Electromagnetic Induction And Alternating Currents – NEET UG PhysicsPractice Questions & PYQs

The time constant of a circuit containing capacitance and resistance characterizes how rapidly the system responds to electrical changes and is defined mathematically as the product of resistance and capacitance. In the transient charge equation $q = q_0 \left( 1 - e^{-t / (C R)} \right)$, the denominator inside the exponential function dictates the time scale, establishing that the characteristic parameter is given by $C R$.

Lenz's law dictates that any induced current in a closed loop will always flow in a direction that opposes the change producing it. As the bar magnet falls through the horizontal metal ring, the changing magnetic flux induces a current in the ring that creates a magnetic field opposing the motion of the magnet. This opposing magnetic force acts upward against gravity, effectively reducing the net downward force on the falling object. Consequently, the downward acceleration of the magnet becomes less than the acceleration due to gravity.

Alternating current varies sinusoidally over time, following the equation $i = i_0 \sin(\omega t)$ where $\omega = 2\pi f$. Setting the instantaneous current equal to the peak value of $5\ \text{A}$ for a frequency of $60\ \text{Hz}$ leads to the trigonometric equation $\sin(120\pi t) = 1$. Solving for the time when the sine function first reaches unity gives $120\pi t = \frac{\pi}{2}$, resulting in a time of $\frac{1}{240}\ \text{seconds}$.

The self-inductance of a long solenoid is determined by the formula $L = \mu_0 n^2 A l$, making the inductance directly proportional to the square of the turn density per unit length. When the number of turns per unit length is doubled, the squared term causes the overall self-inductance to increase by a factor of four.

The total electric charge circulated through a conducting loop during electromagnetic induction is obtained by integrating the induced current with respect to time. Because current equals induced electromotive force divided by resistance, and electromotive force is defined as the negative rate of change of magnetic flux, the time differential cancels out during integration. Consequently, the accumulated charge depends exclusively on the net change in magnetic flux rather than the speed at which that flux changes.

At the exact moment a circuit is energized at $t = 0$, reactive elements behave in contrasting ways because a capacitor offers negligible resistance while an inductor offers large resistance to current flow. Because of this immediate response, current rushes unimpeded through the capacitor branch to produce a maximum reading on its ammeter. In stark contrast, the inductor initially chokes off all current through its branch, causing the corresponding ammeter to read zero.

The induced electromotive force across any inductor is governed by the relation $E = -L \frac{dI}{dt}$, where the magnitude directly relies on the equivalent inductance of the circuit and the rate of change of current.

For configuration $P$, the setup produces an induced emf equivalent to that of a single individual inductor, leading to the expression $E_P = -L \frac{dI}{dt}$.

In configuration $Q$, the identical components are arranged in parallel, which halves the net equivalent inductance to $\frac{L}{2}$ and consequently generates an induced emf of $E_Q = -\frac{L}{2} \frac{dI}{dt}$.

Comparing both scenarios by taking the ratio of $E_P$ to $E_Q$ eliminates the common rate of change of current term, leaving a final ratio of 2.

The self-inductance of a coil relates the induced electromotive force to the time rate of change of current through the equation $\varepsilon = -L \frac{\mathrm{d}i}{\mathrm{d}t}$. Rearranging this expression to solve for inductance gives $L = \frac{-\varepsilon}{\mathrm{d}i / \mathrm{d}t}$. Substituting a voltage magnitude of $5$ volts, a current change from $3$ to $2$ amperes, and a time interval of $1$ millisecond calculates out to a self-inductance of $5$ milli-henry.

The self-inductance of a solenoid is determined by taking the ratio of total magnetic flux linkage to the current flowing through it, expressed as $L = \frac{N\Phi}{I}$. For a solenoid of length $I$ with $n$ turns per unit length, the total number of turns is given by $N = nI$, and the internal magnetic field is uniform with a magnitude of $B = \mu_0 n I$. The magnetic flux passing through each cross-sectional area multiplies with the total turns to yield a total flux linkage of $\mu_0 n^2 I^2 \pi r^2$. Dividing this total flux linkage by the current simplifies the expression to $L = \mu_0 \pi n^2 r^2 I$.

The average electrical power consumed in an alternating current circuit is determined by the formula $P = V_{\text{rms}} I_{\text{rms}} \cos \phi$, where $\phi$ represents the phase angle between the voltage and the current. Comparing the given equations with standard sinusoidal forms reveals that the current lags behind the voltage by a phase difference of $\frac{\pi}{2}$ radians. Substituting this phase angle into the power formula yields a power factor of $\cos(\frac{\pi}{2})$, which evaluates to zero. Thus, the total power consumption in the circuit is completely zero.

The impedance of a series alternating current circuit is evaluated using the formula $Z = \sqrt{R^2 + (X_L - X_C)^2}$, and the phase relationship is given by $\tan\phi = \frac{X_L - X_C}{R}$. Inserting the resistance of 20 ohms, inductive reactance of 45 ohms, and capacitive reactance of 25 ohms gives an impedance of 28.28 ohms. Dividing the supply voltage of 220 V by this impedance yields a current of 7.8 A. Evaluating the tangent expression gives $\frac{45 - 25}{20} = 1$, which indicates a phase angle of 45 degrees.

To find the potential difference across the circuit segment, we apply Kirchhoff's Voltage Law by traversing from point $A$ to point $B$ and accounting for the individual potential drops across each component. Moving along the current, we cross an inductor experiencing a changing current, which produces a potential drop of $L \frac{di}{dt}$, followed by a $5$ volt source, and a resistor that causes an ohmic drop of $i \times 2$. Setting up the traversal equation gives $V_A - L \frac{di}{dt} - 5 - i \times 2 = V_B$, where the inductance term evaluates to $1 \times 1$ given the rate of change of 1 amp per second. Substituting the current of $2 \text{ A}$ and these values into the expression results in $V_A - 1 - 5 - 4 = V_B$, which directly evaluates to a potential difference of 10 volt.

The electrical power in the secondary circuit of an ideal transformer is determined by the product of the secondary voltage and the secondary current using the equation $P = V I$. Because power loss is ignored, the 88 watt output power must equal the product of the 11000 V secondary voltage and the unknown secondary current. Rearranging this relationship gives the current as 88 divided by 11000, which evaluates to $0.008\text{ A}$, or equivalently 8 mA.

The time-dependent decay of a charge in a damped circuit is modeled by the equation $Q = Q_0 e^{-Rt / 2L}$, where the amplitude reduces to half its initial value when $Q = 0.50 Q_0$. Substituting this condition into the decay expression and taking the natural logarithm of both sides simplifies the relationship to $\ln(2) = \frac{Rt}{2L}$. Rearranging to solve for time yields the formula $t = \frac{2L \ln(2)}{R}$. Inserting the given inductance of 12 mH, resistance of 1.5 ohms, and natural log of 2 value calculates to a time of 11.09 ms.

In an alternating current circuit containing a pure inductor, the voltage across the inductor always leads the current by a phase angle of $\pi/2$. Given that the current is expressed as $I = I_0 \sin \omega t$, the voltage drop across the inductor becomes $V_L = V_0 \sin(\omega t + \frac{\pi}{2})$, which simplifies trigonometrically to $V_0 \cos \omega t$. By substituting the peak voltage relation $V_0 = I_0 X_L$ where the inductive reactance is $X_L = \omega L$, the final expression for the voltage drop evaluates to $\omega L I_0 \cos \omega t$.

The operational voltages of a transformer are directly proportional to the number of turns in its windings, governed by the relation $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. Since the primary to secondary turns ratio is given as 1 to 2, the inverse ratio representing secondary to primary turns is 2 to 1. Consequently, the ratio of the secondary voltage to the primary voltage mirrors this turns proportion exactly, resulting in a value of 2 to 1.

To find the peak current in a capacitive alternating current circuit, we first determine the frequency-dependent opposition of the capacitor and then scale the root-mean-square current by the peak factor. The opposition, known as capacitive reactance, is calculated using the formula $X_C = \frac{1}{2 \pi f C}$. Substituting the given source frequency of $50\text{ Hz}$, capacitance of $10 \mu\text{F}$, and $\pi = 3.14$ yields a reactance value of $1000 / 3.14\ \Omega$. Dividing the source voltage of $210\text{ V}$ by this reactance gives the root-mean-square current via the relation $i_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$. Multiplying this resulting root-mean-square current by the square root of two produces the peak current, which evaluates to $0.93\text{ A}$.

The self-inductance of a solenoid depends on its physical dimensions and magnetic properties, expressed through the relation $L = \mu_0 \mu_r n^2 A l$. Since both solenoids share the same relative permeability and cross-sectional area, the inductance is directly proportional to the square of the turns per unit length multiplied by the length, written as $L \propto n^2 l$. Substituting the given conditions where solenoid A has half the turns per unit length and twice the length of solenoid B, the proportionality yields $(0.5)^2 \times 2$ for solenoid A relative to solenoid B. This simplifies to $0.25 \times 2 = 0.5$, which represents a ratio of 1 to 2.

The total impedance in an alternating current network is determined by combining resistance and reactance, keeping in mind that for a direct current source, the frequency is zero and the inductive reactance drops to zero. This reduces the impedance expression $Z_1 = \sqrt{X_L^2 + R^2}$ down to just the ohmic resistance, resulting in a value of 10 ohms.

On the other hand, the second circuit operates with a frequency of 50 hertz and a capacitance of 1000 divided by pi microfarads, creating a capacitive reactance determined by the equation $X_C = \frac{1}{2 \pi f C}$ which evaluates to 10 ohms. Substituting this capacitive reactance and the resistance into the impedance formula $Z_2 = \sqrt{X_C^2 + R^2}$ gives an impedance of 10 multiplied by the square root of 2 ohms, establishing that the first impedance is strictly smaller than the second.

The phase angle in an alternating current circuit is given by the relation $\tan \phi = \frac{X_L}{R}$. The inductive reactance is determined by the equation $X_L = \omega L = 100\pi \times \frac{1}{\pi} = 100$ ohms. Substituting this reactance along with the resistance of 100 ohms gives the ratio $\tan \phi = \frac{100}{100} = 1$. Evaluating this expression results in a phase angle of 45 degrees.