Alternating Current – JEE Main PhysicsPractice Questions & PYQs

The energy $U$ stored in an inductor relates to the instantaneous current $i$ via the expression $U = \frac{1}{2} L i^2$, which implies that the maximum energy $U_{\text{max}}$ is attained at the maximum current $I_{\text{max}}$, or $U_{\text{max}} = \frac{1}{2} L I_{\text{max}}^2$. When the stored energy reaches $1/n$ of its maximum value, we have the condition $\frac{1}{2} L i^2 = \frac{1}{n} (\frac{1}{2} L I_{\text{max}}^2)$, which simplifies to the requirement that $i = \frac{I_{\text{max}}}{\sqrt{n}}$. Substituting this into the standard L-R circuit growth equation $i = I_{\text{max}}(1 - e^{-Rt/L})$ yields $\frac{I_{\text{max}}}{\sqrt{n}} = I_{\text{max}}(1 - e^{-Rt/L})$, which simplifies further to $e^{-Rt/L} = 1 - \frac{1}{\sqrt{n}} = \frac{\sqrt{n}-1}{\sqrt{n}}$. By inverting this expression to solve for the positive exponent and taking the natural logarithm of both sides, we obtain the time $t$ as:

$t = \frac{L}{R} \ln\left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$

Calculating the duration for a resistor to heat up requires determining the electrical power dissipated within the series RLC circuit. The current flow is restricted by the total impedance $Z$, which combines the resistance $R$ and the net reactance of the inductor $L$ and capacitor $C$ according to:
$|Z| = \sqrt{R^2 + (2 \pi f L - \frac{1}{2 \pi f C})^2}$
Substituting the given values of $f = 750 \text{ Hz}$, $R = 100 \, \Omega$, $L = 0.1803 \text{ H}$, and $C = 10 \text{ \mu F}$ into this expression yields a total impedance magnitude of $|Z| \approx 834 \, \Omega$.

The power $P$ dissipated specifically by the resistor is derived from the root-mean-square voltage $V_{\text{rms}} = 20 \text{ V}$ and the calculated impedance:
$P = \frac{V_{\text{rms}}^2 R}{|Z|^2}$
This calculation results in $P \approx 0.0575 \text{ J/s}$. Since the resistor requires a total energy of $H = S \Delta\theta$ to reach the desired temperature change, with a heat capacity $S = 2 \text{ J/}^\circ\text{C}$ and $\Delta\theta = 10^\circ\text{C}$, the total energy needed is $20 \text{ J}$. Dividing this energy by the power rate $P$ gives the time $t = 20 / 0.0575$, which is approximately $348 \text{ s}$.

The current in an inductor-resistor series circuit grows over time according to the expression $i(t) = \frac{E}{R}(1 - e^{-Rt/L})$, where $\frac{E}{R}$ represents the maximum steady-state current reached as time approaches infinity. To determine the ratio of the current at $t = \infty$ to the current at $t = 40 \text{ s}$, we take the reciprocal of the growth factor $(1 - e^{-Rt/L})$, resulting in the expression $\frac{1}{1 - e^{-Rt/L}}$.

Substituting the circuit parameters leads to an exponent of 2 in the term $e^{-Rt/L}$. Using the provided value $e^2 = 7.389$, we determine $e^{-2}$ as $\frac{1}{7.389} \approx 0.135$. The ratio calculation then becomes $\frac{1}{1 - 0.135}$, which simplifies to $\frac{1}{0.865}$. Evaluating this division yields approximately $1.156$.

The quality factor of a series RLC circuit characterizes the sharpness of its resonance and is defined by the formula $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$. Substituting the given values $R=100$, $L=80 \times 10^{-3}$, and $C=2 \times 10^{-6}$ into this expression yields the equation $Q = \frac{1}{100} \sqrt{\frac{80 \times 10^{-3}}{2 \times 10^{-6}}}$. The ratio of the inductance to the capacitance inside the radical simplifies to 40000, and its square root is 200. Dividing this result by 100 gives a final quality factor of 2.

In an AC circuit containing an inductor and a capacitor in series, the current is determined by the net reactance, which is the magnitude of the difference between the capacitive reactance and the inductive reactance. Because the voltage across an ideal capacitor lags the current by $\pi/2$ radians while the voltage across an ideal inductor leads the current by $\pi/2$ radians, the phase relationship of the total current relative to the voltage depends on which component dominates the circuit impedance.

To calculate the reactances, start with the inductive reactance $X_L = 314 \times 40 \times 10^{-3} = 12.56 \Omega$ and the capacitive reactance $X_C = \frac{1}{314 \times 100 \times 10^{-6}} = 31.84 \Omega$. Because the capacitive reactance is greater than the inductive reactance, the circuit behaves capacitively, meaning the current leads the voltage by $\pi/2$ radians. The peak current $I_m$ is determined by the relationship $I_m = \frac{10}{31.84 - 12.56} = 0.52 \text{ A}$. Incorporating this peak current with the phase shift, the current is described by $I(t) = 0.52 \sin(314t + \pi/2)$, which is mathematically equivalent to $I(t) = 0.52 \cos(314t)$.

When an AC signal passes through an inductive coil, the circuit exhibits both resistance and inductive reactance, which combine into a total impedance defined by the equation $Z = \sqrt{R^2 + X_L^2} = 100 \ \Omega$. The phase difference where voltage leads the current by 45 degrees implies that $\tan 45^\circ = \frac{X_L}{R} = 1$, showing that the resistance and inductive reactance are identical in magnitude.

By substituting this condition into the impedance formula, we arrive at $\sqrt{X_L^2 + X_L^2} = 100$, which reduces to $X_L\sqrt{2} = 100$ and yields $X_L = 50\sqrt{2} \ \Omega$. Using the relation for inductive reactance $X_L = 2\pi f L$ with a frequency of 1000 Hz, we calculate $L = \frac{50\sqrt{2}}{2\pi \cdot 1000}$. This evaluates to $L \approx 1.1 \times 10^{-2} \text{ H}$.

The equivalence between an LCR circuit and a mechanical oscillator stems from the fact that both systems are governed by differential equations of identical mathematical form. Kirchhoff's Voltage Law yields the circuit equation $L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = 0$, while the equation of motion for a damped mass-spring system is $m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0$. Comparing these terms demonstrates that mass $m$ is analogous to inductance $L$, the damping coefficient $b$ corresponds to resistance $R$, and the spring constant $k$ maps to the inverse capacitance $1/C$.

Achieving a power factor of unity in an inductive circuit requires resonance, where the capacitive reactance introduced must exactly cancel the existing inductive reactance. We determine the circuit impedance using the power formula $P = \frac{E_{rms}^2}{Z} \cos \phi$. Substituting the provided values $400 = \frac{250^2}{Z} \times 0.8$ reveals the impedance to be $Z = 125 \, \Omega$.

From the power factor of $0.8$, we identify that $\sin \phi = 0.6$, allowing us to calculate the inductive reactance using $X_L = Z \sin \phi$. This yields $X_L = 125 \times 0.6 = 75 \, \Omega$. To achieve unity power factor, we set the capacitive reactance equal to the inductive reactance, so $\frac{1}{2\pi f C} = 75 \, \Omega$. With a frequency of 50 Hz, we solve for capacitance as $C = \frac{1}{2\pi \times 50 \times 75} = \frac{1}{7500 \pi} \, \text{F}$. Converting this value to microfarads gives $C = \frac{10^6}{7500 \pi} \, \mu\text{F} = \frac{400}{3\pi} \, \mu\text{F}$, which corresponds to the form $\frac{n}{3\pi} \, \mu\text{F}$, making $n = 400$.

When a switch closes in a series circuit containing an inductor and resistors, the current rises from zero towards its saturation value defined by the relation $i = i_{max} (1 - e^{-t/\tau})$, where the time constant is $\tau = \frac{L}{R+r}$. Given the combined resistance of $1.0 \, \Omega$ and inductance of $0.01 \, \text{H}$, the circuit time constant is calculated as $\tau = \frac{0.01}{1.0} = 0.01 \, \text{s}$. To find the duration for the current to reach $80\%$ of its saturation limit, we equate $0.8 = 1 - e^{-t/\tau}$, which reduces to $e^{t/\tau} = 5$ or $t = \tau \ln 5$. Using the provided value of $\ln 5 = 1.6$, the required time becomes $t = 0.01 \times 1.6$, which equals $0.016 \, \text{s}$.

The phase difference between currents in parallel branches of an AC circuit is determined by the individual impedance of each branch relative to the source voltage. In an inductive $L-R_1$ path, the current $I_1$ lags the source voltage by a phase angle $\phi_1$, which is determined by the inductive reactance $X_L = \omega L$ and resistance $R_1$, such that $\tan \phi_1 = \frac{X_L}{R_1}$. Given $\omega = 100$ rad/s and $L = \frac{\sqrt{3}}{10}$ H, the reactance is $X_L = 100 \times \frac{\sqrt{3}}{10} = 10\sqrt{3}$ Ω. With $R_1 = 10$ Ω, the relation becomes $\tan \phi_1 = \frac{10\sqrt{3}}{10} = \sqrt{3}$, which establishes a lagging phase angle of $\phi_1 = 60^\circ$.

In the capacitive $C-R_2$ path, the current $I_2$ leads the source voltage by a phase angle $\phi_2$, where $\tan \phi_2 = \frac{X_C}{R_2}$ and the capacitive reactance is $X_C = \frac{1}{\omega C}$. Using the provided capacitance $C = \frac{\sqrt{3}}{2} \times 10^{-6}$ F and $\omega = 100$ rad/s, the reactance is calculated as $X_C = \frac{1}{100 \times \frac{\sqrt{3}}{2} \times 10^{-6}} = \frac{20000}{\sqrt{3}}$ Ω. When considering a resistance $R_2 = 20$ k$\Omega$, the phase angle satisfies $\tan \phi_2 = \frac{20000/\sqrt{3}}{20000} = \frac{1}{\sqrt{3}}$, corresponding to a leading phase angle of $\phi_2 = 30^\circ$. Because one current lags and the other leads the source voltage, the total phase difference between $I_1$ and $I_2$ is the sum of these magnitudes, $60^\circ + 30^\circ = 90^\circ$.

An ideal transformer functions based on the principle of conservation of energy, where the power delivered by the secondary coil equals the power supplied to the primary coil. First, determine the output voltage across the secondary coil using the output power and secondary current. Given an output power of $P = 2200 \text{ W}$ and a secondary current of $I_s = 10 \text{ A}$, the secondary voltage is calculated using $V_s = \frac{P}{I_s}$, which yields $V_s = \frac{2200}{10} = 220 \text{ V}$.

With the secondary voltage established, apply the transformer turns ratio, which states that the ratio of the primary voltage to the secondary voltage is proportional to the ratio of the number of turns in each coil, $\frac{V_p}{V_s} = \frac{N_p}{N_s}$. Substituting the given values of $N_p = 300$ and $N_s = 150$ into this relationship gives $\frac{V_p}{220} = \frac{300}{150}$, simplifying to $\frac{V_p}{220} = 2$, which confirms that the input primary voltage is $V_p = 440 \text{ V}$.

To determine the primary current, utilize the principle that the input power must match the output power, expressed as $P = V_p I_p$. Using the newly calculated input voltage of $440 \text{ V}$ and the total power of $2200 \text{ W}$, the equation $2200 = 440 \times I_p$ reveals the primary current as $I_p = \frac{2200}{440} = 5 \text{ A}$. This relationship is consistent with the inverse proportionality between current and the number of turns, $\frac{I_p}{I_s} = \frac{N_s}{N_p}$, where $\frac{I_p}{10} = \frac{150}{300}$ confirms $I_p = 5 \text{ A}$.

Transformer efficiency describes the effectiveness of power transfer, calculated as the ratio of output power to input power, expressed as $\eta = \frac{V_S I_S}{V_P I_P}$. Given an efficiency of $0.9$ with an input voltage of $2300$ V, a primary current of $5$ A, and an output voltage of $230$ V, the relationship is defined by substituting these known values into the equation $0.9 = \frac{230 \times I_S}{2300 \times 5}$. By simplifying the voltage ratio of $\frac{230}{2300}$ to $\frac{1}{10}$, the expression becomes $0.9 = \frac{I_S}{50}$, which confirms that the output current $I_S$ is $0.9 \times 50$, resulting in $45$ A.

When a circuit with an inductor reaches a steady state long after the switch is closed, the inductor acts as an ideal conductor with zero resistance. This transformation effectively places the two 5 ohms resistors in a parallel configuration across the 15 V battery. The total current follows the relationship $I = \frac{2\varepsilon}{R}$, where $\varepsilon$ is the battery voltage and $R$ is the resistance of each component. Substituting the given values yields $I = \frac{2 \times 15}{5}$, resulting in a total current of 6 A.

In a purely resistive circuit, the current waveform mirrors the applied voltage, meaning both reach their maximum amplitude at a phase angle of $90^\circ$ and their half-maximum values at $30^\circ$. With an angular frequency of $100\pi$ rad/s, the total time period for one full cycle is $T = \frac{2\pi}{100\pi} = \frac{1}{50}$ s. The time required to reach the peak value is a quarter of this period, while reaching the half-peak value occurs at one-twelfth of the period. Subtracting these durations reveals that the time elapsed between half-peak and full peak is $\frac{1}{6}$ of the total period, which calculates to $\frac{1}{300}$ s, or $3.3$ ms.

When switch $S_1$ is closed at $t = 0$, the inductor resists the sudden change in current, leading to an exponential growth phase modeled by $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$ as the system approaches a steady state. At time $t_0$, the source is disconnected and the circuit is closed through $S_2$, forcing the energy stored in the inductor to dissipate, which results in the current decaying exponentially according to $I(t) = I_0 e^{-R(t-t_0)/L}$. This combination of processes creates a continuous current profile that climbs steadily during the charging interval and then drops toward zero once the discharge begins.

The energy dissipated in an AC circuit happens entirely in the resistive component, as inductors and capacitors store and release energy without net consumption. Determining this value requires calculating the average power dissipated, which depends on the circuit's total impedance. Using the frequency of $50 \text{ Hz}$, the angular frequency is $\omega = 100\pi \text{ rad/s}$, leading to an inductive reactance $X_L = 2\pi \text{ \Omega}$ and a capacitive reactance $X_C \approx 26.52 \text{ \Omega}$. The circuit impedance is found using the relation

$Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{60^2 + 20^2} = 20\sqrt{10} \text{ \Omega}$

The power factor is calculated as $\cos \phi = \frac{R}{Z} = \frac{3}{\sqrt{10}}$, allowing for the determination of average power dissipation:

$P_{\text{avg}} = \frac{V^2}{Z} \cos \phi = \frac{24^2}{20\sqrt{10}} \times \frac{3}{\sqrt{10}} = 8.64 \text{ W}$

Multiplying this average power by the time duration of $60 \text{ s}$ yields $518.4 \text{ J}$, which aligns with the value of $5.17 \times 10^2 \text{ J}$.

In an alternating current series circuit consisting of a resistor and a capacitor, the phase difference $\phi$ between the electromotive force and the current is governed by the capacitive reactance $X_C$ and the resistance $R$, expressed as:
$\tan \phi = \frac{X_C}{R}$
Since the given phase difference is $\pi/4$, the equation requires $\tan(\pi/4) = 1$, which means the capacitive reactance must be equal to the resistance, or $X_C = R$.

From the given emf equation $e = e_0 \sin(100t)$, the angular frequency is $\omega = 100 \text{ rad/s}$. Substituting the standard formula for capacitive reactance $X_C = \frac{1}{\omega C}$ into our condition $R = X_C$, we obtain the relationship:
$R = \frac{1}{\omega C}$
Using the provided values for the first circuit option, where $R = 1 \text{ k}\Omega$ or $1000 \Omega$ and $C = 10 \mu\text{F}$ or $10 \times 10^{-6} \text{ F}$, we can verify this condition:
$X_C = \frac{1}{100 \times 10 \times 10^{-6} \text{ F}} = \frac{1}{10^{-3}} = 1000 \Omega$
Because $X_C = 1000 \Omega$ and $R = 1000 \Omega$, the condition $X_C = R$ is satisfied. This confirms that the RC circuit with these specific component values produces the required phase difference of $\pi/4$.

The dynamics of an LR circuit are defined by the gradual rise of current as the inductor resists any sudden change in magnetic flux. When the switch is closed at $t = 0$, the current $i$ as a function of time is described by $i = \frac{E}{R}(1 - e^{-Rt/L})$. The total charge $q$ that flows from the battery is obtained by integrating this current function over the time interval from $0$ to $L/R$. Executing this integral calculation over the duration of one time constant, $\tau = L/R$, results in a total accumulated charge of $\frac{2.7 EL}{R^2}$.

The average power in an alternating current circuit is determined by the product of the root-mean-square voltage, root-mean-square current, and the power factor, $\cos\phi$, while the wattless current is the component of the current that does not contribute to power consumption, defined as $I_{\text{rms}} \sin \phi$. From the given expressions $e = 100 \sin(30t)$ and $i = 20 \sin(30t - \pi/4)$, the peak values are identified as $V_0 = 100$ and $I_0 = 20$, with a phase difference of $\phi = \pi/4$ between them.

Converting these to root-mean-square values yields $V_{\text{rms}} = 100/\sqrt{2}$ and $I_{\text{rms}} = 20/\sqrt{2}$. Substituting these values into the power formula gives the following expression for average power:
$P_{\text{avg}} = \frac{100}{\sqrt{2}} \cdot \frac{20}{\sqrt{2}} \cdot \cos\left(\frac{\pi}{4}\right) = \frac{1000}{\sqrt{2}}$
The wattless current is calculated using the reactive component of the current, which is:
$I_{\text{wattless}} = \frac{20}{\sqrt{2}} \cdot \sin\left(\frac{\pi}{4}\right) = \frac{20}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 10$

The total energy in an LC circuit remains conserved, implying that the initial electrostatic energy stored in the capacitor is equal to the sum of the electrostatic and magnetic energy stored in the capacitor and inductor at any later time. This principle is expressed as $\frac{1}{2} C V_0^2 = \frac{1}{2} C V^2 + \frac{1}{2} L I^2$. Rearranging this equation to solve for the current $I$ yields $I = \sqrt{\frac{C}{L} (V_0^2 - V^2)}$. Substituting the given values $C = 0.2 \times 10^{-6} \, \text{F}$, $L = 0.5 \times 10^{-3} \, \text{H}$, $V_0 = 10 \, \text{V}$, and $V = 5 \, \text{V}$ into the equation gives $I = \sqrt{\frac{0.2 \times 10^{-6}}{0.5 \times 10^{-3}} (10^2 - 5^2)}$. Simplifying this, $I = \sqrt{0.4 \times 10^{-3} \times 75} = \sqrt{0.03} \approx 0.173 \, \text{A}$.