Alternating Current – JEE Main PhysicsPractice Questions & PYQs

The root mean square current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$. Substituting the given current function $i = i_0(t/T)$ into this equation results in $I_{rms}^2 = \frac{1}{T} \int_{0}^{T} \left( \frac{i_0 t}{T} \right)^2 dt = \frac{i_0^2}{T^3} \int_{0}^{T} t^2 dt$. Evaluating the integral yields $I_{rms}^2 = \frac{i_0^2}{T^3} \left[ \frac{t^3}{3} \right]_0^T = \frac{i_0^2}{T^3} \left( \frac{T^3}{3} \right) = \frac{i_0^2}{3}$. Taking the square root of this value produces $I_{rms} = \frac{i_0}{\sqrt{3}}$.

In an alternating current series LCR circuit, the power factor $\cos \phi$ represents the ratio of resistance $R$ to impedance, allowing us to determine the phase angle $\phi$ from $\cos \phi = 0.5$, which yields $\phi = 60^\circ$. The relationship between the net reactance $|X_L - X_C|$ and the resistance is given by the expression $\tan \phi = \frac{|X_L - X_C|}{R}$. Substituting the values $R = 100 \Omega$ and $\phi = 60^\circ$ into this formula leads to $\sqrt{3} = \frac{|X_L - X_C|}{100}$, which implies the net reactance difference is $|X_L - X_C| = 100\sqrt{3}$. Comparing this calculated value to the expression provided in the problem, $\sqrt{3} \alpha$, demonstrates that $\alpha = 100$.

In an LCR series circuit, the phase angle $\phi$ between the current and the source voltage is governed by the relationship $\tan \phi = \frac{|X_L - X_C|}{R}$, where $X_L$ represents inductive reactance, $X_C$ represents capacitive reactance, and $R$ is the resistance. We begin by calculating the capacitive reactance using the angular frequency $\omega = 300 \text{ rad/s}$ and capacitance $C = 100 \times 10^{-6} \text{ F}$, which yields $X_C = \frac{1}{300 \times 100 \times 10^{-6}} = \frac{100}{3} \Omega$.

When switch $S_1$ is closed, the circuit operates with inductive reactance $X_{L1}$ and a phase angle of $30^\circ$, leading to the equation $\tan 30^\circ = \frac{X_{L1} - \frac{100}{3}}{R}$, which simplifies to $\frac{1}{\sqrt{3}} = \frac{X_{L1} - \frac{100}{3}}{R}$. Alternatively, when $S_2$ is closed, the circuit uses $X_{L2}$ with a phase angle of $60^\circ$, giving $\tan 60^\circ = \frac{X_{L2} - \frac{100}{3}}{R}$, or $\sqrt{3} = \frac{X_{L2} - \frac{100}{3}}{R}$.

To find the desired value, we divide the second equation by the first to eliminate $R$, resulting in $3 = \frac{X_{L2} - \frac{100}{3}}{X_{L1} - \frac{100}{3}}$. Multiplying across gives $3(X_{L1} - \frac{100}{3}) = X_{L2} - \frac{100}{3}$, which simplifies to $3X_{L1} - 100 = X_{L2} - \frac{100}{3}$. Rearranging these terms produces $3X_{L1} - X_{L2} = 100 - \frac{100}{3} = \frac{200}{3}$. Since reactance is defined as $X_L = \omega L$, we can write this as $3\omega L_1 - \omega L_2 = \frac{200}{3}$. Substituting $\omega = 300 \text{ rad/s}$, we obtain $3L_1 - L_2 = \frac{200}{3 \times 300} = \frac{2}{9}$.

In a series RC circuit, the total impedance is defined by $Z = \sqrt{R^2 + X_C^2}$, where the capacitive reactance is inversely proportional to frequency, given by $X_C = \frac{1}{\omega C}$. With the voltage $V$ kept constant, the initial current $I$ is expressed as:
$I = \frac{V}{\sqrt{R^2 + X_C^2}}$
When the angular frequency is reduced to $\omega / 4$, the new capacitive reactance increases to $X'_C = \frac{1}{(\omega / 4)C} = 4X_C$. Consequently, the new impedance becomes $Z' = \sqrt{R^2 + (4X_C)^2} = \sqrt{R^2 + 16X_C^2}$. Given the new current is $I' = I/3$, we can relate the two states by dividing the initial current expression by the new one:
$\frac{I}{I/3} = \frac{\sqrt{R^2 + 16X_C^2}}{\sqrt{R^2 + X_C^2}} = 3$
Squaring both sides of this equation eliminates the square roots and allows for the isolation of the variables:
$\frac{R^2 + 16X_C^2}{R^2 + X_C^2} = 9 \implies R^2 + 16X_C^2 = 9R^2 + 9X_C^2$
Rearranging the terms yields $8R^2 = 7X_C^2$, which simplifies to the ratio of resistance to reactance at frequency $\omega$ as $\frac{R}{X_C} = \sqrt{\frac{7}{8}}$.

In an LCR circuit, maximum current occurs at resonance where the inductive reactance equals the capacitive reactance, satisfying the condition $\omega = \frac{1}{\sqrt{LC}}$. The angular frequency derived from the source voltage $V = 5 \sin(100t)$ is $\omega = 100 \text{ rad/s}$. Given the inductance $L = 2 \text{ H}$ from the circuit figure, the resonance condition $\omega^2 = \frac{1}{LC}$ determines the required capacitance.

Substituting the known values into the rearranged equation $C = \frac{1}{\omega^2 L}$ results in:
$C = \frac{1}{(100)^2 \times 2} = \frac{1}{20000} \text{ F}$
Converting this value to microfarads yields:
$C = 5 \times 10^{-5} \text{ F} = 50 \text{ \mu F}$

The magnetic energy $U$ stored in an inductor and the power $P$ dissipated by its resistance $R$ are expressed as $U = \frac{1}{2} L I^2$ and $P = I^2 R$, where $L$ is the inductance and $I$ is the current. The ratio of inductive reactance $X_L = 2 \pi f L$ to resistance $R$ is derived by substituting $L = \frac{2U}{I^2}$ and $R = \frac{P}{I^2}$ into the ratio expression:

$\frac{X_L}{R} = \frac{2 \pi f L}{R} = \frac{2 \pi f (2U / I^2)}{(P / I^2)} = \frac{4 \pi f U}{P}$

Substituting the given values $U = 16$, $P = 32$, $f = 50$, and $\pi = 3.14$:

$\frac{X_L}{R} = \frac{4 \times 3.14 \times 50 \times 16}{32} = 314$

In an LCR series circuit, the power factor is defined as the ratio of resistance to impedance, representing the cosine of the phase angle between voltage and current. To find this value, we first determine the total impedance of the circuit using the relationship $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Substituting the given resistance of 80 ohms, inductive reactance of 20.0 ohms, and capacitive reactance of 80.0 ohms into this expression, we calculate the impedance as the square root of 80 squared plus the square of the difference between 20.0 and 80.0, which results in a total impedance of 100 ohms. Dividing the resistance by this total impedance yields the power factor, calculated as 80 divided by 100, which equals 0.8.

Resonance in a series LCR circuit occurs when the inductive reactance equals the capacitive reactance, forcing the circuit to behave as a purely resistive load. The angular frequency at which this condition is met is determined by the formula $\omega = \frac{1}{\sqrt{LC}}$. Given the circuit parameters of inductance $L = 1.6 \text{ H}$ and capacitance $C = 40 \mu\text{F}$, we first calculate the resonant angular frequency by substituting these values into the equation:

$\omega = \frac{1}{\sqrt{1.6 \times 40 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125 \text{ rad/s}$

With the angular frequency established, we can easily find the inductive reactance using the relationship $X_L = \omega L$. Multiplying the calculated angular frequency of $125 \text{ rad/s}$ by the given inductance of $1.6 \text{ H}$ yields an inductive reactance of $200 \Omega$.

The power factor of a series LCR circuit is given by the formula $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}$, where $X_L = 2\pi f L$ is the inductive reactance and $X_C = \frac{1}{2\pi f C}$ is the capacitive reactance. Calculating these reactances using the source frequency $f = 50 \, Hz$ and the specific component values $R, L,$ and $C$ from the circuit allows for the determination of the total impedance $Z$.

Substituting the calculated impedance into the power factor expression yields $\cos \phi = 0.6$. Given that the power factor is represented as $\frac{\alpha}{10}$, equating $\frac{\alpha}{10} = 0.6$ results in $\alpha = 6$.

When an alternating current is expressed as a superposition of sine and cosine components such as $I=I_A \sin \omega t + I_B \cos \omega t$, it can be rewritten as a single sinusoidal function with a peak amplitude equal to $\sqrt{I_A^2 + I_B^2}$. The root mean square value of any sinusoidal current is determined by dividing its peak amplitude by $\sqrt{2}$. By applying this relationship to the combined amplitude, we find $I_{rms} = \frac{\sqrt{I_A^2 + I_B^2}}{\sqrt{2}}$, which simplifies to $\sqrt{\frac{I_A^2 + I_B^2}{2}}$.

The standard equation for alternating current is $i = I_0 \sin(\omega t)$, where $I_0$ represents the peak current and $\omega$ represents the angular frequency. Comparing this form to the given equation $i = 100 \sqrt{2} \sin(100 \pi t)$, we identify the peak current $I_0 = 100 \sqrt{2}$ A and the angular frequency $\omega = 100 \pi$ rad/s.

The RMS value of the current is calculated using the relation $I_{rms} = \frac{I_0}{\sqrt{2}}$. Substituting the peak current value gives $I_{rms} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100$ A.

Frequency is related to angular frequency by the formula $\omega = 2 \pi f$. Solving for $f$ yields $f = \frac{\omega}{2 \pi} = \frac{100 \pi}{2 \pi} = 50$ Hz.

In a series LCR circuit, current reaches its maximum value at the resonance frequency where the inductive reactance balances the capacitive reactance, satisfying the relationship $\omega = \frac{1}{\sqrt{LC}}$. Substituting the circuit parameters into this formula yields $\omega = \frac{1}{\sqrt{100 \times 10^{-3} \times 25 \times 10^{-9}}}$. This expression simplifies to $\frac{1}{\sqrt{2500 \times 10^{-12}}}$, which further reduces to $\frac{1}{50 \times 10^{-6}}$. Performing the final division results in an angular frequency of $2 \times 10^4 \, \text{rad s}^{-1}$.

When a current waveform is composed of distinct DC and AC parts, the overall root mean square value is calculated using the property that the square of the total RMS value equals the sum of the squares of the RMS values of each individual component. The DC term $5\sqrt{2}$ contributes a squared RMS value of $(5\sqrt{2})^2 = 50$. The AC component is $10 \cos(650\pi t + \frac{\pi}{6})$, which has a peak amplitude of 10, meaning its squared RMS contribution is $(\frac{10}{\sqrt{2}})^2 = 50$. Adding these individual squared values results in $50 + 50 = 100$, and the square root of this sum gives a final RMS value of 10 Amp.

The power rating and voltage of a device determine its internal resistance, while the peak current in an alternating current system is calculated by multiplying the root-mean-square current by $\sqrt{2}$. Determining the internal resistance requires the formula $R = \frac{V^2}{P}$, which with 220 V and 100 W yields $R = 484$ ohms. Calculating the RMS current involves dividing the supply voltage by this resistance using $I_{\text{rms}} = \frac{V_{\text{rms}}}{R}$, which gives $I_{\text{rms}} = \frac{5}{11}$ A. Finally, multiplying this value by $\sqrt{2}$ yields the peak current $I_0 = I_{\text{rms}}\sqrt{2}$, resulting in approximately 0.64 A.

The maximum current in an LR series circuit is determined by the ratio of the peak source voltage $V_0$ to the circuit impedance $Z$, given by $I_0 = \frac{V_0}{Z}$. The inductive reactance is calculated as $X_L = 2 \pi f L = 100\pi \, \Omega$, and the total impedance of the circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\pi \sqrt{2} \, \Omega$. Assuming the provided AC supply voltage is the root mean square value $V_{rms} = 100\pi \, \text{V}$, the peak voltage is $V_0 = V_{rms} \sqrt{2} = 100\pi \sqrt{2} \, \text{V}$, which yields a maximum current of $I_0 = \frac{100\pi \sqrt{2}}{100\pi \sqrt{2}} = 1 \, \text{A}$.

Average power dissipation in an AC circuit occurs only across the resistance, as the inductor and capacitor store energy without dissipating it. To find the power, first calculate the total impedance of the series circuit using the formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$ where $R$ is the resistance, $X_L$ is the inductive reactance, and $X_C$ is the capacitive reactance. Substituting the provided values $R = 50$, $X_L = 100$, and $X_C = 50$ gives $Z = \sqrt{50^2 + (100 - 50)^2} = 50\sqrt{2} \text{ }\Omega$. Next, determine the root-mean-square current \in the circuit using the relation $I_{rms} = \frac{V}{Z}$ where $V = 10 \text{ V}$ is the source voltage. This yields $I_{rms} = \frac{10}{50\sqrt{2}} = \frac{1}{5\sqrt{2}} \text{ A}$. Finally, calculate the average power dissipated by the resistor using $P = I_{rms}^2 R$. Substituting the values results \in $P = \left(\frac{1}{5\sqrt{2}}\right)^2 \times 50 = \frac{1}{50} \times 50 = 1 \text{ W}$.

In a series LCR circuit, resonance occurs when the inductive and capacitive reactances cancel each other out, leaving the total impedance $Z$ equal to the resistance $R$. At this state, the current amplitude is defined by Ohm's law as $I_0 = \frac{E}{R}$. When the resistance is doubled to $2R$ while keeping the emf $E$ constant, the new current amplitude becomes $I' = \frac{E}{2R}$. By substituting the initial current amplitude into this equation, the new current simplifies to $\frac{I_0}{2}$.

In circuits consisting of identical symmetric branches, a phase difference of 90 degrees between the input voltage and the differential output voltage across those branches indicates that the circuit is operating at a frequency where the magnitude of the resistance matches the capacitive reactance. When these two impedances are equal, the voltage drops across the components create the specific phase shift required to satisfy the physical condition of the system.

The capacitive reactance is determined by the formula $X_C = \frac{1}{\omega C}$ and to satisfy the circuit condition, we set this reactance equal to the resistance, resulting \in the equation $R = \frac{1}{\omega C}$. With the resistance equal to 100,000 Ohms and the capacitance equal to 0.0000000001 Farads, we rearrange the relationship to isolate the angular frequency as $\omega = \frac{1}{R \times C}$. Substituting the values into this equation yields an angular frequency of 100,000 rad/s. Comparing this result to the given frequency expression $10^x$ rad/s confirms that the value of x is 5.

The total energy in an LC circuit oscillates between the electric field of the capacitor and the magnetic field of the inductor, and at the point of maximum current, the initial electrical energy is entirely converted into magnetic energy. By setting the initial potential energy, $\frac{1}{2}CV^2$, equal to the peak magnetic energy, $\frac{1}{2}LI_{\text{max}}^2$, we can determine the maximum current using the relationship $I_{\text{max}} = V \sqrt{\frac{C}{L}}$.

Substituting the provided values of 100 $\mu$F for capacitance, 6.4 mH for inductance, and 12 V for potential, the expression for the maximum current becomes $12 \sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}$. The value inside the square root simplifies to $\frac{1}{64}$, and taking its square root results in 1/8. Multiplying 12 by 1/8 yields a maximum current of 1.5 A.

The efficiency of a transformer represents the ratio of useful output power to total input power, indicating how effectively the device manages energy conversion. First, calculate the input power by multiplying the primary voltage of 2300 V by the primary current of 5 A, which results in 11500 W. Given the efficiency of 90%, the useful output power becomes 10350 W. Applying the relation $P_{out} = V_s \times I_s$, we set up the equation $10350 = 230 \times I_s$, which solves to an output current of 45 A.