Alternating Current – JEE Main PhysicsPractice Questions & PYQs

In an AC circuit where a lamp and a capacitor are connected in series, the source voltage is related to the individual voltage drops across the components by the vector sum $V_{\text{source}}^2 = V_R^2 + V_C^2$. Since the lamp is operating at its rated power of $50 \text{ W}$ at a voltage of $100 \text{ V}$, we determine the resistance of the lamp as $R = \frac{V^2}{P} = \frac{100^2}{50} = 200 \Omega$. With a source voltage of $200 \text{ V}$, the voltage across the capacitor must satisfy the relation $V_C^2 = 200^2 - 100^2 = 30000$, which gives $V_C = 100\sqrt{3} \text{ V}$.

The current flowing through the circuit is dictated by the lamp's operation at its rated voltage, calculated as $I = \frac{V_R}{R} = \frac{100}{200} = 0.5 \text{ A}$. Using this current and the capacitor voltage, the capacitive reactance is $X_C = \frac{V_C}{I} = \frac{100\sqrt{3}}{0.5} = 200\sqrt{3} \Omega$. The formula for capacitive reactance is $X_C = \frac{1}{2\pi f C}$, where $f = 50 \text{ Hz}$. Substituting the reactance value into this equation results in $200\sqrt{3} = \frac{1}{2\pi \cdot 50 \cdot C} = \frac{1}{100\pi C}$, which simplifies to $C = \frac{1}{20000\pi\sqrt{3}} \text{ F}$. Converting this capacitance into microfarads yields $C = \frac{10^6}{20000\pi\sqrt{3}} \mu\text{F} = \frac{50}{\pi\sqrt{3}} \mu\text{F}$. By comparing this result to the expression $\frac{50}{\pi\sqrt{x}} \mu\text{F}$, we find that $x = 3$.

Wattless current signifies a condition where the average power consumption of an AC circuit is zero. The average power is calculated using the relation $P = V_{rms} I_{rms} \cos \phi$, implying that the power factor $\cos \phi$ must be zero, which occurs only when the phase difference between voltage and current is $90^\circ$. Such a phase shift is characteristic of a purely inductive circuit, where the absence of resistance prevents any real power dissipation.

Impedance in a series LCR circuit acts as the total opposition to alternating current, defined by the relationship $Z = \sqrt{R^2 + (X_L - X_C)^2}$. First, determine the inductive reactance $X_L = 2 \pi f L = 2 \pi (50)(0.1) = 10 \pi \Omega$ and the capacitive reactance $X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (50)(10^{-4})} = \frac{100}{\pi} \Omega$. As these values are very similar, their difference $(X_L - X_C)$ becomes negligible, allowing the circuit impedance to be approximated as $Z \approx R = 10 \Omega$. With a source voltage of $V = 220 \text{ V}$, applying Ohm's law $I = \frac{V}{Z}$ yields the current $I = \frac{220}{10} = 22 \text{ A}$.

In a series LCR circuit, current flows in phase with the electromotive force when the circuit reaches resonance. At resonance, the inductive reactance and capacitive reactance cancel each other out, which is defined by the equation $\omega L = \frac{1}{\omega C}$. This allows us to determine the resonance angular frequency using the equation $\omega = \frac{1}{\sqrt{LC}}$.

With an inductor of 0.5 mH and a capacitor of 200 µF, we express these values in SI units as $L = 5 \times 10^{-4}$ H and $C = 2 \times 10^{-4}$ F. Substituting these into the formula, we find $\omega = \frac{1}{\sqrt{5 \times 10^{-4} \times 2 \times 10^{-4}}}$. Simplifying the denominator results in $\sqrt{10} \times 10^{-4}$, leading to an angular frequency of $\omega = \frac{10^4}{\sqrt{10}}$ rad/s. Converting this to standard frequency using the equation $f = \frac{\omega}{2\pi}$ yields $f = \frac{10^4}{2\pi \sqrt{10}}$ Hz. Calculating this value results in approximately 500 Hz, which is equal to 5 × 10² Hz, identifying the required multiplier as 5.

The voltage across an inductor in an alternating current circuit is determined by the inductive reactance, defined as $V_L = \omega L I$, where the angular frequency $\omega = 2 \pi f$. Because the magnitude of $V_L$ is twice the resistive voltage drop $V_R = R I$, we equate $\omega L I = 2 R I$, which simplifies to $\omega L = 2 R$. Using the given frequency of 50 Hz and the resistance of 5 ohms, the relationship becomes $2 \pi \times 50 \times L = 10$, resulting in $100 \pi L = 10$.

Calculating the inductance yields $L = \frac{1}{10 \pi}$ H, which converts to $\frac{100}{\pi}$ mH. Equating this to the expression $\frac{1}{K \pi}$ mH shows that $K = 0.01$, which rounds to 0.

In a purely inductive circuit, the opposition to current flow, known as inductive reactance, is defined by the formula $X_L = 2\pi f L$. With the frequency at 50 Hz and inductance at 200 mH, which is equivalent to 0.2 H, the reactance is calculated as $X_L = 2\pi \times 50 \times 0.2 = 20\pi \text{ } \Omega$. The root mean square current is determined using the relation $I_{rms} = V_{rms} / X_L$, which yields $I_{rms} = 220 / (20\pi) = 11/\pi \text{ A}$. To find the peak current, the rms value is multiplied by $\sqrt{2}$, resulting in the equation $I_{peak} = \sqrt{2} I_{rms} = 11\sqrt{2} / \pi \text{ A}$. Expressing the integer 11 as the square root of 121 allows the numerator to be written as $\sqrt{121 \times 2} = \sqrt{242}$, leading to a peak current of $\sqrt{242} / \pi \text{ A}$, which confirms that $a$ is 242.

The growth of current in a circuit containing an inductor and a resistor is governed by the time constant $\tau = L/R$, which determines the rate at which current approaches its maximum steady-state value $I_0 = E/R$. Since the inductance is $1 \text{ H}$ and the resistance is $100 \Omega$, the time constant is $\tau = 1/100 = 0.01 \text{ s}$, or $10 \text{ ms}$. To find when the current reaches half its steady-state value, we set the current equation $i = I_0(1 - e^{-t/\tau})$ equal to $I_0/2$. This simplifies to the condition $1 - e^{-t/\tau} = 0.5$, or $e^{-t/\tau} = 0.5$, which requires taking the natural logarithm to solve for time. Using the provided $\ln 2 = 0.693$, the time elapsed is $t = \tau \ln 2 = 10 \times 0.693 = 6.93 \text{ ms}$, which is approximately $7 \text{ ms}$.

To determine the energy stored in the magnetic field at $15 \text{ ms}$, we first calculate the instantaneous current at that moment. Given the time constant $\tau = 10 \text{ ms}$ and the elapsed time $t = 15 \text{ ms}$, the exponent is $-t/\tau = -15/10 = -1.5$. The current is then $i = (E/R)(1 - e^{-1.5})$. With $E = 6 \text{ V}$ and $R = 100 \Omega$, and using the given value $e^{-3/2} = 0.25$, we find $i = (6/100)(1 - 0.25) = 0.06 \times 0.75 = 0.045 \text{ A}$. The energy stored in the magnetic field is defined by $U = \frac{1}{2} L i^2$. Substituting the circuit values, we calculate $U = \frac{1}{2} \times 1 \times (0.045)^2 = 0.5 \times 0.002025 \approx 0.001 \text{ J}$, which is equivalent to $1 \text{ mJ}$.

The impedance of a series LCR circuit is governed by the driving frequency because the inductive and capacitive components offer reactances that oppose each other. At the resonant frequency, these reactances perfectly cancel out, leaving the circuit impedance at its minimum value equal to the resistance. When the driving frequency shifts away from this resonant point, the net reactance increases, which elevates the total impedance and consequently suppresses the amplitude of the current.

The resonant angular frequency is determined by the formula $\omega_0 = \frac{1}{\sqrt{LC}}$. Substituting the given values $L = 0.01\,\text{H}$ and $C = 10^{-6}\,\text{F}$ results in $\omega_0 = \frac{1}{\sqrt{10^{-8}}} = 10^4\,\text{rad/s}$. Because the circuit operates at a frequency $60\%$ lower than resonance, the effective angular frequency is $\omega' = 0.4 \omega_0 = 4000\,\text{rad/s}$. At this frequency, the inductive reactance becomes $X_L' = \omega' L = 4000 \times 0.01 = 40\,\Omega$, while the capacitive reactance is $X_C' = \frac{1}{\omega' C} = \frac{1}{4000 \times 10^{-6}} = 250\,\Omega$. The total circuit impedance is calculated as $Z = \sqrt{R^2 + (X_L' - X_C')^2} = \sqrt{10^2 + (40 - 250)^2} = \sqrt{100 + (-210)^2} = \sqrt{44200} \approx 210.2\,\Omega$. Dividing the source voltage amplitude by this calculated impedance provides the peak current amplitude $I_m = \frac{V_m}{Z} = \frac{50}{210.2} \approx 0.238\,\text{A}$, which corresponds to $238\,\text{mA}$.

The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt{2}}$ times its maximum value correspond to the half-power points of the system. The bandwidth of this circuit is defined as the difference between these two frequencies and relates to the circuit parameters via $\Delta\omega = \omega_2 - \omega_1 = \frac{R}{L}$. By subtracting the given values, the bandwidth is calculated as $232 - 212 = 20$ rad/s. Given that the resistance is $R = 5$ $\Omega$, we rearrange the relationship to solve for self-inductance as $L = \frac{R}{\Delta\omega}$. Substituting the values results in $L = \frac{5}{20} = 0.25$ H, which is equal to 250 mH.

Identifying the nature of each element is the first step. Because the current is in phase with the voltage for element X, it acts as a pure resistor where $R = 100/5 = 20$ Ohms. Element Y causes the current to lag by $\pi/2$, which identifies it as a pure inductor with inductive reactance $X_L = 100/5 = 20$ Ohms.

When both elements are connected in series, the total impedance of the circuit is calculated using the Pythagorean relationship for AC circuits:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}$
The peak current flowing through the series combination is given by the ratio of the peak voltage to this total impedance:
$I_{peak} = 100 / (20\sqrt{2}) = 5/\sqrt{2}$
Amperes. The rms value is found by dividing this peak current by $\sqrt{2}$:
$I_{rms} = (5/\sqrt{2}) / \sqrt{2} = 5/2$
Amperes.

The energy dissipated as heat in a resistor carrying an alternating current is described by the equation $H = I_{\text{rms}}^2 R t$, where $H$ is the heat energy, $I_{\text{rms}}$ is the root-mean-square current, $R$ is the resistance, and $t$ is the time duration. To determine the current, first evaluate the impedance of the series circuit using the inductive reactance of 10 ohms and the capacitive reactance of 100 ohms. Given the resistance of 120 ohms, the total impedance is calculated using the formula $Z = \sqrt{120^2 + (100 - 10)^2} = 150 \text{ }\Omega$.

With a source voltage of 20 V, the circuit current is $I_{\text{rms}} = 20 / 150 = 2/15 \text{ A}$. The thermal energy required to raise the resistor temperature by 16 degrees Celsius, given a thermal capacity of 2 joules per degree Celsius, is 32 joules. Substituting these values into the expression $32 = (2/15)^2 \times 120 \times t$ provides the required duration, which results in $t = 15 \text{ s}$.

An ideal transformer maintains constant power between its primary and secondary sides, meaning the total power consumed by the primary circuit equals the power delivered to the load. Given the system operates at a unity power factor with a load of 80,000 W, we can determine the current and resistance for each side using $P = VI$ and $R = \frac{V}{I}$. On the primary side, dividing the power of 80,000 W by the voltage of 8,000 V results in a current of 10 A. Calculating the resistance by dividing 8,000 V by 10 A gives 800 Ohms.

For the secondary side, dividing the same power of 80,000 W by the secondary voltage of 160 V yields a current of 500 A. Finally, dividing the secondary voltage of 160 V by the calculated current of 500 A results in a secondary resistance of 0.32 Ohms.

Alternating current varies sinusoidally over time, following a functional relationship such as $I = I_0 \cos(\omega t)$. The current reaches its maximum peak when the phase angle is 0, and it transitions to the root mean square value of $I_0 / \sqrt{2}$ when the phase angle advances to $\pi/4$. Given the supply frequency of $50 Hz$, the angular frequency is calculated as $\omega = 2\pi f = 100\pi rad/s$. Solving for the specific time interval between these two states yields $\Delta t = \frac{\pi / 4}{\omega} = \frac{\pi / 4}{100\pi} s$. This calculation results in $\frac{1}{400} s$, which is equivalent to $2.5 ms$.

Maximum average power in a series LCR circuit is achieved at resonance, where the power factor is unity. This condition occurs when the inductive reactance and capacitive reactance are equal, ensuring the circuit's net reactance is zero and its impedance is purely resistive.

To find the required capacitance, we equate the given inductive reactance of $X_L = \frac{70}{11} \, \Omega$ to the capacitive reactance formula $X_C = \frac{1}{2\pi f C}$ using the supply frequency $f = 50 \, \text{Hz}$. This provides the relationship:
$\frac{70}{11} = \frac{1}{2 \pi \times 50 \times C}$
Simplifying the expression leads to:
$\frac{70}{11} = \frac{1}{100 \pi C}$
Rearranging for $C$ gives:
$C = \frac{11}{7000 \pi}$
Substituting the value $\pi = \frac{22}{7}$ into the equation, we find $C = \frac{11}{7000 \times (\frac{22}{7})} = \frac{11}{22000} = \frac{1}{2000} \, \text{F}$. Converting this result into microfarads yields $500 \, \mu\text{F}$.

To operate the lamp at its rated capacity, it acts as a fixed resistance determined by its voltage and power ratings. We first determine this resistance using $R_L = \frac{V^2}{P}$, which calculates to $R_L = \frac{25^2}{5} = 125$ ohm. The lamp requires a specific current to function correctly, defined by $I = \frac{P}{V}$, resulting in $I = \frac{1}{5}$ A. Because the lamp and the external resistor $R$ are connected in series, the same current flows through both, so the total circuit resistance must satisfy $\frac{220}{R + 125} = \frac{1}{5}$. Solving this equation gives $R + 125 = 1100$, which indicates that the required resistance is $R = 975$ ohm.

When an alternating current circuit operates at extremely high frequencies, inductive components behave as open circuits while capacitive components act as short circuits. This specific circuit configuration simplifies significantly under these conditions, effectively isolating a resistance of 5 ohms connected to the 220 V power source. Applying Ohm's law to this reduced structure allows us to calculate the effective current $I$ using the equation $I = \frac{220}{5}$, which yields a final value of 44 A.

In a series $LCR$ circuit, the phase relationship between the source voltage and the resultant current is determined by the interplay between the circuit's resistance and its net reactance. This phase difference $\phi$ is defined by the relationship $\tan \phi = \frac{X_L - X_C}{R}$, which represents the ratio of the net reactive effect to the ohmic resistance.

To find the phase difference, we first extract the angular frequency $\omega = 3000$ rad/s from the voltage expression $V(t) = 210 \sin(3000t)$. The inductive reactance is then calculated as $X_L = \omega L = 3000 \times 10 \times 10^{-3} = 30 \text{ } \Omega$, while the capacitive reactance is determined by $X_C = \frac{1}{\omega C} = \frac{1}{3000 \times 25 \times 10^{-6}} = \frac{40}{3} \text{ } \Omega$. Subtracting these values provides the net reactance, $X_L - X_C = 30 - \frac{40}{3} = \frac{50}{3} \text{ } \Omega$. Substituting this into the tangent equation gives $\tan \phi = \frac{50/3}{100} = \frac{1}{6}$, which equals approximately $0.17$. Consequently, the phase difference is expressed as $\tan^{-1}(0.17)$.

The resonance frequency $\omega_r$ is defined by the condition where inductive reactance $X_L = \omega L$ exactly equals capacitive reactance $X_C = \frac{1}{\omega C}$, resulting in a net reactance of zero. Because the impedance of the circuit is calculated as $Z = \sqrt{R^2 + (X_L - X_C)^2}$, this condition implies that the total impedance simplifies to just the resistance $R$ at resonance. Conversely, when the angular frequency is lower than $\omega_r$, the capacitive reactance dominates the inductive reactance, which causes the circuit to behave as a capacitive component.

A sinusoidal current reaches its peak value when its argument corresponds to $\frac{\pi}{2}$, which occurs at one-quarter of the total time period. For the given expression $I = 5 \sin(120 \pi t)$, we identify the angular frequency as $\omega = 120 \pi$. The time period $T$ is determined by the equation $T = \frac{2\pi}{\omega}$, which simplifies to $T = \frac{1}{60}$ s. Since the time taken to reach the peak is $t = \frac{T}{4}$, we can compute the result as $t = \frac{1}{240}$ s.

Metal detectors operate based on the concept of electrical resonance within alternating current circuits, where the system is tuned to a precise resonant frequency defined by $f = \frac{1}{2\pi\sqrt{LC}}$. As long as the surrounding environment remains unchanged, this frequency is consistently maintained by the device's internal oscillator. When a metallic object enters the detection field, it disrupts the inductance of the circuit, effectively shifting the system away from its resonant state and triggering the alarm.