the resistance of 8 ohms. This condition implies the circuit behaves as a purely resistive load, so the power dissipation relies entirely on the root-mean-square voltage. Although the inductance of 24 mH and capacitance of 60 microfarads define the resonance frequency, they do not influence the power dissipated at that frequency.

The RMS voltage is determined using $V_{rms} = \frac{V_0}{\sqrt{2}}$. Substituting the peak voltage of 250 volts, the power is found using the formula $P = \frac{V_{rms}^2}{R}$. This leads to the following calculation:
$P = \frac{(250/\sqrt{2})^2}{8} = \frac{62500}{16} = 3906.25$
This value is 3906.25 watts, which is approximately 3.906 kW, and rounding to the nearest integer yields $x = 4$.

We start by determining the total current flowing through the circuit by applying Ohm's law to the resistor $R$. With a voltage drop of $V_R = 15 \, \text{V}$ across a resistance of $60 \, \Omega$, the main current is $I = 15 / 60 = 0.25 \, \text{A}$. Since the capacitor carries this main current, its reactance is calculated as $X_C = \frac{V_C}{I} = \frac{10}{0.25} = 40 \, \Omega$. Using the relation $X_C = \frac{1}{\omega C}$ with an angular frequency $\omega = 100 \, \text{rad/s}$, we find the capacitance as $C = \frac{1}{100 \times 40} = 0.25 \times 10^{-3} \, \text{F}$, which is $250 \, \mu\text{F}$.

To find the inductance, we examine the parallel branch containing the coil and the resistor $R'$. The potential difference across this branch is $20 \, \text{V}$, so the current flowing through the resistor $R'$ is $I_{R'} = \frac{20}{40} = 0.5 \, \text{A}$. By applying Kirchhoff's current law at the junction, the current through the inductor is the magnitude of the difference between the main current and the current through $R'$, resulting in $|I_L| = |0.25 - 0.5| = 0.25 \, \text{A}$. The inductive reactance is then $X_L = \frac{V_L}{|I_L|} = \frac{20}{0.25} = 80 \, \Omega$, which leads to an inductance value of $L = \frac{X_L}{\omega} = \frac{80}{100} = 0.8 \, \text{H}$.

In an RL series circuit, the combined opposition to current, known as impedance, arises from both the resistor and the inductor, while the phase relationship between the voltage and current depends on the ratio of inductive reactance to resistance. The inductive reactance is $X_L = 2\pi f L = 2 \times \frac{22}{7} \times 50 \times 0.07 = 22 \Omega$. With a resistance of $12 \Omega$, the total impedance is $Z = \sqrt{R^2 + X_L^2} = \sqrt{12^2 + 22^2} = \sqrt{628} \approx 25.06 \Omega$. The current flowing through the circuit is $I = \frac{V}{Z} = \frac{220}{25.06} \approx 8.77 A$, which rounds to $8.8 A$. The phase angle between the voltage and current is given by $\phi = \tan^{-1}(\frac{X_L}{R})$, which yields $\phi = \tan^{-1}(\frac{22}{12}) = \tan^{-1}(\frac{11}{6})$.

A hot wire ammeter registers the root mean square value of an alternating current, which quantifies the effective heating effect produced by the signal. To determine this reading for the expression $I=I_{1} \sin \omega t+I_{2} \cos \omega t$, we calculate the square root of the mean value of the squared current over one complete time period.

Expanding the square of the current yields the following expression:
$I^2 = I_{1}^{2} \sin^2 \omega t + I_{2}^{2} \cos^2 \omega t + 2 I_{1} I_{2} \sin \omega t \cos \omega t$
When averaging this expression over a full cycle, the cross-product term $2 I_{1} I_{2} \sin \omega t \cos \omega t$ becomes zero because it oscillates symmetrically about the axis. The average value for both $\sin^2 \omega t$ and $\cos^2 \omega t$ over a full period is $\frac{1}{2}$. Consequently, the mean of the squared current simplifies to $\frac{I_{1}^{2}}{2} + \frac{I_{2}^{2}}{2}$. The ammeter reading is the square root of this average, which results in $\sqrt{\frac{I_{1}^{2}+I_{2}^{2}}{2}}$.

The inductive reactance of an AC circuit is defined by the formula $X_{L} = 2 \pi f L$, which demonstrates that reactance is directly proportional to the supply frequency. When the frequency is reduced by half, this linear relationship dictates that the inductive reactance must also decrease to half of its original value. Because the current in the circuit is governed by the relationship $I = \frac{V}{X_{L}}$, halving the reactance in the denominator effectively doubles the resulting current. Consequently, a decrease in the opposition offered by the inductor results in a two-fold increase in the magnitude of the current flowing through the circuit.

When a current consists of both an alternating sinusoidal component and a steady direct current, the total root mean square (RMS) value is determined by the square root of the sum of the squares of the RMS values of each independent part, expressed as $I_{\text{rms}}^2 = I_{1,\text{rms}}^2 + I_{2,\text{rms}}^2$. For the alternating term $\sqrt{42} \sin (\frac{2\pi}{T} t)$, the peak amplitude is $\sqrt{42}$, yielding an RMS value of $\frac{\sqrt{42}}{\sqrt{2}}$, which simplifies to $\sqrt{21}$ A. The constant component of 10 A maintains an RMS value equal to its magnitude of 10 A. Substituting these individual values into the superposition equation provides an RMS result of $\sqrt{(\sqrt{21})^2 + 10^2}$, which simplifies to $\sqrt{21 + 100}$ or $\sqrt{121}$ A, resulting in a final current of 11 A.

The power factor of a series AC circuit is governed by the ratio $\cos \phi = \frac{R}{Z}$, where the impedance $Z$ is calculated using $Z = \sqrt{R^2 + (X_L - X_C)^2}$. For the initial circuit containing only the inductor and resistor, the capacitive reactance is zero, yielding an impedance of $Z_1 = \sqrt{R^2 + (3R)^2} = R\sqrt{10}$ and an initial power factor of $\cos \phi_1 = \frac{1}{\sqrt{10}}$.

Introducing a capacitor with a reactance of $2R$ creates a net reactance of $3R - 2R = R$, modifying the impedance to $Z_2 = \sqrt{R^2 + R^2} = R\sqrt{2}$. This results in a new power factor of $\cos \phi_2 = \frac{1}{\sqrt{2}}$. Dividing the new power factor by the old one produces the ratio $\frac{1/\sqrt{2}}{1/\sqrt{10}} = \sqrt{\frac{10}{2}} = \sqrt{5}$. As the problem defines this ratio as $\sqrt{5}:x$, it follows that $x$ is $1$.

Bandwidth in a series L-C-R circuit serves as a measure of the frequency range around resonance, determined by the relationship $\beta = \frac{R}{L}$. Since the bandwidth is directly proportional to the resistance, increasing the resistance inevitably causes the resonance circuit's bandwidth to widen. Meanwhile, the resonance frequency $\omega = \frac{1}{\sqrt{LC}}$ remains independent of resistance, and the quality factor $Q = \frac{\omega L}{R}$ decreases as the resistance grows.

Average power dissipation in an AC circuit is determined by $P_{\text{avg}} = \frac{V_{\text{rms}}^2 R}{Z^2}$, where $Z$ is the impedance and $R$ is the resistance. For the power dissipated in both circuits to be identical, the reactive circuit must function at resonance, where the total impedance simplifies to the value of the resistance because the inductive and capacitive reactances cancel each other out. Equating the inductive reactance $\omega L$ to the capacitive reactance $\frac{1}{\omega C}$ establishes the resonant condition $x_C = x_L$. Rearranging the resonant frequency formula to $\omega^2 = \frac{1}{LC}$ and substituting the provided values results in $\omega^2 = \frac{10^6}{0.1 \times 40}$. Solving this equation yields $\omega = 500$ rad/s.

The quality factor of a series $L-C-R$ circuit, which represents the sharpness of resonance, is defined by the ratio of inductive reactance to resistance as $Q = \frac{X_L}{R}$. Doubling the inductance increases the inductive reactance by a factor of two, while halving the resistance reduces the denominator by a factor of two, causing the overall quality factor to increase four-fold. This relationship is represented as:

$Q_{new} = \frac{2X_L}{R / 2} = 4 \left( \frac{X_L}{R} \right)$

With an initial quality factor of 100, applying this four-fold increase results in a final value of 400.

When an LCR circuit reaches its resonant frequency, the inductive reactance and capacitive reactance effectively cancel each other out. This causes the total impedance of the circuit to become purely resistive, meaning the impedance is equal to the resistance of the resistor. Because the circuit behaves as a purely resistive load at this state, the power dissipated can be calculated using the formula $P = \frac{V^2}{R}$. Substituting the values of 250 V for voltage and 5 Ω for resistance into this equation yields $P = \frac{250^2}{5}$. Performing this calculation results in 12500 W, which can be expressed as $125 \times 10^2 \text{ W}$.

Peak current and resonance frequency in an LCR series circuit depend on the balance between impedance and frequency. First, calculate the inductive reactance $X_L = \omega L = 100 \times 100 \times 10^{-3} = 10$ Ω and the capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = 100$ Ω. The total impedance of the circuit is found using the relation $Z = \sqrt{(X_C - X_L)^2 + R^2} = \sqrt{(100 - 10)^2 + 120^2} = 153.3$ Ω. By dividing the peak voltage by this impedance, the peak current is determined as $I_0 = \frac{V_0}{Z} = \frac{30}{153.3} \approx 0.2$ A. The resonant frequency is calculated using the formula $f = \frac{1}{2 \pi \sqrt{LC}}$, which, when substituting the given inductance and capacitance values, simplifies to $f = \frac{1}{2 \pi \sqrt{100 \times 10^{-3} \times 100 \times 10^{-6}}} = 50$ Hz.

The quality factor of a resonant circuit serves as a measure of the circuit's selectivity or the sharpness of its resonance, defined as the ratio of inductive reactance to the circuit resistance. For a series resonant circuit, this relationship is expressed by the formula $Q = \frac{\omega L}{R}$, where $\omega$ represents the angular frequency. Because the angular frequency relates to the frequency $f$ through the expression $\omega = 2 \pi f$, the equation for the quality factor becomes:

$Q = \frac{2 \pi f L}{R}$

By substituting the provided values $L = 2 \times 10^{-4}$ H, $R = 6.28$ $\Omega$, $f = 10 \times 10^6$ Hz, and $\pi = 3.14$ into this expression, we can determine the circuit's performance. Since the product of $2$ and $3.14$ is exactly $6.28$, the term in the numerator effectively cancels out the resistance value in the denominator. This simplification leaves the equation as $Q = 10 \times 10^6 \times 2 \times 10^{-4}$, which yields $Q = 2000$.

In power supply circuits, specific components are used to transform and regulate electrical energy to ensure it is suitable for various applications. A transformer serves the primary purpose of adjusting the AC voltage level, effectively stepping it up or down as needed, while the rectifier subsequently converts this AC into DC. Since the raw DC output from a rectifier contains fluctuations known as ripples, a filter is employed to smooth the signal. Finally, the stabiliser provides a stable, constant output voltage regardless of any changes in the input or load current, completing the chain of power management.

A transformer is classified by its capacity to either increase or decrease electrical potential based on the relationship between its input and output voltages. By dividing the power delivered to the load by the current flowing through it, we calculate the secondary voltage using the formula $V_{2} = \frac{P}{I_{L}}$ which gives $V_{2} = \frac{60}{0.11} \approx 545.45 \text{ V}$. Because the calculated secondary voltage of 545.45 V exceeds the primary voltage of 220 V, specifically $V_{2} > V_{1}$, the device functions as a step-up transformer.

Resonance in an LC circuit occurs when the natural frequency of the inductor-capacitor system perfectly matches the frequency of the incoming electromagnetic wave. This resonant frequency $f$ relates to the wave speed $v$ and wavelength $\lambda$ as $f = \frac{v}{\lambda}$, and it is also determined by the circuit parameters through the relation $f = \frac{1}{2 \pi \sqrt{LC}}$.

Equating these expressions leads to $\frac{v}{\lambda} = \frac{1}{2 \pi \sqrt{LC}}$, and squaring both sides allows for solving the inductance $L$ with the following formula:
$L = \frac{\lambda^2}{4 \pi^2 v^2 C}$

By substituting the known values $\lambda = 960$ m, $v = 3 \times 10^8$ m/s, and $C = 2.56 \times 10^{-6}$ F, and applying the common approximation $\pi^2 \approx 10$, the calculation proceeds as follows:
$L = \frac{960^2}{4 \times 10 \times (3 \times 10^8)^2 \times 2.56 \times 10^{-6}}$
$L = \frac{921600}{40 \times 9 \times 10^{16} \times 2.56 \times 10^{-6}}$
$L = \frac{921600}{921.6 \times 10^{10}} = 1000 \times 10^{-10} = 10 \times 10^{-8} \text{ H}$

The instantaneous current in an AC circuit follows the equation $i = i_0 \sin(\omega t)$. The current reaches its maximum peak when the phase angle $\omega t$ equals $\pi/2$ radians, and it hits its root mean square value of $i_0 / \sqrt{2}$ when the phase angle equals $\pi/4$ radians. The time elapsed between these two states is calculated by the difference in phase divided by the angular frequency, represented by $\Delta t = (\pi/2 - \pi/4) / \omega = \pi / (4\omega)$. By substituting the angular frequency $\omega = 2 \pi f$ with the supply frequency of 50 Hz, the time duration becomes 1/400 seconds. Converting this calculation into milliseconds results in 2.5 ms.

Resonance in a series circuit is achieved when the inductive reactance and capacitive reactance are equal, causing the total impedance of the circuit to be at its minimum. This phenomenon occurs at a resonant frequency defined by the relationship $f = \frac{1}{2\pi\sqrt{LC}}$.

Rearranging this expression to isolate the inductance results in the equation $L = \frac{1}{4\pi^2 f^2 C}$. Substituting the resonant frequency of 60 Hz and the capacitance of 0.1 µF, which is equivalent to 10^-7 F, into the equation provides the necessary value. Performing the computation yields an inductance of 70.3 H.

An RC circuit inherently resists instantaneous changes in charge, meaning the capacitor takes time to adjust whenever the input square wave switches between high and low levels. As the voltage shifts, the capacitor accumulates charge according to $Q = Q_{max}(1 - e^{-\frac{t}{RC}})$ during the charging phase and releases it following $Q = Q_{max}e^{-\frac{t}{RC}}$ during the discharging phase. This continuous, time-dependent cycle creates smooth, exponential curves at the edges of the waveform instead of the sharp corners present in the input signal.