Alternating Current – JEE Main PhysicsPractice Questions & PYQs

In a series LCR circuit, the voltage across the inductor $V_L$ and the voltage across the capacitor $V_C$ are exactly $\pi$ radians out of phase, meaning they operate in direct opposition to one another. The net potential difference across this LC combination is determined by the formula $|V_L - V_C|$. Substituting the given values of 50 V for both components, the calculation becomes 50 - 50, resulting in 0 V.

The resonant frequency of an LCR circuit is governed by the relation $f = \frac{1}{2\pi \sqrt{LC}}$. For this frequency to remain constant, the product $LC$ must also remain invariant, meaning that any change in capacitance must be balanced by an inverse change in inductance. Since the capacitance is increased to $2C$, we equate the original product to the new one using the expression $LC = L' \cdot 2C$. Solving for $L'$, we find that the inductance must be $\frac{L}{2}$ to keep the resonant frequency unchanged.

A D.C. ammeter is calibrated to calculate and display the average value of the current flowing through it. Since alternating current oscillates symmetrically, its net value over any complete cycle is mathematically zero, which we can express as $I_{avg} = 0$. Consequently, the meter pointer simply remains at zero because it cannot track the rapid, instantaneous changes in direction that cancel each other out over time.

The total energy in an oscillating $LC$ circuit is constant and initially resides entirely within the capacitor as
$E = \frac{Q^2}{2C}$
where $Q$ represents the maximum charge. When the energy is distributed equally between the electric field of the capacitor and the magnetic field of the inductor, the capacitor holds exactly half of this total maximum energy. Setting the energy stored in the capacitor at an arbitrary charge $q$ equal to half of the total energy yields
$\frac{q^2}{2C} = \frac{1}{2} \left( \frac{Q^2}{2C} \right)$
By canceling the common factor of $\frac{1}{2C}$ from both sides, the equation simplifies to $q^2 = \frac{Q^2}{2}$, which indicates that the charge on the capacitor is $q = \frac{Q}{\sqrt{2}}$.

When an alternating magnetic flux passes through a transformer core, it induces unwanted parasitic currents known as eddy currents that lead to energy waste. By laminating the core, we effectively break the continuous metal path and reduce the cross-sectional area available for these circulating currents to flow. This reduction in area increases the electrical resistance, which inherently limits the magnitude of the eddy currents and minimizes the energy lost as heat.

Ideal transformers operate on the principle that power remains constant between coils, resulting in an inverse relationship between current and the number of turns, expressed by the relationship $\frac{I_s}{I_p} = \frac{N_p}{N_s}$. With 140 turns in the primary coil and 280 turns in the secondary coil, the turns ratio is 0.5. Since the primary current is 4 A, substituting these values into the equation $\frac{I_s}{4} = \frac{140}{280}$ shows that the secondary current is 2 A.

In an AC circuit containing a resistor $R$ and an inductor $L$ connected in series, the power factor is defined as the ratio of the circuit's resistance to its total impedance. Because the total impedance results from the vector combination of the resistance and the inductive reactance $X_L = \omega L$, it forms the hypotenuse of a right-angled triangle with sides $R$ and $\omega L$, leading to an impedance magnitude of $Z = \sqrt{R^2 + \omega^2 L^2}$. Consequently, the power factor $\cos \phi$ is determined by the ratio of the resistance to the impedance, expressed as $R / (R^2 + \omega^2 L^2)^{1/2}$.

The power factor in a series L-R circuit is defined by the phase angle $\phi$ between the source voltage and current, where $\cos \phi = \frac{R}{\sqrt{R^2 + X_L^2}}$ and $\tan \phi = \frac{\omega L}{R}$. Given an initial power factor of $1/\sqrt{2}$, the phase angle is $45^\circ$, which indicates the inductive reactance equals the resistance, such that $\frac{\omega_1 L}{R} = 1$. With the initial angular frequency $\omega_1$ of 1000 rad/s, the ratio $\frac{L}{R}$ corresponds to $1/1000$ s.

When the source frequency increases to $\omega_2$ of 2000 rad/s, the new inductive reactance shifts the phase angle to $\phi_2$. The new tangent of the phase angle is:
$\tan \phi_2 = \frac{\omega_2 L}{R} = 2000 \times \left(\frac{1}{1000}\right) = 2$
Using the identity $\cos \phi = \frac{1}{\sqrt{1 + \tan^2 \phi}}$, the resulting power factor becomes $\frac{1}{\sqrt{1 + 2^2}}$, which equals $1/\sqrt{5}$.