Alternating Current – JEE Main PhysicsPractice Questions & PYQs

A fundamental principle in electromagnetic theory is that the displacement current arising within a capacitor is physically identical to the conduction current flowing through the external wires connected to it. This continuity ensures that charge is effectively transferred across the capacitor, meaning that both the conduction current and the displacement current share the same numerical value.

The conduction current in this capacitive circuit is given by $I = V / X_C$, and since the capacitive reactance is $X_C = 1 / (\omega C)$, the current can be expressed as $I = V \omega C$. Substituting the given values of 230 V, 300 rad/s, and 200 pF, the calculation proceeds as $I = 230 \times 300 \times 200 \times 10^{-12} \text{ A}$. Evaluating this equation results in $I = 13.8 \times 10^{-6} \text{ A}$, which is 13.8 μA. Therefore, both the conduction current and the displacement current are 13.8 μA.

The resonance frequency of a series LCR circuit is governed by the relationship $\omega = \frac{1}{\sqrt{LC}}$. For the circuit to maintain the same resonant frequency when the capacitance is modified, the product of the inductance and capacitance must remain constant. When the capacitance is increased from $C$ to $4C$, we set the new product equal to the initial value by solving $L' \times 4C = L \times C$, which simplifies to $L' = \frac{L}{4}$. Since the new inductance must be one-fourth of the original value, the inductance has been reduced by $L - \frac{L}{4}$, resulting in a decrease of $\frac{3}{4}L$.

In a series circuit containing a resistor and an inductor, the current remains constant throughout, while the total supply voltage is divided based on the individual impedance of each component. Because the current flows equally through the entire circuit, we can determine its magnitude by dividing the voltage drop across the resistor by its resistance value. Once the current is established, we can calculate the total impedance of the entire circuit using the source voltage, which then allows us to isolate the inductive reactance and determine the inductance of the coil.

The current flowing through the circuit is calculated using the voltage across the 90 Ohms resistor, giving $I = 36 / 90 = 0.4$ A. Given the total supply voltage of 120 V, the total impedance is $Z = 120 / 0.4 = 300$ Ohms. Using the relation $Z = \sqrt{R^2 + X_L^2}$ where $R = 90$ Ohms, we find $300^2 = 90^2 + X_L^2$, which yields $90000 = 8100 + X_L^2$. This simplifies to $X_L^2 = 81900$, resulting \in an inductive reactance of $X_L \approx 286.18$ Ohms. Finally, using the equation $X_L = 2 \pi f L$ with a frequency of 60 Hz and taking $\pi = 3.14$, the inductance is $L = 286.18 / (2 × 3.14 × 60) \approx 0.76$ H.

In a series LCR circuit, the current reaches its maximum value at resonance, a state where the inductive reactance exactly balances the capacitive reactance, leaving only the resistance to oppose current flow. This state happens at a specific resonant frequency determined by the values of the inductor and the capacitor.

The resonant frequency is defined by the equation:
$f = \frac{1}{2\pi \sqrt{LC}}$
Given inductance L is 100 mH and capacitance C is 2.5 nF, the product LC equals $25 \times 10^{-11}$ s². Taking the square root gives $5 \times 10^{-6} \times \sqrt{10}$. Since $\pi^2 = 10$, we know $\pi$ is equal to $\sqrt{10}$. Substituting these into the denominator of the resonant frequency formula, we get $2 \times \sqrt{10} \times 5 \times 10^{-6} \times \sqrt{10}$, which simplifies to $10 \times 10 \times 10^{-6}$, or $10^{-4}$. Dividing 1 by this value results in 10,000 Hz, which is $10 \times 10^3$ Hz.

In an LCR circuit operating at resonance, the inductive and capacitive reactances cancel each other out, which simplifies the total impedance to equal the resistance, represented by $Z = R$. Given that the current amplitude follows the relationship $I = \frac{V}{R}$, the current is inversely proportional to the resistance. If the resistance is halved, the denominator decreases by a factor of two, forcing the overall current amplitude to become double its original value.

The average power dissipated in an alternating current circuit is determined by the product of the root mean square voltage, the root mean square current, and the cosine of the phase difference between them, represented by the formula $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(\phi)$. For the voltage expression, the peak voltage is 100 V, which gives a root mean square voltage of $V_{\text{rms}} = 100 / \sqrt{2}$ V. Similarly, the peak current of 100 mA converts to 0.1 A, resulting in a root mean square current of $I_{\text{rms}} = 0.1 / \sqrt{2}$ A. With the phase difference $\phi = \pi/3$, the power factor is $\cos(\pi/3) = 0.5$. Substituting these values into the calculation provides the average power:

$P_{\text{avg}} = \left( \frac{100}{\sqrt{2}} \right) \times \left( \frac{0.1}{\sqrt{2}} \right) \times 0.5 = 2.5$

This yields a final value of 2.5 W.

The behavior of an inductor in an AC circuit is defined by its inductive reactance, which acts as the frequency-dependent opposition to current flow. This reactance is determined by the formula $X_L = 2 \pi f L$, where the voltage drop across the component relates to the current through the relationship $V_L = I X_L$.

Determining the specific current requires substituting the frequency of $50\text{Hz}$ and the inductance of $10\text{mH}$ into the reactance formula, resulting in $X_L = 2 \times 3.14 \times 50 \times 10 \times 10^{-3}\Omega$. This calculation simplifies to $3.14\Omega$ for the reactance. With the voltage across the inductor specified as $31.4\text{V}$, the current is found by dividing this voltage by the reactance, yielding $I = 31.4 / 3.14$, which results in $10\text{A}$.

Power dissipation in an AC circuit occurs exclusively across the resistor, as capacitors do not consume real power. To calculate this, determine the circuit's total impedance $Z$ using the relation $Z = \sqrt{R^2 + X^2}$. By substituting the given resistance $R = 4 \Omega$ and reactance $X = 4\sqrt{3} \Omega$ into this expression, the impedance is found to be $Z = \sqrt{16 + 48} = 8 \Omega$.

Next, determine the root-mean-square voltage by dividing the peak voltage of $8\sqrt{2}$ V by $\sqrt{2}$, which results in $V_{\text{rms}} = 8$ V. With these values, the rms current flowing through the series circuit is obtained using Ohm's law as $I_{\text{rms}} = V_{\text{rms}} / Z = 8 / 8 = 1$ A. Finally, the power dissipated is calculated using the formula $P = I_{\text{rms}}^2 R$, which results in $1^2 \times 4 = 4$ W.

In a purely capacitive circuit, the voltage lags behind the current by $90^\circ$, while in a purely inductive circuit, the voltage leads the current by $90^\circ$ due to the opposing nature of inductive reactance. When an LCR series circuit is at resonance, the inductive reactance $X_L$ cancels out the capacitive reactance $X_C$, causing the circuit to behave as if it were purely resistive. Any general LCR series circuit operating away from this resonant frequency will have a net reactance, which results in the voltage and current being out of phase by some angle.

Determining the potential difference across a capacitor in a series LCR circuit requires first finding the total current, which is driven by the circuit's overall impedance. We start by calculating the reactances for the inductor and capacitor, where the inductive reactance is $X_L = \omega L = 100 \times 1 = 100 \, \Omega$ and the capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 20 \times 10^{-6}} = 500 \, \Omega$. Next, we find the total impedance $Z$ of the series circuit using these reactances and the given resistance of $300 \, \Omega$, resulting \in $Z = \sqrt{(X_L - X_C)^2 + R^2} = \sqrt{(100 - 500)^2 + 300^2} = 500 \, \Omega$. With the impedance established, the rms current flowing through the circuit components is calculated as $i_{rms} = V_{rms} / Z = 50 / 500 = 0.1 \, A$. Since the same current passes through the capacitor, its rms potential difference is determined by the product of the capacitive reactance and this current, which yields $V_C = X_C × i_{rms} = 500 × 0.1 = 50 \, V$.

The behavior of an LCR series circuit is fundamentally determined by the impedance formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance, the condition $X_L = X_C$ causes the reactances to cancel out, which minimizes the impedance to $R$ and results \in the maximum possible amplitude current $I_m = V_m/R$. Because the total impedance $Z$ \in an LCR circuit is always greater than or equal to the resistance $R$ of a purely resistive circuit, the current $I_m = V_m/Z$ can never exceed the current $I_m = V_m/R$ found in the resistive setup. This confirms that the current in a purely resistive circuit is always greater than or equal to the current in the series LCR circuit when connected to the same voltage source.

Inserting a dielectric material between the plates of a capacitor increases its capacitance, which subsequently reduces the capacitive reactance given by the formula $X_C = \frac{1}{\omega C}$. This decrease in reactance leads to a drop in the total impedance of the series circuit, expressed as $Z = \sqrt{R^2 + X_C^2}$. Since the current flowing through the bulb is inversely related to the total impedance, this reduction in $Z$ results in a higher current, causing the bulb to glow with greater intensity.

The current in an AC circuit containing a capacitor is determined by the ratio of the rms voltage to the capacitive reactance. By observing the emf equation $E = 110\sqrt{2} \sin(100t)$, we identify the peak voltage, which corresponds to an rms voltage of 110 volts and an angular frequency $\omega$ of 100 rad/s. The capacitive reactance is calculated using $X_C = \frac{1}{\omega C}$, where the capacitance is 2 $\mu$F, resulting in a value of 5000 ohms. Dividing the rms voltage by this reactance yields an rms current of $\frac{110}{5000}$ amperes, which simplifies to 22 mA.

The power factor of a series LCR circuit is given by the ratio of resistance to total impedance, expressed as $\cos \phi = \frac{R}{Z}$. To find this, calculate the inductive reactance $X_L = 2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} = 10$ ohms. Then, determine the capacitive reactance $X_C = \frac{1}{2 \pi \times 50 \times (10^{-3}/\pi)} = 10$ ohms. Since both reactances are equal, the circuit operates at resonance, making the total impedance equal to the resistance, $Z = R$. Consequently, the power factor is $\frac{R}{Z} = 1$.

A coil behaves differently when exposed to direct current versus alternating current. In a DC circuit, an inductor acts as a simple resistor because the frequency is zero, meaning it offers no inductive reactance. According to Ohm's law, the resistance of the coil is $R = \frac{V}{I} = \frac{20}{5} = 4 \Omega$.

When an AC voltage is applied, the coil introduces an inductive reactance $X_L$ in addition to its resistance, resulting in a total impedance $Z$. Since the AC current is $4 \text{ A}$ under $20 \text{ V}$, the impedance is $Z = \frac{20}{4} = 5 \Omega$. These components are related by the equation $Z^2 = R^2 + X_L^2$
Substituting the values $Z=5$ and $R=4$ yields $25 = 16 + X_L^2$, which determines the reactance $X_L = 3 \Omega$.

The inductive reactance is defined by the formula $X_L = 2 \pi f L$
Using the values $f = 50 \text{ Hz}$ and $\pi = 3$, the equation becomes $3 = 2 \times 3 \times 50 \times L$. This simplifies to $3 = 300 L$, giving an inductance of $0.01 \text{ H}$, which is equivalent to $10 \text{ mH}$.

In a purely resistive circuit, the current follows the same sinusoidal variation as the voltage, expressed as $I(t) = I_0 \sin(\omega t)$. The current reaches half its peak value at a phase angle of $\frac{\pi}{6}$ and reaches the full peak value at a phase angle of $\frac{\pi}{2}$. The time elapsed between these two points is determined by the phase difference, $\Delta \theta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$, divided by the angular frequency $\omega$. Given $\omega = 100 \pi \text{ rad/s}$, the time calculation is $\Delta t = \frac{\pi / 3}{100 \pi} = \frac{1}{300} \text{ s}$, which results in approximately $3.3 \text{ ms}$.

The displacement current between the plates of a capacitor is effectively equivalent to the conduction current flowing through the rest of the circuit. We determine the capacitive reactance using the formula $X_C = \frac{1}{2 \pi f C}$ with a frequency of 4 kHz and a capacitance of 12 \mu F, which yields a reactance of approximately 3.317 \Omega . Dividing the peak voltage of 40 V by this reactance using the relation $I = \frac{V}{X_C}$ gives the maximum displacement current, which is nearly 12 A.

Power in an inductor-resistor circuit is dissipated exclusively by the resistive component, which acts as a constant parameter regardless of the voltage supply type. Because an inductor behaves as a simple conductor under steady direct current, the internal resistance is determined by the ratio of the applied direct voltage to the steady current, expressed as $100 / 5 = 20$ ohms. When the circuit is connected to an alternating voltage source, the total opposition to current, known as impedance, arises from the combination of this resistance and the inductive reactance, calculated as $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = 40$ ohms. We then find the root-mean-square current by dividing the rms voltage, $V_{\text{rms}} = 200 / \sqrt{2}$, by the impedance, which gives an rms current of $5 / \sqrt{2}$. Squaring this current and multiplying it by the resistance, $P = I_{\text{rms}}^2 R$, yields a final power dissipation of 250 W.

Transformers operate on the principle of mutual induction where the ratio of primary voltage $V_1$ to secondary voltage $V_2$ equals the ratio of the number of turns $N_1$ and $N_2$ in the coils, expressed by the relationship $\frac{V_1}{V_2} = \frac{N_1}{N_2}$. Substituting the primary voltage of 220 V and the turns ratio of 100 to 10 into this equation results in a secondary voltage $V_2 = 22$ V. Given that the secondary coil is connected to a circuit with series resistances, a current $I = 1$ mA produces the specified output voltage $V_0 = 7$ V.

The root-mean-square value of a current composed of both a direct component and an alternating sinusoidal component is determined by finding the square root of the sum of the square of the direct current and the mean squared value of the alternating current. For the given expression $i = 6 + \sqrt{56} \sin (100 \pi t + \frac{\pi}{3})$, the constant term represents the direct component of 6 A, while the amplitude of the alternating term is $\sqrt{56}$ A.

This calculation follows the relationship $I_{\text{rms}} = \sqrt{I_{\text{dc}}^2 + \frac{I_{\text{peak}}^2}{2}}$ to account for both parts effectively. Substituting the direct current component of 6 A and the peak alternating amplitude of $\sqrt{56}$ A, the equation becomes $I_{\text{rms}} = \sqrt{6^2 + \frac{(\sqrt{56})^2}{2}}$. Simplifying the squares within the expression yields $\sqrt{36 + 28}$, which further reduces to $\sqrt{64}$, resulting in a final value of 8 A.