Alternating Current – JEE Main PhysicsPractice Questions & PYQs

Inductive reactance is determined by $X_L = 2\pi fL$, which shows that reactance increases linearly as the frequency rises. In contrast, capacitive reactance is governed by $X_C = \frac{1}{2\pi fC}$, illustrating an inverse relationship where reactance drops as frequency increases. Observing these mathematical trends against the graph, curve $A$ represents the capacitive reactance and curve $B$ represents the inductive reactance.

Energy stored in an inductor is calculated as $U = \frac{1}{2}LI^2$. The rate at which this energy is stored is the time derivative of this quantity, which simplifies to $P = LI\frac{dI}{dt}$. For a series LR circuit connected to an emf $E$, the current evolves over time as $I = \frac{E}{R}(1 - e^{-t/\tau})$, where $\tau = \frac{L}{R}$ is the time constant. Calculating the derivative of the current results in $\frac{dI}{dt} = \frac{E}{L}e^{-t/\tau}$. By substituting the expressions for current and its rate of change into the power formula, we obtain P = L \cdot \[\frac{E}{R}(1 - e^{-t/\tau})]$ \cdot [\frac{E}{L}e^{-t/\tau}]$, which simplifies to$P = \frac{E^2}{R}(e^{-t/\tau} - e^{-2t/\tau})$.

To determine the maximum rate of energy storage, we find the extremum of this power expression by setting its time derivative to zero, a condition met when $e^{-t/\tau} = \frac{1}{2}$. Substituting this value back into the power equation yields $P_{max} = \frac{E^2}{R}(\frac{1}{2} - \frac{1}{4}) = \frac{E^2}{4R}$. Comparing this result to the given expression $\frac{E^a}{2b}$, we equate $a = 2$ and $2b = 4R$, which leads to the relation $b = 2R$. Dividing $b$ by $a$ results in the value $\frac{2R}{2} = R$. Given the provided numerical result, the effective value is 25.

For an ideal transformer supplying a purely resistive load, the power consumed is governed by the relationship between the voltage across the load and its resistance. The power delivered to the secondary circuit is 60 kW, or 60,000 W, and this energy must dissipate across the resistive load connected to the 120 V secondary output. We define this power consumption using the equation $P = \frac{V_s^2}{R_s}$.

By rearranging this expression to solve for the secondary resistance, we have $R_s = \frac{V_s^2}{P}$. Inserting the given values provides $R_s = \frac{120^2}{60,000}$, which simplifies to $\frac{14,400}{60,000} \Omega$. Dividing these numbers results in 0.24 $\Omega$. Since the required unit is milliohms, we multiply this result by 1,000 to obtain a final value of 240 m$\Omega$.

An AC generator produces its maximum electromotive force when the magnetic flux through the coil changes at the fastest rate, which is governed by the relation $\varepsilon_{\text{max}} = NAB\omega$. We begin by converting the rotational frequency of 360 revolutions per minute into angular velocity by calculating $\omega = 2\pi \times (\frac{360}{60}) = 12\pi \text{ rad/s}$. Using the provided value of $\pi = \frac{22}{7}$, the angular frequency is determined to be $12 \times \frac{22}{7} = \frac{264}{7} \text{ rad/s}$. Substituting the known values of $N=100$, $A=14 \times 10^{-2} \text{ m}^2$, and $B=3.0 \text{ T}$ into the equation $NAB\omega$ gives $100 \times 14 \times 10^{-2} \times 3 \times \frac{264}{7}$, which simplifies to $42 \times \frac{264}{7}$, yielding a maximum emf of 1584 V.

The magnitude of rms current in an alternating circuit is inversely dependent on its total impedance, as higher impedance restricts the flow of charge. Because the impedance of circuit (b) is calculated as $Z = \sqrt{R^2 + (X_L - X_C)^2}$, the condition $X_L \neq X_C$ ensures that this impedance is strictly greater than the resistance $R$ found in the resonant circuit (a). Consequently, because circuit (b) inherently possesses higher impedance due to its unequal reactances, the rms current in (b) can never be larger than the current in (a).

Inductive reactance is the opposition offered by an inductor to alternating current, defined by the relationship $X_L = \omega L$. Identifying the necessary components from the source voltage expression, the angular frequency is 628 rad/s and the inductance is 5 mH, which converts to 0.005 H. Substituting these values into the calculation $X_L = 628 \times 0.005$ yields the final reactance of $X_L = 3.14 \Omega$.

The power factor in an alternating current circuit represents the ratio of resistive impedance to total impedance, denoted by the cosine of the phase angle. In a series LCR circuit, this value is defined by $\cos \phi = \frac{R}{Z}$, where the impedance $Z$ is calculated using the resistance $R$ and the net reactance derived from the difference between capacitive reactance $X_C$ and inductive reactance $X_L$.

The net reactance is $X_C - X_L = 130 - 70 = 60$ ohms. Substituting the resistance $R = 80$ ohms and this net reactance into the impedance formula $Z = \sqrt{R^2 + (X_C - X_L)^2}$ provides $Z = \sqrt{80^2 + 60^2} = 100$ ohms. With $Z$ determined, the power factor becomes $\frac{80}{100}$, or $\frac{8}{10}$. Since this ratio is equivalent to $\frac{x}{10}$, it follows that $x = 8$.

In an LCR series circuit, the current reaches its maximum amplitude at resonance, a state where the inductive reactance exactly balances the capacitive reactance. This condition occurs at the resonant frequency, which is governed by the relationship $f = \frac{1}{2 \pi \sqrt{LC}}$.

Rearranging this formula to solve for the inductance involves squaring both sides to isolate the term for L, which yields the expression $L = \frac{1}{4 \pi^2 f^2 C}$. Inserting the given values where the frequency is 2000 Hz, the capacitance is 62.5 nF, and taking \pi ^2 as 10, we obtain the calculation $L = \frac{1}{4 \times 10 \times (2000)^2 \times 62.5 \times 10^{-9}}$. Evaluating the denominator simplifies the calculation to 10, resulting in an inductance of 0.1 H, which is equivalent to 100 mH.

To determine the peak current flowing through a series LCR circuit, we use the generalized version of Ohm's law, where the current amplitude equals the ratio of peak input voltage to total circuit impedance. This impedance is calculated using the relationship $Z = \sqrt{R^2 + (X_L - X_C)^2}$. By substituting the resistance of 80 $\Omega$, the inductive reactance of 100 $\Omega$, and the capacitive reactance of 40 $\Omega$ into this formula, we find that $Z = \sqrt{80^2 + (100 - 40)^2} = 100$ $\Omega$. With a peak input voltage of 2500 V, the current amplitude is simply the peak voltage divided by the total impedance, which results in 25 A.

In electrical circuits, the product $RC$ and the ratio $\frac{L}{R}$ both serve as characteristic time constants, defining the rate at which transient processes occur in capacitive and inductive systems. The resonant behavior of an $LC$ circuit involves the term $\sqrt{LC}$, which mathematically corresponds to the reciprocal of angular frequency and therefore possesses the dimension of time. In contrast, examining the relationship between energy storage, $\frac{1}{2}LI^2 = \frac{1}{2}CV^2$, reveals that the ratio $\frac{L}{C}$ is equivalent to $R^2$, a quantity representing resistance squared rather than a temporal duration.

The total reactance of an AC circuit is determined by the formula $X = |X_C - X_L|$, which confirms that the reactance becomes zero when the capacitive reactance equals the inductive reactance, a condition entirely possible in circuits containing both elements. The second statement is incorrect because the average power delivered in an AC circuit depends on the phase angle, and for a purely reactive circuit where the phase difference is $90^\circ$, the average power is zero. Thus, the condition described in the first statement is physically achievable, whereas the absolute claim made in the second statement regarding power is false.

In an AC circuit where the inductor and capacitor are arranged in parallel, resonance occurs when $X_L = X_C$. Under these specific conditions, the equation $\frac{1}{Z} = \sqrt{(\frac{1}{X_L} - \frac{1}{X_C})^2}$ results in an admittance of zero, which implies that the total circuit impedance $Z \to \infty$. Given the 110 V and 50 Hz source, this infinite opposition to current flow ensures that no current can traverse the circuit branches. Consequently, the current through the 55 ohm resistance is 0 A.

The resonant frequency of a series RLC circuit is determined by the expression $f = \frac{1}{2\pi\sqrt{LC}}$, while the quality factor is expressed as $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$. When the inductance is doubled to $2L$ and the capacitance is doubled to $2C$, the product under the square root in the frequency formula becomes $(2L)(2C) = 4LC$, causing the new resonant frequency to be $f_2 = \frac{1}{2\pi\sqrt{4LC}}$. This simplifies mathematically to $f_2 = \frac{1}{2} \cdot \frac{1}{2\pi\sqrt{LC}}$, confirming that the new frequency is exactly half of the original frequency $f_1$. Simultaneously, the ratio of inductance to capacitance within the quality factor formula transforms into $\sqrt{\frac{2L}{2C}}$, which reduces back to the original ratio $\sqrt{\frac{L}{C}}$. Therefore, the quality factor remains unchanged, resulting in $Q_2 = Q_1$.

The principle of energy conservation governs the behavior of an LC circuit, where the maximum electrostatic energy initially stored in the capacitor fully converts into the maximum magnetic energy within the inductor during oscillation. Initially, the capacitor acts as the source of all energy, holding a charge determined by its capacitance and the supply voltage. Using the relation $Q = CV$ with the provided capacitance of $500 \times 10^{-6} \text{ F}$ and the voltage of $100 \text{ V}$, the total charge on the capacitor is $5 \times 10^{-2} \text{ C}$.

At the point of peak current, the electrostatic energy, represented as $\frac{Q^2}{2C}$, is equal to the magnetic energy in the inductor, $\frac{1}{2} L I_{\max}^2$. Equating these values allows us to solve for the maximum current using the formula:
$I_{\max} = \frac{Q}{\sqrt{LC}}$
By substituting the inductance $L = 50 \times 10^{-3} \text{ H}$ and the capacitance, the denominator becomes $\sqrt{50 \times 10^{-3} \times 500 \times 10^{-6}}$. Simplifying the expression inside the square root gives $\sqrt{25 \times 10^{-6}}$, which evaluates to $5 \times 10^{-3}$. Dividing the total charge of $5 \times 10^{-2} \text{ C}$ by this result leads to a maximum current of $10 \text{ A}$.

Minimum impedance in an alternating current circuit occurs at resonance, a state where the inductive reactance and capacitive reactance are equal and opposite. Adjusting the series inductance to meet this condition allows the line to effectively balance out its distributed capacitance, as defined by the relationship:

$\omega L = \frac{1}{\omega C}$

The total capacitance of the line is determined by multiplying its total length of 100 km by 0.01 $\mu$F/km, resulting in 1 $\mu$F, or $1 \times 10^{-6}$ F. With a signal frequency of 0.5 kHz, which is 500 Hz, the angular frequency $\omega$ is calculated as $2\pi \times 500$, resulting in $1000\pi$ rad/s. Rearranging the resonance condition to solve for inductance gives the following expression:

$L = \frac{1}{\omega^2 C}$

Substituting the known values leads to $L = 1 / ((1000\pi)^2 \times 10^{-6})$. Simplifying the denominator, $(1000\pi)^2$ becomes $10^6 \pi^2$, and multiplying by $10^{-6}$ leaves just $\pi^2$. Since the approximation $\pi^2 = 10$ is given, the calculation simplifies to $L = 1/10$ H, which is equal to 100 mH.

The resonant frequency of a series LCR circuit is determined by the equation $\omega_0 = \frac{1}{\sqrt{LC}}$. To obtain a higher resonant frequency, the value of the product $LC$ must be reduced, which requires lowering the effective capacitance of the circuit. Since adding another capacitor in series decreases the total equivalent capacitance, this configuration directly results in an increase in the resonant frequency.

An AC generator functions by converting mechanical energy into electrical energy, often through the rotation of a coil within a magnetic field. Meanwhile, a galvanometer is a highly sensitive instrument specifically designed to detect the presence of current in a circuit. Transformers are essential devices that utilize electromagnetic induction to change an alternating voltage to a smaller or greater value depending on the requirements of the circuit. Finally, metal detectors operate by utilizing the principle of resonance in an AC circuit to identify the presence of metal objects, which completes the matching of these common electrical systems.

The power factor in a series $AC$ circuit is defined as the ratio of resistance to impedance, expressed as $\cos \phi = \frac{R}{Z}$. For an $LR$ circuit where $X_L = R$, the total impedance is $Z = \sqrt{R^2 + X_L^2} = R\sqrt{2}$, which results in an initial power factor of $P_1 = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$. When a capacitor is introduced such that $X_C = X_L$, the circuit reaches resonance, canceling the reactances and leaving the circuit purely resistive with $Z = R$. This results in a resonance power factor of $P_2 = \frac{R}{R} = 1$, and the required ratio is $\frac{P_1}{P_2} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$.

In a purely inductive circuit, the alternating voltage across the inductor consistently leads the current by a phase angle of 90 degrees. This indicates that the peak voltage is calculated using the relationship $V_m = I_m \omega L$, while the phase angle of the voltage waveform is the current's phase angle shifted ahead by 90 degrees.

With a peak current of 5 and an angular frequency of $49 \pi$, we determine the peak voltage by substituting the inductance of 0.03 H and the value 22/7 for $\pi$. The calculation follows the expression $V_m = 5 \times 49 \times \frac{22}{7} \times 0.03$, which results in 23.1. Applying the phase shift, the starting angle of $-30^\circ$ becomes $-30^\circ + 90^\circ = +60^\circ$. Combining these parameters yields the final voltage expression $v(t) = 23.1 \sin(49 \pi t + 60^\circ)$.

The production of heat in a resistor follows Joule's law, where the thermal energy dissipated is governed by the formula $H = I^2 Rt$. In a direct current circuit, the current remains constant, meaning the heat generated is calculated by squaring the current and multiplying it by the resistance and time. For the first case, using a current of 4 A and a resistance of 3 Ω, the heat generated is $H_1 = 4^2 \times 3 \times t$, which simplifies to $48t$.

In an alternating current circuit, heating is determined by the root-mean-square current, which is derived from the peak current $I_0$ as $I_{\text{rms}} = I_0 / \sqrt{2}$. Substituting the peak value of 4 A into this relation gives an RMS current whose square is $I_{\text{rms}}^2 = 4^2 / 2 = 8$. When this current flows through a resistance of 2 Ω, the heat produced is $H_2 = 8 \times 2 \times t$, resulting in $16t$. By comparing the two quantities through the ratio $H_1 / H_2 = 48t / 16t$, the final result is 3:1.