Alternating Current – JEE Main PhysicsPractice Questions & PYQs

The efficiency ($\eta$) of a transformer is the ratio of output power to input power, defined by the formula $\eta = \frac{V_s I_s}{V_p I_p}$, where $V_p$ and $I_p$ are the primary voltage and current, and $V_s$ and $I_s$ are the secondary voltage and current. Substituting the given values $\eta = 0.9$, $V_p = 2300 \text{ V}$, $V_s = 230 \text{ V}$, and $I_p = 5 \text{ A}$ into the equation yields $0.9 = \frac{230 \times I_s}{2300 \times 5}$. Rearranging the expression to isolate the secondary current gives $I_s = \frac{0.9 \times 2300 \times 5}{230}$, which simplifies to $I_s = 45 \text{ A}$.

The quality factor represents the sharpness of resonance in an RLC circuit, quantifying the efficiency of energy storage relative to energy dissipation. At the resonant frequency $\omega_0$, where inductive reactance perfectly balances capacitive reactance, this factor is determined by the ratio of the inductive reactance to the circuit's total resistance. Since the inductive reactance is calculated as the product of the angular frequency and the inductance, the relationship is defined as $Q = \frac{\omega_0 L}{R}$.

The total impedance $Z$ in a series LCR circuit is calculated using the formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$, where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$ are the inductive and capacitive reactances, respectively. Given the angular frequency $\omega = 320 \text{ rad/s}$, resistance $R = 5 \Omega$, inductance $L = 25 \text{ mH}$, and capacitance $C = 1000 \text{ \mu F}$, the reactances are determined as $X_L = 320 \times 25 \times 10^{-3} = 8 \Omega$ and $X_C = \frac{1}{320 \times 10^{-3}} = 3.125 \Omega$. Substituting these values into the impedance formula yields $Z = \sqrt{5^2 + (8 - 3.125)^2} = \sqrt{25 + (4.875)^2} \approx \sqrt{25 + 23.77} \approx 6.98 \Omega$, which is approximately $7 \Omega$.

The phase difference $\phi$ between the source voltage and the current is determined by the equation $\tan \phi = \frac{X_L - X_C}{R}$, which yields $\tan \phi = \frac{8 - 3.125}{5} = \frac{4.875}{5} = 0.975$. Because $\tan 45^{\circ} = 1$, the value $0.975$ indicates that the phase difference $\phi$ is approximately $45^{\circ}$.

An arc lamp operates as a resistor because its fixed voltage and current requirements of $80 \text{ V}$ and $10 \text{ A}$ allow us to calculate its resistance as $R = 80 / 10 = 8 \ \Omega$. When this lamp is connected to a $220 \text{ V}$ AC supply while maintaining the same $10 \text{ A}$ current, the total impedance of the circuit must be $Z = 220 / 10 = 22 \ \Omega$.

Since the circuit consists of a series resistor and inductor, the impedance follows the relationship $Z^2 = R^2 + X_L^2$, where $X_L = \omega L$. Substituting the known values leads to $22^2 = 8^2 + (\omega L)^2$, which simplifies to $484 = 64 + (\omega L)^2$, or $(\omega L)^2 = 420$. With the angular frequency defined as $\omega = 2\pi f = 2 \cdot \pi \cdot 50 = 100\pi \text{ rad/s}$, solving for $L$ yields $L = \sqrt{420} / (100\pi)$, resulting in an inductance of approximately $0.065 \text{ H}$.

When an alternating current source is connected to a series circuit containing a resistor and an inductor, the system eventually reaches a steady state once the transient effects subside after a sufficiently long duration. In this regime, the current oscillates at the same frequency as the driving voltage, but the inductor introduces a phase lag due to its reactance. This creates a current profile defined by $I(t) = I_0 \sin(\Omega t - \phi)$, which manifests as a sinusoidal wave shifted in time relative to the voltage source.

When the first key remains closed for a long time, the inductor acts as a simple wire, allowing the steady current to be determined by the electromotive force divided by the total resistance. With a battery of 15 V and resistance of 150 $\Omega$, the initial current is $I_0 = 15 / 150 = 0.1$ A.

Opening the first key while simultaneously closing the second key transforms the setup into an isolated inductor-resistor loop, where the stored magnetic energy dissipates and the current decays exponentially according to the expression $i(t) = I_0 e^{-Rt/L}$

Calculating the exponent factor for a time of 0.001 s gives $Rt/L = (150 \times 0.001) / 0.03 = 5$. Substituting this value into the decay equation yields $i = 0.1 \times e^{-5}$. Using the given approximation $e^5 \approx 150$, the current becomes $i = 0.1 / 150$ A, which results in 0.67 mA.

An LCR circuit undergoing damped oscillations acts analogously to a physical pendulum losing energy due to friction, where the resistance $R$ serves as the damping agent and the inductance $L$ provides the electrical inertia. Because $L_1 > L_2$, the circuit with the larger inductance effectively resists the decay of current and charge more strongly, resulting in a slower rate of energy dissipation compared to the circuit with the smaller inductance $L_2$. Consequently, the square of the maximum charge $Q_{\max}^2$ on the capacitor, which is directly proportional to the stored energy at each oscillation peak, decreases more gradually over time for $L_1$ than for $L_2$. This means the decay curve for $L_1$ will remain consistently above the curve for $L_2$ throughout the process, reflecting the slower rate of energy dissipation in the higher-inductance circuit.

When point C is moved to point B, the circuit effectively places the resistor and the inductor in a parallel arrangement. Because these components are connected in parallel, the instantaneous potential difference across the resistor must exactly match the potential difference across the inductor. This equality holds at all times, so the ratio of the voltage across the resistance to the voltage across the inductor is 1, even at the specific time $t = L/R$.

When switch $S_1$ is closed in an RC circuit, the capacitor charges according to the standard exponential function $q(t) = CV(1 - e^{-t/\tau})$, where $\tau$ denotes the time constant $RC$. To determine the charge at $t = 2\tau$, we substitute this value for the time variable $t$ within the charging equation. This substitution simplifies the exponent to $-2\tau/\tau$, which evaluates to $-2$. Consequently, the charge stored on the capacitor at this specific instant is calculated as $q = CV(1 - e^{-2})$.

The behavior of a charging capacitor in a series RC circuit is governed by the equation $V = V_0(1 - e^{-t/RC})$, where $V$ represents the capacitor voltage at time $t$, $V_0$ is the constant supply voltage, $R$ is the series resistance, and $C$ is the capacitance. Given that the neon bulb ignites at $120 V$ when the supply is $200 V$, the capacitor must reach $120 V$ at $t = 5 s$. Inserting these values into the charging equation gives $120 = 200(1 - e^{-5 / (R \times 2 \times 10^{-6})})$, which simplifies to $0.6 = 1 - e^{-5 / (R \times 2 \times 10^{-6})}$ and further to $e^{-5 / (R \times 2 \times 10^{-6})} = 0.4$.

Inverting the exponential term results in $e^{5 / (R \times 2 \times 10^{-6})} = 2.5$. Taking the natural logarithm on both sides provides the expression
$\frac{5}{R \times 2 \times 10^{-6}} = \ln(2.5)$
Using the conversion $\ln(x) = 2.303 \log_{10}(x)$, the natural logarithm becomes $2.303 \times 0.4$, which is approximately $0.9212$. Isolating $R$ in the equation yields $R = 5 / (0.9212 \times 2 \times 10^{-6})$, resulting in a resistance of approximately $2.71 \times 10^6 \Omega$.

In an LC circuit, total energy continuously oscillates between the electric field of the capacitor and the magnetic field of the inductor. When these energies are shared equally, the electric energy accounts for half of the system's total initial energy, leading to the condition $\frac{1}{2} \frac{q^2}{C} = \frac{1}{2} \left( \frac{1}{2} \frac{q_0^2}{C} \right)$. Simplifying this relationship yields $q = \frac{q_0}{\sqrt{2}}$, which, when substituted into the oscillation formula $q = q_0 \cos(\omega t)$, requires $\cos(\omega t) = \frac{1}{\sqrt{2}}$. With the angular frequency defined as $\omega = \frac{1}{\sqrt{LC}}$, solving the phase $\omega t = \frac{\pi}{4}$ results in the time $t = \frac{\pi}{4} \sqrt{LC}$.

An inductor acts as an open circuit immediately after a switch is closed because it resists any instantaneous change in current, while it behaves as a simple conducting wire once the circuit reaches a steady state. At $t=0$, the path containing the inductor is effectively blocked, ensuring all current passes solely through $R_2$ with a value of $\frac{V}{R_2}$. By the time $t=\infty$, the inductor functions as a perfect conductor, connecting resistors $R_1$ and $R_2$ in parallel to create an equivalent resistance of $\frac{R_1 R_2}{R_1+R_2}$, resulting in a battery current of $\frac{V (R_1+R_2) }{R_1 R_2}$.

The power dissipated in an alternating current circuit containing a resistor, inductor, and capacitor depends on the total impedance of the network, as the resistor is the only component that consumes real power. By analyzing the circuit conditions when components are removed, the individual reactances can be determined. When the capacitor is taken out, the circuit becomes an LR series combination where the current lags behind the voltage by $30^{\circ}$, a relationship defined by $\tan(30^{\circ}) = \frac{X_L}{R}$, which gives the inductive reactance $X_L = 200 \tan(30^{\circ}) = \frac{200}{\sqrt{3}} \Omega$. Similarly, removing the inductor creates a CR series combination where the current leads the voltage by $30^{\circ}$, leading to $\tan(30^{\circ}) = \frac{X_C}{R}$ and yielding a capacitive reactance $X_C = \frac{200}{\sqrt{3}} \Omega$.

In the complete LCR circuit, the impedance $Z$ is defined by $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Since the reactances $X_L$ and $X_C$ are equal, the reactive term cancels out, leaving the total impedance equal to the resistance, $Z = 200 \Omega$. The power dissipated is determined by the applied voltage and the impedance, allowing the formula to be simplified as follows:

$P = \frac{V^2 R}{Z^2} = \frac{V^2}{R} = \frac{220^2}{200} = 242 \text{ W}$

When a battery is connected to an inductor and a resistor in series, the inductor initially acts as an open circuit to oppose the rapid surge of current, creating a back electromotive force that makes the potential drop across it equal to the battery's total voltage at the instant the switch is closed. As time progresses, the current through this inductor-resistor branch increases following the standard transient growth equation, which causes the voltage across the inductor to decay exponentially toward zero as the current approaches a steady state.

The current in the branch containing $L$ and $R_2$ follows the relationship $i = \frac{E}{R_2} (1 - e^{-(R_2/L)t})$. By differentiating this expression with respect to time, the rate of change is determined to be $\frac{di}{dt} = \frac{E}{L} e^{-(R_2/L)t}$. Multiplying this rate by the inductance $L$ yields the potential drop across the inductor, expressed as $V_L = E e^{-(R_2/L)t}$. Substituting the provided values, where the electromotive force is 12 V, the resistance is 2 Ω, and the inductance is 0.4 H, the exponent term becomes $-(2 / 0.4)t$, which simplifies to $-5t$. Consequently, the potential drop as a function of time is $12 e^{-5t} \text{ V}$.

The average power consumed in an AC circuit is determined by the expression $P = E_{\text{rms}} I_{\text{rms}} \cos \phi$, where $\phi$ represents the phase difference between the voltage and the current. Comparing the provided expressions $E = E_0 \sin \omega t$ and $I = I_0 \sin (\omega t - \frac{\pi}{2})$, it is clear that the phase difference is $\phi = \frac{\pi}{2}$. Since the power factor term $\cos \phi$ evaluates to $\cos (\frac{\pi}{2}) = 0$, the total power consumption in the circuit is zero.

The behavior of a series resonant LCR circuit is governed by the condition where inductive reactance perfectly cancels capacitive reactance, making the total impedance purely resistive. Given the voltage across the resistor is 100 V and the resistance is 1000 ohms, the circuit current is determined by $I = \frac{100}{1000} = 0.1$ A. At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance, which is defined as $X_L = \frac{1}{\omega C}$. By substituting the given angular frequency of 200 rad/s and capacitance of 2 $\times 10^{-6}$ F, the reactance becomes $X_L = \frac{1}{200 \times 2 \times 10^{-6}} = 2500$ ohms. Consequently, the voltage across the inductor is calculated by multiplying the circuit current by this inductive reactance, $V_L = 0.1 \times 2500 = 250$ V.

Faraday's Law of Induction dictates that the electromotive force in a rotating coil depends on the time rate of change of the magnetic flux, which is defined as $\phi = N B A \cos(\omega t)$. Differentiating this flux with respect to time gives the instantaneous induced electromotive force as $e = N B A \omega \sin(\omega t)$. Since the maximum possible value of the sine term is one, the peak induced voltage produced by the generator is $N A B \omega$.

The phase difference in an alternating current circuit is dictated by the combination of its resistive and reactive elements. Since an R-L circuit inherently contains resistance, the phase difference is constrained to lie strictly between $0$ and $\frac{\pi}{2}$, making it impossible for this configuration to achieve a shift of exactly $\frac{\pi}{2}$. Pure inductive, pure capacitive, and LC circuits lack this resistive component, allowing them to produce the specific phase difference of $\frac{\pi}{2}$.

Maximum power in an AC circuit is achieved at electrical resonance, where the inductive reactance of the motor is exactly countered by the capacitive reactance. These reactances are defined as $X_L = 2 \pi f L$ and $X_C = \frac{1}{2 \pi f C}$ respectively, where $f$ is frequency and $L$ is inductance. Equating $X_L = X_C$ results in the resonance frequency condition, which can be rearranged to solve for capacitance:

$C = \frac{1}{4 \pi^2 f^2 L}$

Substituting the provided values where frequency is 50 Hz and inductance is 10 H gives $C = \frac{1}{4 \pi^2 \times 50^2 \times 10}$. Computing the denominator leads to $4 \pi^2 \times 2500 \times 10$, which simplifies to $100,000 \pi^2$. By applying the common approximation $\pi^2 \approx 10$, the calculation results in $1 \times 10^{-6}$ F, which corresponds to a capacitance of 1 µF.

In alternating current circuits, the power factor describes the phase relationship between voltage and current and is calculated as the ratio of the circuit's resistance to its total impedance. This relationship is defined by the following equation:

$\cos \phi = \frac{R}{Z}$

By substituting the known values, where the resistance is 12 ohm and the impedance is 15 ohm, the calculation requires dividing 12 by 15. Performing this division yields a power factor of 0.8.