Alternating Current – JEE Main PhysicsPractice Questions & PYQs

An a.c. ammeter is designed to display the root mean square (RMS) value of the current flowing through a circuit. In an inductive circuit, the current is restricted by the inductive reactance, which is defined as the product of the angular frequency and the inductance. Given the electromotive force equation $E = 440 \sin 100 \pi t$, we can identify the peak voltage as $440 \text{ V}$ and the angular frequency as $100 \pi \text{ rad/s}$ by comparing it to the standard form $E = E_0 \sin \omega t$.

By substituting the angular frequency and the given inductance $L = \frac{\sqrt{2}}{\pi} \text{ H}$ into the reactance formula, we find the inductive reactance:
$X_L = 100 \pi \cdot \frac{\sqrt{2}}{\pi} = 100 \sqrt{2} \, \Omega$

Using this reactance, we calculate the peak current $I_0$ by dividing the peak voltage by the inductive reactance:
$I_0 = \frac{440}{100 \sqrt{2}} = 2.2 \sqrt{2} \text{ A}$

Finally, because the ammeter measures the RMS value of the current, we determine this by dividing the peak current by $\sqrt{2}$, which results in an ammeter reading of:
$I_{rms} = \frac{2.2 \sqrt{2}}{\sqrt{2}} = 2.2 \text{ A}$

A transformer operates on the principle that the ratio of turns between the primary and secondary coils is directly proportional to the ratio of their respective voltages, represented by $\frac{N_p}{N_s} = \frac{V_p}{V_s}$. By substituting the known values for a domestic supply of 220 V converted to 12 V with 24 secondary turns, we establish the equation $\frac{N_p}{24} = \frac{220}{12}$. Solving for the number of turns in the primary coil, $N_p$, requires multiplying both sides by 24 to get $N_p = 220 \times \frac{24}{12}$. Since 24 divided by 12 equals 2, the calculation simplifies to $N_p = 220 \times 2$, confirming there are 440 primary turns.

When a direct current circuit containing an inductor reaches a steady state, the inductor ceases to oppose current changes and functions as a simple conducting wire with zero resistance. This characteristic allows us to treat the inductive branch as a short circuit, which simplifies the entire resistor network.

The three resistors of 3 ohms connected in parallel result in an equivalent resistance of 1 ohm. Incorporating the additional 2 ohms of series resistance brings the total resistance of the circuit to 3 ohms. With a battery voltage of 30 V, the final current is found using $I = \frac{30}{3} = 10$ A.

To maximize the average power in an L-C-R series circuit, the circuit must be in resonance, which implies the power factor is one. Under this condition, the impedance is at its minimum and is purely resistive because the inductive reactance $X_L$ is exactly cancelled by the capacitive reactance $X_C$. With a given inductive reactance $X_L = 250\pi \text{ }\Omega$ and a frequency $f = 50 \text{ Hz}$, we set $X_L$ equal to the capacitive reactance formula $X_C = \frac{1}{2\pi f C}$. Substituting the known values leads to the equation $250\pi = \frac{1}{2\pi(50)C}$. Isolating $C$ results in the following expression:

$C = \frac{1}{2\pi(50) \cdot 250\pi} = \frac{1}{2 \cdot 50 \cdot 250 \cdot \pi^2}$

Using the given approximation $\pi^2 = 10$, the calculation becomes $C = \frac{1}{100 \cdot 250 \cdot 10} = \frac{1}{250000} \text{ F}$. Converting this result into microfarads yields a final value of $4 \text{ }\mu\text{F}$.

At resonance, the inductive reactance and capacitive reactance in an L-C-R circuit completely cancel one another, reducing the circuit's total impedance to just its ohmic resistance $R$. Since the circuit behaves as a purely resistive load, the power formula simplifies to
$P = \frac{V^2}{R}$
We rearrange this equation to isolate the resistance, giving us $R = \frac{V^2}{P}$. Substituting the given power of 16 watts and the source voltage of 120 volts into the expression $\frac{120^2}{16}$ yields a final resistance of 900 ohms.

In parallel AC circuits, the total current is determined by the phasor sum of branch currents rather than their algebraic sum, due to the phase differences introduced by reactive components. For the upper branch containing a resistor and a capacitor, the capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = 100 \Omega$, resulting in an impedance of $Z_1 = \sqrt{100^2 + 100^2} = 100\sqrt{2} \Omega$. The current in this branch is $I_1 = \frac{200}{100\sqrt{2}} = \sqrt{2} \text{ A}$, which leads the voltage by $45^\circ$.

The lower branch consists of a resistor and an inductor with an inductive reactance of $X_L = \omega L = 100 \times 0.5 = 50 \Omega$. This creates an impedance of $Z_2 = \sqrt{50^2 + 50^2} = 50\sqrt{2} \Omega$, resulting in a current of $I_2 = \frac{200}{50\sqrt{2}} = 2\sqrt{2} \text{ A}$ that lags the voltage by $45^\circ$. Because the two currents are $90^\circ$ out of phase, the magnitude of the total current is calculated using vector addition as $I = \sqrt{I_1^2 + I_2^2} = \sqrt{(\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{2 + 8} = \sqrt{10} \approx 3.16 \text{ A}$. Since this result is not represented among the provided choices, none of the options are correct.

Inductors exhibit transient behavior that changes fundamentally between the moment a circuit is energized and when it reaches steady state. At $t=0$, an inductor behaves as an open circuit, blocking current flow through its branch. In this condition, the remaining resistive paths combine to an equivalent resistance of $R_{eq} = \frac{18}{5} \, \Omega$, and by applying Ohm's law, the current is $I = \frac{5E}{18}$.

After the circuit remains connected long enough to reach $t=\infty$, the inductor acts as a short circuit, offering no resistance and allowing current to bypass its original constraints. This rearrangement of the resistive network results in a new equivalent resistance of $R_{eq} = \frac{33}{10} \, \Omega$. Consequently, the steady-state current flowing through the circuit is $I = \frac{10E}{33}$.

Phase relationships in AC circuits are dictated by how voltage and current respond to individual components. In a purely resistive circuit, the voltage and current are perfectly in phase, meaning there is no phase difference. When dealing with a pure inductor, the back electromotive force causes the current to lag behind the voltage by $\pi/2$, whereas in a pure capacitor, the charging process allows the current to lead the voltage by $\pi/2$. Combining these elements into an L-C-R series circuit results in a net impedance characterized by a phase angle defined by $\phi = \tan^{-1} \left(\frac{X_C - X_L}{R}\right)$, which accounts for the interplay between inductive reactance $X_L$ and capacitive reactance $X_C$.

An inductor stores energy in its magnetic field and resists changes in current. When the battery is removed from an RL circuit, the current does not drop to zero immediately but decays exponentially according to the relationship:

$I(t) = I_{\text{max}} e^{-Rt/L}$

First, determine the initial current before the switch is opened using Ohm's law, which gives $I_{\text{max}} = 20 / 10000$ A, or 2 mA.

Next, evaluate the decay by calculating the exponent with time $t = 1 \times 10^{-6}$ s, resistance $R = 10 \times 10^3 \, \Omega$, and inductance $L = 10 \times 10^{-3}$ H. Substituting these into the expression $-Rt/L$ yields an exponent of $-1$, and multiplying the initial current by the decay factor $e^{-1} = 0.37$ gives $I = 2 \times 0.37 = 0.74$ mA. Since the current is defined as $x / 100$ mA, the value of $x$ is 74.

In a circuit containing only a capacitor, the flow of current must precede the potential difference across the plates because electrical charge needs to accumulate before the voltage can rise. This phase shift dictates that the alternating current $I_C$ leads the voltage $V_C$ by a phase angle of $\frac{\pi}{2}$. Consequently, any accurate phasor diagram for this system must depict the current vector rotated $\frac{\pi}{2}$ radians counter-clockwise from the voltage vector.

The displacement current existing between the plates of a parallel plate capacitor is fundamentally equal to the conduction current flowing in the external wires, as this continuity is required by Kirchhoff's current law. To determine the amplitude of this current, we first calculate the capacitance using the formula $C = \frac{\epsilon_0 A}{d}$. With the permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ F/m, plate area $A = 1$ m$^2$, and plate separation $d = 2 \times 10^{-3}$ m, the capacitance is $C = \frac{8.85 \times 10^{-12} \times 1}{2 \times 10^{-3}} = 4.425 \times 10^{-9}$ F.

Once the capacitance is known, we can relate the peak voltage $V_0$ to the amplitude of the current $I_0$ using the capacitive reactance $X_C$, where $I_0 = \frac{V_0}{X_C}$. Since $X_C = \frac{1}{\omega C}$ and the angular frequency is $\omega = 2\pi f$, the current amplitude is given by $I_0 = V_0 \cdot 2\pi f \cdot C$. Substituting the given peak voltage $V_0 = 20$ V and frequency $f = 50$ Hz, we calculate $I_0 = 20 \times 2\pi \times 50 \times 4.425 \times 10^{-9}$. Using $\pi \approx 3.14$, this calculation yields $I_0 \approx 2000 \times 3.14 \times 4.425 \times 10^{-9}$, resulting in approximately $27.79 \times 10^{-6}$ A, or $27.79$ $\mu$A.

At extremely high frequencies, the behavior of reactive components is defined by their frequency response, where capacitors act as short circuits due to negligible reactance, while inductors act as open circuits due to effectively infinite reactance. Removing the open-circuited components and replacing the shorted components with direct paths allows the circuit to be reduced to a purely resistive network.

In this resulting configuration, a 1 Ω resistor sits in series with two 2 Ω resistors arranged in parallel. The equivalent resistance for the parallel portion is $\frac{2 \times 2}{2+2} = 1$ Ω, and summing this with the series 1 Ω resistor results in a final effective impedance of 2 Ω.

In an LCR series circuit, the phase relationship between voltage and current is dictated by the net reactance of the circuit. When the current leads the voltage by $45^\circ$, the circuit is capacitive, meaning the capacitive reactance exceeds the inductive reactance according to the relationship $\tan \phi = \frac{X_C - X_L}{R}$. With a phase angle of $45^\circ$, the tangent value is $1$, which simplifies the equation to $X_C - X_L = R$. Given the resistance is $1 \Omega$, this becomes $X_C - X_L = 1$.

The inductive reactance is calculated as $X_L = \omega L$, resulting in $X_L = 300 \times 0.03 = 9 \Omega$. Substituting this value into the equation $X_C - X_L = 1$, we find $X_C - 9 = 1$, which gives $X_C = 10 \Omega$. Since the capacitive reactance is defined as $X_C = \frac{1}{\omega C}$, we set $\frac{1}{300 \times C} = 10$, leading to $C = \frac{1}{3000} \text{ F}$. Expressing this in the required form gives $C = \frac{1}{3} \times 10^{-3} \text{ F}$, and identifying the term $x$ in $\frac{1}{x} \times 10^{-3}$ reveals that the value of $x$ is 3.

In an AC circuit containing an inductor, capacitor, and resistor, the phase relationship between voltage and current depends on the net reactance, which is determined by comparing the inductive reactance $\omega L$ and the capacitive reactance $1/(\omega C)$. When $\omega L > 1/(\omega C)$, the inductive reactance dominates, causing the current to lag behind the applied emf, while the condition $\omega L < 1/(\omega C)$ indicates that capacitive reactance dominates, resulting in the current leading the emf. At the specific condition where $\omega L = 1/(\omega C)$, the inductive and capacitive effects cancel out, bringing the current and emf into phase. Finally, the resonant frequency corresponds to the state where total impedance is at its minimum, which ensures that the maximum current flows through the system.

In a series LCR circuit, the overall behavior is determined by the interplay between inductive reactance $X_L = 50\pi \times 10^{-3}$ ohms and capacitive reactance $X_C = \frac{10^{11}}{100\pi}$ ohms. Comparing these values reveals that $X_C \gg X_L$, indicating that the capacitor plays the dominant role in the circuit. Furthermore, since the net reactance is significantly larger than the resistance, satisfying $|X_C - X_L| \gg R$, the circuit's impedance is effectively dictated by the capacitive reactance. As a result, the phase angle $\phi$ determined by $\tan \phi = \frac{X_C - X_L}{R}$ approaches $90^\circ$, causing the current to lead the voltage and classifying the system as a predominantly capacitive circuit.

In a series RLC circuit, the total opposition to current, known as impedance, arises from the combination of resistance and the net reactance resulting from the inductor and capacitor. Because inductive and capacitive reactances act in opposition to one another, the net reactance is calculated as the difference between them, which determines the overall impedance via the formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Given the condition that $X_L = R = X_C$, the reactive component effectively cancels out to zero, leaving only the resistance to dictate the magnitude of the circuit's impedance. Consequently, substituting these values into the formula yields $Z = \sqrt{R^2 + (R - R)^2}$, simplifying directly to $Z = R$.

The root mean square current is determined by calculating the square root of the average value of the squared current over a complete cycle. Expanding the expression $i^2 = (i_{1} \sin \omega t + i_{2} \cos \omega t)^2$ yields $i_{1}^2 \sin^2 \omega t + i_{2}^2 \cos^2 \omega t + 2 i_{1} i_{2} \sin \omega t \cos \omega t$. Since the average values of $\sin^2 \omega t$ and $\cos^2 \omega t$ over a full cycle are both $\frac{1}{2}$, and the cross-term $\sin \omega t \cos \omega t$ averages to zero, the mean of the squared current simplifies to $\frac{1}{2}(i_{1}^2 + i_{2}^2)$. Taking the square root of this mean value results in an rms current of $\frac{1}{\sqrt{2}}(i_{1}^2 + i_{2}^2)^{1/2}$.

In an L-C-R circuit, the phase difference between current and voltage is determined by the relative magnitudes of inductive and capacitive reactance. When the capacitor is removed, the remaining R-L series circuit exhibits a phase lag of 45 degrees, which implies that the inductive reactance equals the resistance, or $X_L = R = 110$ ohms. Removing the inductor instead leaves an R-C circuit, where a phase lead of 45 degrees confirms that the capacitive reactance must also equal the resistance, so $X_C = R = 110$ ohms.

Since the inductive and capacitive reactances are equal, the circuit is operating in a state of resonance, where the reactive components cancel each other out. In this state, the total impedance simplifies to the resistance:
$Z = \sqrt{R^2 + (X_L - X_C)^2} = R$
With the total impedance equal to 110 ohms, the rms current is found by dividing the supply voltage of 220 V by the resistance, yielding 2 A.

Alternating current follows a sinusoidal oscillation, where the peak value is the amplitude $I_0$ and the effective level is the root mean square value defined as $I_0 / \sqrt{2}$. With a source frequency of 50 Hz, we calculate the angular frequency using $\omega = 2 \pi f$, which equals $100 \pi \, \text{rad/s}$.

To determine the time required to transition from the peak to the rms value, we model the current using the function $I = I_0 \cos(\omega t)$ so that the maximum occurs at time $t = 0$. Setting this instantaneous expression equal to the rms value leads to the following equation:

$\cos(100 \pi t) = \frac{1}{\sqrt{2}}$

This equality holds when the phase angle equals $\pi/4$ radians. Solving for time, we find $100 \pi t = \pi/4$, which results in $t = 1/400$ seconds. Converting this fraction to decimal form yields $0.0025$ seconds, which is equivalent to 2.5 ms.

The power factor of a series L-C-R circuit is determined by the ratio of the circuit resistance to its total impedance, given by the expression $\cos \phi = \frac{R}{Z}$. To find the total impedance, use the relation $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$ with the values $R = 6 \Omega$, $X_{L} = 10 \Omega$, and $X_{C} = 4 \Omega$. The net reactance is $10 \Omega - 4 \Omega = 6 \Omega$, which results in an impedance of $\sqrt{6^{2} + 6^{2}}$, simplifying to $6\sqrt{2} \Omega$. Dividing the resistance by this impedance yields $\frac{6}{6\sqrt{2}}$, which results in a final power factor of $\frac{1}{\sqrt{2}}$.