Class 12 Physics (Part 1) Chapter 12 Alternating Current

Key Concept: Mean value of AC for one cycle and half-cycle, RMS value

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For an alternating current $I = I_0 \sin \omega t$, the mean value over one complete cycle is calculated as:
$I_{m} = \frac{1}{T} \int_{0}^{T} I_0 \sin \omega t \, dt = 0$
This is because the integral of $\sin \omega t$ over a full period $T$ is zero, confirming that the Assertion is true. The Reason correctly explains why the mean value is zero: the positive and negative half-cycles indeed cancel each other out, leading to zero net displacement of charge. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Inductive reactance, Phase difference

c) $26.57^\circ$
[Solution Description]
Step 1: Calculate the inductive reactance $X_L = \omega L = 300 \times 0.05 = 15 \, \Omega$.
Step 2: The phase difference $\phi$ in an RL circuit is given by:
$\tan \phi = \frac{X_L}{R} = \frac{15}{30} = 0.5$
Step 3: Thus, $\phi = \tan^{-1}(0.5) \approx 26.57^\circ$.

Key Concept: Phase Difference, Inductive Reactance

d) Current lags the voltage by $90^\circ$
[Solution Description]
For a pure inductor, the current lags behind the voltage by $90^\circ$. The phase difference $\phi$ is given by:
$\phi = 90^\circ$
Therefore, the current lags the voltage by $90^\circ$.

Key Concept: Frequency and Periodic Time Relationship

b) 0.02 seconds
[Solution Description]
Frequency (f) is related to periodic time (T) by the formula $T = \frac{1}{f}$. Given $f = 50$ Hz, $T = \frac{1}{50} = 0.02$ seconds or 20 milliseconds.

Key Concept: Instantaneous voltage induced in a rotating coil

c) 17.68 V
[Solution Description]
The instantaneous voltage induced in the coil is given by:
$V = N B A \omega \sin \omega t$
Given, $N = 100$, $B = 0.1 \, \text{T}$, $A = 0.05 \, \text{m}^2$, $\omega = 50 \, \text{rad/s}$, and $\theta = 45^\circ$.
Since $\theta = \omega t$, we substitute $\omega t = 45^\circ$.
Now, $\sin 45^\circ = \frac{1}{\sqrt{2}}$.
Plugging in the values:
$V = 100 \times 0.1 \times 0.05 \times 50 \times \frac{1}{\sqrt{2}} \\ = 25 \times \frac{1}{\sqrt{2}} \\ \approx 17.68 \, \text{V}$

Key Concept: Phasor Analysis, Power Factor

c) $0.8$
[Solution Description]
The power factor is defined as $\cos\phi$, where $\phi$ is the phase angle between the current and voltage. A lagging power factor means the current lags the voltage. Here, the power factor is given directly as $0.8$, so $\cos\phi = 0.8$.

Key Concept: Phase Difference

c) $-90^\circ$ (current leads voltage)
[Solution Description]
In a purely capacitive circuit, the current leads the voltage by $90^\circ$. This means the phase difference $\phi$ is $-90^\circ$.

Key Concept: Instantaneous Current and Voltage

b) Current leads voltage by 90°
[Solution Description]
In a purely capacitive circuit, the current leads the voltage by 90°. This means that when the voltage is at its peak, the current is zero, and vice versa. The phase relationship is such that the current reaches its maximum value before the voltage does.

Key Concept: Mean Value for Half-Cycle

c) 6.37 A
[Solution Description]
Given the peak current $I_0 = 10 A$, the mean current for a positive half-cycle is calculated using the formula:
$I_m = \frac{2 I_0}{\pi}$
Substituting the given value:
$I_m = \frac{2 \times 10}{\pi} \approx 6.37 A$

Key Concept: Instantaneous Power, L-R Circuit

b) 550 W
[Solution Description]
The average power in an L-R circuit over one complete cycle is given by:
$P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi$
Substituting the given values:
$P = 220 \times 5 \times \cos 60^\circ = 220 \times 5 \times 0.5 = 550 \text{ W}$

Key Concept: Resistive Circuit, RMS Current

a) $I_{rms} = 3.535$ A, $V_0 = 50$ V
[Solution Description]
Given current $I = 5 \sin (120\pi t)$.
Peak current ($I_0$) = 5 A.
RMS current ($I_{rms}$) = $\frac{I_0}{\sqrt{2}} = \frac{5}{\sqrt{2}} = 3.535$ A.
Resistance ($R$) = 10 ohms.
Peak voltage ($V_0$) = $I_0 R = 5 \times 10 = 50$ V.

Key Concept: Peak Value, RMS Value

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the RMS value of an alternating current is always less than its peak value. This is true because the RMS value for a sinusoidal waveform is given by $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$, and $\frac{1}{\sqrt{2}} \approx 0.707 < 1$. The reason correctly explains why the assertion is true by providing the formula that shows the RMS value is derived from the peak value divided by $\sqrt{2}$, which ensures it is always less than the peak value.

Key Concept: Phasor Representation in C-R Circuit

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a C-R circuit, the voltage across the capacitor ($V_C$) lags behind the current by $90^\circ$, while the voltage across the resistor ($V_R$) is in phase with the current. The resultant applied voltage ($V$) is the vector sum of $V_R$ and $V_C$, leading to a phase difference $\phi$ between the current and the applied voltage. The tangent of this phase angle is given by $\tan \phi = \frac{V_C}{V_R} = \frac{X_C}{R}$, where $X_C$ is the capacitive reactance. Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Direction of Induced Current

b) From B to A
[Solution Description]
Fleming's right-hand rule states that if the thumb points in the direction of motion (upwards), the forefinger points in the direction of the magnetic field (North to South), then the middle finger gives the direction of the induced current (into the plane for AB).

Key Concept: Eddy Current Losses, Hysteresis Losses

b) Increasing the thickness of the core
[Solution Description]
Step 1: Copper losses are reduced by using thick wires to minimize resistance.
Step 2: Eddy current losses are minimized by laminating the core.
Step 3: Hysteresis losses are reduced by using magnetic materials with small hysteresis loops.
Step 4: Using a solid iron core increases eddy current losses as it does not limit circulating currents.
Here, increasing the thickness of the core would increase eddy currents, not reduce them.

Key Concept: Pure Resistor Circuit

a) $0^\circ$
[Solution Description]
For a pure resistor circuit, the current and voltage are always in phase. This follows directly from Ohm's law:
Given voltage: $V = V_0 \sin(\omega t)$
Current: $I = \frac{V_0}{R} \sin(\omega t) = I_0 \sin(\omega t)$
Thus, the phase difference is $0$.

Key Concept: Power Dissipation in AC

c) 250 W
[Solution Description]
The average power dissipated in a resistor for an AC circuit is given by $P = I_{\text{rms}}^2 R$. Here, $I_{\text{rms}} = 5$ A and $R = 10 \Omega$. Thus, $P = 5^2 \times 10 = 250$ W.

Key Concept: Power dissipation in choke coil

a) Nearly zero
[Solution Description]
The power dissipated in the choke coil is given by:
$P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi$
where $\cos \phi$ is the power factor. For a choke coil with negligible resistance ($R \approx 0$) and high inductance ($L$), the power factor is:
$\cos \phi = \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \approx 0$
Thus, the power dissipated ($P$) is nearly zero.

Key Concept: Capacitive Reactance

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that in a purely capacitive circuit, the current leads the voltage by $90^\circ$, which is correct as per the phase relationship in capacitive circuits. The Reason states that capacitive reactance $X_C$ decreases with increasing frequency, which is also true because $X_C = \frac{1}{2\pi f C}$. However, the Reason does not explain why the current leads the voltage by $90^\circ$; it merely describes the behavior of capacitive reactance with frequency. Therefore, both statements are true, but the Reason is not the correct explanation for the Assertion.

Key Concept: Ohm's Law in AC Circuit

b) 1.414 A
[Solution Description]
First, find the peak current using Ohm's law: $I_0 = \frac{V_0}{R} = \frac{100}{50} = 2 \text{ A}$. Then, calculate the rms current: $I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} \approx 1.414 \text{ A}$.

Key Concept: Q-Factor Formula

c) $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
[Solution Description]
The Q-factor (Quality Factor) in a series resonant circuit is given by $Q = \frac{\omega_0}{\Delta \omega} = \frac{1}{R} \sqrt{\frac{L}{C}}$, where $\omega_0$ is the resonant frequency, $\Delta \omega$ is the bandwidth, $R$ is the resistance, $L$ is the inductance, and $C$ is the capacitance.

Key Concept: Capacitive Reactance, Frequency Dependence

b) It halves
[Solution Description]
Capacitive reactance is inversely proportional to frequency ($X_C = \frac{1}{2\pi fC}$). Doubling the frequency will halve the reactance.
Calculation:
1. Original reactance at 50 Hz: $X_{C1} = \frac{1}{2\pi \times 50 \times 10^{-5}} = 318.31 \Omega$
2. New reactance at 100 Hz: $X_{C2} = \frac{1}{2\pi \times 100 \times 10^{-5}} = 159.15 \Omega$
3. Ratio: $\frac{X_{C2}}{X_{C1}} = \frac{159.15}{318.31} = 0.5$ (halved)

Key Concept: Impedance at Resonance

b) $5 \, \Omega$
[Solution Description]
At resonance in a series L-C-R circuit, the reactances $X_L$ and $X_C$ cancel each other out, so the impedance $Z$ is purely resistive and equal to $R$.
$Z = R = 5 \, \Omega$
Therefore, the impedance at resonance is $5 \, \Omega$.

Key Concept: Resonant Frequency, Bandwidth

c) 5
[Solution Description]
First, calculate the resonant frequency ($\omega_0$) using $\omega_0 = \frac{1}{\sqrt{LC}}$.
Given:
$L = 50 \times 10^{-3} \text{ H}$,
$C = 20 \times 10^{-6} \text{ F}$.
Plugging in the values:
$\omega_0 = \frac{1}{\sqrt{(50 \times 10^{-3})(20 \times 10^{-6})}}$,
$\omega_0 = \frac{1}{\sqrt{1000 \times 10^{-9}}}$,
$\omega_0 = \frac{1}{\sqrt{10^{-6}}}$,
$\omega_0 = 1000 \text{ rad/s}$.
Next, calculate bandwidth ($\Delta\omega$) using $\Delta\omega = \frac{R}{L}$.
$\Delta\omega = \frac{10}{50 \times 10^{-3}}$,
$\Delta\omega = 200 \text{ rad/s}$.
Now, the Q-factor is given by $Q = \frac{\omega_0}{\Delta\omega}$,
$Q = \frac{1000}{200} = 5$.
Alternatively, one can also compute Q-factor directly using $Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{10} \sqrt{\frac{50 \times 10^{-3}}{20 \times 10^{-6}}} = \frac{1}{10} \times 50 = 5$.

Key Concept: Alternating Voltage and Current, Faraday's Law, Lenz's Law

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the induced emf is maximum when the plane of the coil is perpendicular to the magnetic field. However, according to Faraday’s law, the induced emf (\$V = -N \frac{d\Phi_B}{dt}\$) depends on the rate of change of magnetic flux (\$\frac{d\Phi_B}{dt}\$). When the plane of the coil is perpendicular to the magnetic field (\$\theta = 0^\circ\$ or \$180^\circ\$), the flux (\$\Phi_B = BA \cos \omega t\$) is maximum, but its rate of change (\$\frac{d\Phi_B}{dt} = -BA \omega \sin \omega t\$) is zero because \$\sin 0^\circ = 0\$ and \$\sin 180^\circ = 0\$. Thus, the induced emf is zero at these positions. The emf is maximum when the plane of the coil is parallel to the magnetic field (\$\theta = 90^\circ\$ or \$270^\circ\$), where \$\sin \omega t = \pm 1\$. Therefore, the Assertion is false. The Reason correctly states that the rate of change of flux is zero when the plane is perpendicular to the field, which is true. Hence, the correct answer is (d).

Key Concept: Power Factor

c) 1
[Solution Description]
In a purely resistive AC circuit, the phase difference between voltage and current is zero ($\phi = 0$). Hence, the power factor is:
$\cos \phi = \cos 0^\circ = 1$.
Therefore, the power factor is 1.

Key Concept: Peak Voltage Calculation

c) 169.7 V
[Solution Description]
The peak voltage is related to the RMS voltage by:
$V_0 = \sqrt{2} \, V_{\text{rms}}$
Given $V_{\text{rms}} = 120$ V,
$V_0 = 1.414 \times 120 \approx 169.7\, \text{V}.$

Key Concept: Transformer Construction

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to the syllabus, in a step-up transformer, the primary coil (input side) has fewer turns of thick wire compared to the secondary coil (output side), which has more turns of thin wire. This is because a step-up transformer increases the voltage from the primary to the secondary side. The assertion is correct as it accurately describes the construction of a step-up transformer.
The reason states that in a step-up transformer, the input voltage is less than the output voltage, which is also true. However, this does not explain why the primary coil has fewer turns; rather, it is a consequence of the transformer's function. Therefore, the reason is true but does not correctly explain the assertion.

Key Concept: Transformation Ratio

c) 880 V
[Solution Description]
The transformation ratio $r$ for a transformer is given by:
$r = \frac{N_s}{N_p} = \frac{V_s}{V_p}$
Given $N_p = 200$, $N_s = 800$, and $V_p = 220 \, \text{V}$. Substituting these values:
$\frac{V_s}{220} = \frac{800}{200} \implies V_s = 4 \times 220 = 880 \, \text{V}$
Therefore, the output voltage is 880 V.

Key Concept: Bandwidth

c) $\Delta \omega = \frac{R}{L}$
[Solution Description]
Bandwidth is the difference between the upper and lower half power frequencies. It is given by:
$\Delta \omega = \omega_2 - \omega_1 = \frac{R}{L}$

Key Concept: Energy in L-C Circuit

b) 10 $\mu$J
[Solution Description]
The initial energy stored in the capacitor is given by:
$E = \frac{q_0^2}{2C}$
Substituting $q_0 = 10 \times 10^{-6} \, \text{C}$ and $C = 5 \times 10^{-6} \, \text{F}$:
$E = \frac{(10 \times 10^{-6})^2}{2 \times 5 \times 10^{-6}} = \frac{100 \times 10^{-12}}{10 \times 10^{-6}} = 10 \times 10^{-6} \, \text{J} = 10 \, \mu\text{J}$

Key Concept: Peak Value, Periodic Time

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given: Angular frequency $\omega = 100\pi$ rad/s and peak current $I_0 = 10$ A.
First, calculate the frequency using the formula $f = \frac{\omega}{2\pi}$. Substituting $\omega = 100\pi$, we get:
$f = \frac{100\pi}{2\pi} = 50 \text{ Hz}.$
Since frequency represents the number of cycles per second, the assertion states that the current completes 50 cycles in one second, which matches the calculated frequency.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Power Factor in L-R Circuit

b) 0.6
[Solution Description]
The power factor ($\cos \phi$) in an L-R circuit is given by $\cos \phi = \frac{R}{Z}$, where $Z$ is the impedance. The impedance of the circuit is calculated as $Z = \sqrt{R^2 + X_L^2} = \sqrt{(30\,\Omega)^2 + (40\,\Omega)^2} = 50\,\Omega$. Thus, the power factor is $\cos \phi = \frac{30\,\Omega}{50\,\Omega} = 0.6$.

Key Concept: Induced EMF, Angular Velocity

c) 100 V
[Solution Description]
The maximum emf generated by an AC generator is given by:
$V_0 = N B A \omega$
where $N$ is the number of turns (100), $B$ is the magnetic field (0.4 T), $A$ is the area of the coil (0.05 m\textsuperscript{2}), and $\omega$ is the angular velocity (50 rad/s).
Substituting the values:
$V_0 = 100 \times 0.4 \times 0.05 \times 50 = 100 \text{ V}$
So, the maximum emf generated is 100 V.

Key Concept: Phasor Representation, Phase Difference

b) $I = 120 \sin\left(100\pi t - \frac{\pi}{4}\right)$
[Solution Description]
Given: Voltage $V = 120 \sin(100\pi t)$. Current lags voltage by $\phi = \frac{\pi}{4}$ radians.
The general form of current in an AC circuit when it lags the voltage is $I = I_0 \sin(\omega t - \phi)$, where $\omega = 100\pi$ rad/s (from the voltage equation).
Substituting $\phi = \frac{\pi}{4}$, we get:
$I = I_0 \sin\left(100\pi t - \frac{\pi}{4}\right)$. However, since $I_0$ is not provided, we represent it as $I = k \sin\left(100\pi t - \frac{\pi}{4}\right)$ where $k$ is the amplitude.

Key Concept: Maximum Value of Alternating Current

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the peak value of alternating current is denoted by $I_0$, which is true as per the syllabus definition. The reason states that the peak value represents the maximum current produced in one complete cycle, which is also true and correctly explains why the peak value is denoted by $I_0$.

Key Concept: Electromagnetic Induction, Fleming's Right-Hand Rule

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the direction of the induced current reverses every half rotation of the coil in an AC generator. This is true because, as the coil rotates, the arms AB and CD change their direction of motion relative to the magnetic field. According to Fleming’s right-hand rule, this causes the direction of the induced current to reverse. The Reason correctly explains the Assertion by linking the reversal of current direction to the change in relative motion of the coil arms and the application of Fleming’s right-hand rule. Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Impedance Calculation

b) $20 \sqrt{2} \Omega$
[Solution Description] The impedance $Z$ of a series L-C-R circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Substituting the given values:
$Z = \sqrt{20^2 + (30 - 10)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \Omega$

Key Concept: Electromagnetic Induction - AC Generator

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that in an AC generator, the induced emf is maximum when the plane of the coil is perpendicular to the magnetic field. This is true because the emf induced in the coil is given by $V = V_0 \sin \omega t$, where $V_0 = N B A \omega$. The emf is maximum ($V_0$) when $\sin \omega t = 1$, which occurs when the plane of the coil is parallel to the magnetic field (not perpendicular). Therefore, the assertion is false.
The reason states that the rate of change of magnetic flux linked with the coil is maximum when the plane of the coil is perpendicular to the magnetic field. This is also incorrect because the rate of change of magnetic flux ($d\phi/dt$) is maximum when the plane of the coil is parallel to the magnetic field (when $\cos \omega t = 0$). Thus, the reason is false as well.
However, the correct explanation for the maximum emf is indeed related to the rate of change of flux, but both statements are inaccurate in their current form. Hence, the correct answer is:

Key Concept: Frequency and Time Period

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The frequency of an alternating current is defined as the number of cycles completed per second. For a domestic alternating current, the frequency is given as 50 Hz (hertz), which means the current completes 50 cycles in one second. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion. Therefore, the correct answer is option (a).

Key Concept: Average power in pure inductive and capacitive circuits, phase difference

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a purely inductive circuit, the instantaneous power is given by $P = V_{\text{rms}} I_{\text{rms}} \cos(\omega t) \sin(\omega t)$. Integrating this over one full cycle results in zero because the product $\cos(\omega t)\sin(\omega t)$ has an average value of zero over a period. This confirms that the Assertion is true. The Reason states the correct explanation — the $90^\circ$ phase difference between voltage and current causes the power oscillations to cancel out over a cycle. Hence, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Resonance Condition

c) $X_L = X_C$
[Solution Description] Resonance occurs when the inductive reactance $X_L$ equals the capacitive reactance $X_C$. This means:
$\omega L = \frac{1}{\omega C}$
or equivalently:
$f_0 = \frac{1}{2\pi\sqrt{LC}}$
At resonance, the impedance is purely resistive ($Z = R$) and the current is maximum.

Key Concept: Impedance at resonance condition

d) It becomes minimum and equals the resistance
[Solution Description]
At resonance, the inductive reactance $X_L$ equals the capacitive reactance $X_C$, making the net reactance zero ($X_L - X_C = 0$). Thus, the impedance $Z$ of the circuit becomes purely resistive and is minimized:
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$
Therefore, the impedance is minimum and equal to the resistance $R$.

Key Concept: Transformer Efficiency, Power Conversion

c) 95\%
[Solution Description]
The efficiency $\eta$ of the transformer is given by:
$\eta = \frac{\text{Output Power}}{\text{Input Power}} = \frac{9500}{10000} = 0.95$ or 95\%.

Key Concept: Inductive Reactance

c) 15.7 $\Omega$
[Solution Description]
The formula for inductive reactance is $X_L = 2 \pi f L$.
Given:
Inductance $L = 50$ mH = $50 \times 10^{-3}$ H,
Frequency $f = 50$ Hz.
Substituting the values:
$X_L = 2 \pi \times 50 \times 50 \times 10^{-3} = 15.7 \, \Omega$.

Key Concept: Impedance at Resonance

c) It becomes purely resistive and equal to R
[Solution Description]
At resonant frequency, the inductive reactance ($X_L$) and capacitive reactance ($X_C$) cancel each other out, as $X_L = X_C$. Therefore, the total impedance ($Z$) of the circuit becomes purely resistive, equal to the resistance ($R$) in the circuit:
$Z = R$

Key Concept: Basic Concept

c) All of the above factors
[Solution Description]
The voltage induced in a coil is given by $V = -N \frac{d\Phi_B}{dt}$. Here, $\Phi_B$ is the magnetic flux through the coil, which changes as the coil rotates. The rate of change of flux determines the induced voltage.

Key Concept: Capacitive Reactance

b) 318.47 Ω
[Solution Description]
The capacitive reactance $X_C$ is given by the formula:
$X_C = \frac{1}{2\pi f C}$
Here, $f = 50$ Hz and $C = 10 \times 10^{-6}$ F.
Plugging in the values:
$X_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = \frac{1}{3.14 \times 10^{-3}} \approx 318.47 \, \Omega$

Key Concept: Frequency of Oscillations in L-C Circuit, Resonant Frequency

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The resonant frequency $f_r$ of an L-C-R series circuit is given by:
$f_r = \frac{1}{2\pi \sqrt{LC}}$
This formula clearly shows that $f_r$ depends only on $L$ and $C$ and not on $R$. Thus, Assertion (A) is true.
At resonance, $X_L = X_C$:
$\omega L = \frac{1}{\omega C}$
This condition leads to minimum impedance ($Z = R$) and maximum current. Hence, Reason (R) is also true.
However, Reason (R) describes the phenomena at resonance but does not explain why the resonant frequency is independent of $R$. Therefore, Reason (R) is not the correct explanation for Assertion (A).

Key Concept: Phase Angle in L-R Circuit

b) $37^\circ$
[Solution Description]
Given:
$R = 10 \Omega$, $L = 0.02 \text{ H}$, $f = 60 \text{ Hz}$
Inductive reactance, $X_L = 2\pi fL = 2 \times \pi \times 60 \times 0.02 = 7.54 \Omega$
Phase angle, $\phi = \tan^{-1}\left(\frac{X_L}{R}\right) = \tan^{-1}\left(\frac{7.54}{10}\right) = \tan^{-1}(0.754) = 37^\circ$

Key Concept: Phasors, Resultant Voltage

c) $20 \text{ V}$
[Solution Description]
In a series R-L circuit, $V_R$ is in phase with the current, and $V_L$ leads the current by $90^\circ$. The resultant voltage $V$ is the vector sum of $V_R$ and $V_L$:
$V = \sqrt{V_R^2 + V_L^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \text{ V}$

Key Concept: Wattless Current Condition

b) When resistance is zero and circuit is purely reactive
[Solution Description]
Wattless current flows when the phase difference between current and voltage is $90^\circ$, which occurs in purely inductive or purely capacitive circuits. Since $\cos 90^\circ = 0$, the average power dissipation is zero, making the current "wattless."

Key Concept: Power factor calculation, L-R circuit

Correct Answer option b) 0.691
[Solution Description]
First calculate impedance $Z$:
$\omega = 2\pi f = 2\pi \times 50 = 314 \, \text{rad/s}$
$X_L = \omega L = 314 \times 0.1 = 31.4 \, \Omega$
$Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 31.4^2} = \sqrt{900 + 985.96} = \sqrt{1885.96} = 43.42 \, \Omega$
Power factor $\cos \phi = R/Z = 30/43.42 = 0.691$

Key Concept: Frequency of AC

c) 50 Hz
[Solution Description]
The frequency ($f$) of an alternating current is the reciprocal of its periodic time ($T$). Given $T = 0.02 \text{ s}$, the frequency is calculated as:
$f = \frac{1}{T} = \frac{1}{0.02} = 50 \text{ Hz}$

Key Concept: Damped Oscillatory Current, Energy Dissipation

c) The amplitude decreases exponentially with time.
[Solution Description]
In a damped L-C-R circuit, the amplitude of oscillations decreases exponentially with time due to energy dissipation as heat in the resistor. The decay rate depends on the ratio $\frac{R}{2L}$. Here, since $R$ is non-zero, the amplitude decreases exponentially but not instantaneously.

Key Concept: Half Power Points, Q-Factor

d) Assertion is false, but Reason is true.
[Solution Description]
The bandwidth of a series L-C-R circuit is given by $\Delta \omega = \frac{R}{L}$. Hence, increasing R increases the bandwidth, so the Assertion (A) is false. The Q-factor is defined as $Q = \frac{\omega_0}{\Delta \omega}$, which means Q and $\Delta \omega$ are inversely related. Thus, Reason (R) is true. Since the Assertion is false and the Reason is true, the correct option is (d).

Key Concept: Phasor Diagram

c) Phase difference between the two signals
[Solution Description]
The angle between two phasors (e.g., current and voltage phasors) represents the phase difference ($\phi$) between them. For example, if the current lags the voltage by $30^\circ$, the angle between their phasors will be $30^\circ$.

Key Concept: Frequency of Oscillations

c) It remains the same
[Solution Description]
The frequency of oscillations in an L-C circuit is given by:
$f = \frac{1}{2\pi \sqrt{LC}}$
If $L$ is doubled ($L' = 2L$) and $C$ is halved ($C' = \frac{C}{2}$), the new frequency $f'$ becomes:
$f' = \frac{1}{2\pi \sqrt{2L \cdot \frac{C}{2}}} = \frac{1}{2\pi \sqrt{LC}}$
Thus, the frequency remains unchanged.

Key Concept: Basic Concept: Alternating Current

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that alternating current changes its direction periodically, which is true because AC reverses direction at regular intervals. The reason states that the magnitude of AC varies with time, which is also true as it follows a sinusoidal pattern. However, the variation in magnitude does not explain why AC changes direction, so while both statements are true, the reason does not correctly explain the assertion.

Key Concept: Faraday’s Law of Induction, Transformers

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that in a step-up transformer, the output voltage increases due to the secondary coil having more turns than the primary coil. This is true as the ratio of output voltage to input voltage is given by $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
The reason states that the induced emf in the secondary coil is directly proportional to the number of turns in the secondary coil according to Faraday's law. This is also true because Faraday's law states that the induced emf is proportional to the rate of change of flux linkage ($\frac{d\Phi_B}{dt}$), and for a transformer with $N_s$ turns, the emf is $V_s = -N_s \frac{d\Phi_B}{dt}$.
However, while both statements are true, the reason does not fully explain why the output voltage increases in a step-up transformer. The increase in voltage is specifically due to the ratio of turns between the secondary and primary coils, not just the number of turns in the secondary coil alone.

Key Concept: Half Power Points, Bandwidth

b) 50 rad/s
[Solution Description]
The Q-factor $Q$ is given by:
$Q = \frac{\text{Resonant Frequency}}{\text{Bandwidth}}$
Given $\omega_0 = 1000$ rad/s and $Q = 20$, we can find the bandwidth $\Delta \omega$ as:
$\Delta \omega = \frac{\omega_0}{Q} = \frac{1000}{20} = 50 \text{ rad/s}$

Key Concept: Definitions and Properties of AC

c) $\frac{I_0}{\sqrt{2}}$
[Solution Description]
The RMS value of an alternating current is given by the formula:
$I_{rms} = \frac{I_0}{\sqrt{2}}$
Here, $I_0$ is the peak value of the current.

Key Concept: Impedance, High-Frequency Behavior of Inductor

c) Remains almost the same
[Solution Description]
At 100 Hz:
\$X_L = 2 \pi f L = 2 \times \pi \times 100 \times 0.01 = 6.28 \Omega\$.
Impedance \$Z_1 = \sqrt{R^2 + X_L^2} = \sqrt{(10^4)^2 + (6.28)^2} \approx 10 \text{ k}\Omega\$.
At 10 kHz:
\$X_L = 2 \pi f L = 2 \times \pi \times 10^4 \times 0.01 = 628 \Omega\$.
Impedance \$Z_2 = \sqrt{R^2 + X_L^2} = \sqrt{(10^4)^2 + (628)^2} \approx 10.02 \text{ k}\Omega\$.
The impedance increases slightly but remains close to \$10 \text{ k}\Omega\$ because \$R\$ dominates at high frequencies.

Key Concept: Basic definition of AC

c) It changes both magnitude and direction periodically.
[Solution Description]
Alternating current (AC) changes its magnitude and direction periodically, unlike direct current (DC), which remains constant in magnitude and direction.

Key Concept: Energy transfer in LC circuit, Energy conservation

d) When $\omega t = \frac{\pi}{3}$
[Solution Description]
The total energy in the LC circuit remains constant and is given by $E_{total} = \frac{1}{2}\frac{q_0^2}{C}$. When 75\% of the energy is magnetic, we have:
$\frac{1}{2}Li^2 = 0.75 \times \frac{1}{2}\frac{q_0^2}{C}$
Simplifying:
$i^2 = \frac{0.75q_0^2}{LC}$
The current in an LC circuit is given by $i = -\omega q_0 \sin(\omega t)$ where $\omega = \frac{1}{\sqrt{LC}}$. Squaring this gives:
$i^2 = \omega^2 q_0^2 \sin^2(\omega t) = \frac{q_0^2}{LC} \sin^2(\omega t)$
Setting equal to our previous expression:
$\frac{q_0^2}{LC} \sin^2(\omega t) = \frac{0.75q_0^2}{LC}$
Solving gives $\sin(\omega t) = \pm \frac{\sqrt{3}}{2}$, so $\omega t = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$ etc.

Key Concept: Power Loss in AC Circuits - LCR Circuit

d) It becomes 1
[Solution Description]
At resonance in an LCR circuit, the inductive reactance ($\omega L$) equals the capacitive reactance ($\frac{1}{\omega C}$). The phase angle $\phi$ becomes zero because $\tan \phi = \frac{\omega L - \frac{1}{\omega C}}{R} = 0$. Hence, the power factor $\cos \phi = \cos 0^\circ = 1$, indicating maximum power dissipation.

Key Concept: Resistive Circuit, Power Dissipation

b) 100 W
[Solution Description]
The given voltage is $V = 100 \sin (100\pi t)$.
Peak voltage ($V_0$) = 100 V.
RMS voltage ($V_{rms}$) = $\frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 70.71$ V.
Resistance ($R$) = 50 ohms.
Power dissipated ($P$) = $\frac{V_{rms}^2}{R} = \frac{(70.71)^2}{50} = \frac{5000}{50} = 100$ W.

Key Concept: Power efficiency, Energy loss analysis

d) Efficiency: 95\%; Loss: Copper windings
[Solution Description]
First, calculate the input power ($P_{\text{in}}$) and output power ($P_{\text{out}}$):
$P_{\text{in}} = V_p \times I_p = 480\,\text{V} \times 10\,\text{A} = 4800\,\text{W}$
$P_{\text{out}} = V_s \times I_s = 120\,\text{V} \times 38\,\text{A} = 4560\,\text{W}$
Now, compute the efficiency ($\eta$):
$\eta = \left(\frac{P_{\text{out}}}{P_{\text{in}}}\right) \times 100\% = \left(\frac{4560}{4800}\right) \times 100\% = 95\%$
Most energy losses occur due to resistance in the windings (copper loss) and eddy currents/hysteresis in the core (iron loss). Given high current flow, copper loss is significant here.

Key Concept: Electromagnetic Induction, Working Principle

b) The plane of the coil is perpendicular to the magnetic field
[Solution Description]
The emf induced in a rotating coil is given by:
$V = V_0 \sin(\omega t)$
When $V = 0$, it implies:
$\sin(\omega t) = 0$
This occurs when:
$\omega t = n\pi \quad \text{where} \quad n = 0, 1, 2, \dots$
At these instants, the plane of the coil is perpendicular to the magnetic field, i.e., the flux linkage is either maximum or minimum, but the rate of change of flux is zero, resulting in zero emf. Hence, the coil is vertical (perpendicular to the magnetic field) at this instant.

Key Concept: Phase Relationship in Resistive AC Circuit

d) $0^\circ$
[Solution Description]
In a purely resistive AC circuit, the voltage and current are in phase, meaning there is no phase difference between them.
Therefore, the phase difference between the alternating voltage and current is zero.

Key Concept: Capacitive Reactance, Phase Difference

c) $90^\circ$
[Solution Description]
In a purely capacitive circuit, the current always leads the voltage by $90^\circ$. The capacitance value and frequency don't affect this fundamental phase relationship in an ideal capacitor.
Step by step:
1. For any pure capacitor, the phase angle between current and voltage is always $+90^\circ$ (current leading).
2. This is because $I = C\frac{dV}{dt}$, making current proportional to the rate of change of voltage.
3. The values of capacitance and frequency only affect the magnitude of reactance ($X_C$) but not the phase relationship.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a step-up transformer, the transformation ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ is greater than 1. This means $N_s$ (number of turns in the secondary) must be greater than $N_p$ (number of turns in the primary). Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Damped Oscillations

b) Presence of resistance $R$
[Solution Description]
Damped oscillations occur in an L-C circuit due to the presence of resistance ($R$), which causes energy loss in the form of heat. This reduces the amplitude of oscillations over time until they eventually stop.

Key Concept: Half-Power Points

b) $\omega L - \frac{1}{\omega C} = \pm R$
[Solution Description]
At the half-power points, the power delivered is half of the maximum power. The condition for this is $\omega L - \frac{1}{\omega C} = \pm R$. This gives two frequencies $\omega_1$ and $\omega_2$ which define the bandwidth.

Key Concept: Series L-R Circuit, Impedance, Power Dissipation

b) 693.8 W
[Solution Description]
First, calculate the inductive reactance ($X_L$):
$X_L = 2\pi f L = 2 \times \pi \times 50 \times 0.1 = 31.42\ \Omega$
Next, find the impedance ($Z$) of the circuit:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{50^2 + 31.42^2} = \sqrt{2500 + 987.22} = \sqrt{3487.22} = 59.05\ \Omega$
Now, compute the rms current ($I_{\text{rms}}$):
$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{220}{59.05} = 3.725\ \text{A}$
The average power dissipated ($P$) is given by:
$P = I_{\text{rms}}^2 R = (3.725)^2 \times 50 = 13.876 \times 50 = 693.8\ \text{W}$

Key Concept: Potential difference across components, Resonance

c) 60 V
[Solution Description]
Steps to solve:
1. Calculate current at resonance: $I = \frac{V}{R} = \frac{10}{5} = 2$ A.
2. Potential difference across capacitor: $V_C = I \times X_C = 2 \times 30 = 60$ V.

Key Concept: Electromagnetic Induction - Induced EMF

a) 0 degrees
[Solution Description]
The induced emf in a rotating coil is given by:
$V = V_0 \sin \omega t$
The emf is zero when $\sin \omega t = 0$. This occurs at angles where $\omega t = 0^\circ, 180^\circ, 360^\circ$, etc. Thus, the emf is zero when the plane of the coil is perpendicular to the magnetic field.

Key Concept: Phase Difference in AC Circuits

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a purely inductive circuit, when an alternating voltage is applied, the current does not rise immediately due to the back emf generated by the inductor. This results in the current reaching its maximum value later than the voltage, leading to a phase lag of $90^\circ$. The inductive reactance ($X_L = \omega L$) opposes the change in current, which directly causes this phase difference. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Frequency and Angular Frequency Relationship

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given the frequency $f = 50$ Hz, we can find angular frequency $\omega$ using the formula $\omega = 2\pi f$. Substituting the value of $f$, we get $\omega = 2\pi \times 50 = 100\pi$ rad/s. Therefore, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Phase Difference

b) $30°$
[Solution Description]
When current leads the voltage, the phase difference ($\phi$) is positive. Here, the phase difference is given as $30°$.

Key Concept: Moving Coil Galvanometer in AC Circuit

d) The needle's inertia prevents it from following rapid oscillations
[Solution Description]
In an AC circuit, current reverses direction periodically. At 50 Hz, the current changes direction 100 times per second. A moving coil galvanometer's needle would theoretically move left and right rapidly due to alternating torque. However, due to its inertia, the needle cannot respond to such rapid changes and remains stationary, showing no deflection.

Key Concept: Phase Difference in Inductive Circuit

c) $90^\circ$
[Solution Description]
In a purely inductive circuit, the current lags behind the voltage by $\pi/2$ radians or $90^\circ$. This is derived from the solution of the differential equation for the inductor: $I = \frac{V_0}{\omega L} \sin(\omega t - \pi/2)$.

Key Concept: Flux Linkage, Time Variation

b) $t = \frac{1}{2f}$
[Solution Description]
The emf of an AC generator varies as $V = V_0 \sin(\omega t)$, where $\omega = 2\pi f$.
The emf becomes zero when $\sin(\omega t) = 0$. This occurs at $\omega t = n\pi$, where $n$ is an integer (0, 1, 2, ...).
For the first zero crossing after $t = 0$, $n = 1$:
$\omega t = \pi \implies t = \frac{\pi}{\omega} = \frac{\pi}{2\pi f} = \frac{1}{2f}$
So, the emf first becomes zero at $t = \frac{1}{2f}$.

Key Concept: RMS Value

c) 10 A
[Solution Description]
The RMS value of an alternating current is calculated using the formula: $I_{rms} = \frac{I_0}{\sqrt{2}}$. Here, the peak current $I_0$ is given as 14.14 A.
Step 1: Write down the formula for RMS current.
$I_{rms} = \frac{I_0}{\sqrt{2}}$
Step 2: Substitute the given peak current value.
$I_{rms} = \frac{14.14}{\sqrt{2}}$
Step 3: Calculate the numerical value.
$I_{rms} = \frac{14.14}{1.414} = 10$ A

Key Concept: AC Current Output

c) $I = I_0 \sin \omega t$
[Solution Description]
The instantaneous alternating current $I$ produced by an AC generator is given by $I = I_0 \sin \omega t$, where $I_0$ is the peak current and $\omega$ is the angular frequency.

Key Concept: Half-Power Points, Bandwidth, and Q-Factor

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the bandwidth $\Delta \omega$ is equal to the difference between the upper and lower half-power frequencies, which is true by definition. The reason provides the formula for calculating the bandwidth in terms of resistance $R$ and inductance $L$, which is also correct. However, the reason does not directly explain why the bandwidth equals the frequency difference; it merely gives a method to calculate it. Therefore, both are true, but the reason is not the correct explanation of the assertion.

Key Concept: Periodic Time, Persistence of Vision

b) 12000 times
[Solution Description]
Given periodic time $T = 0.01$ s, frequency $f = \frac{1}{T} = \frac{1}{0.01} = 100$ Hz. In one cycle, the bulb turns off twice (at zero crossings). Therefore, in one second, it turns off $2 \times 100 = 200$ times. In one minute (60 seconds), it turns off $200 \times 60 = 12000$ times.

Key Concept: Transformers

d) 1000 V
[Solution Description]
The transformation ratio for a transformer is:
$\frac{V_s}{V_p} = \frac{N_s}{N_p} = r$
Given $r = 5$ and $V_p = 200 \text{ V}$:
$V_s = r \times V_p = 5 \times 200 = 1000 \text{ V}$
Hence, the secondary voltage is 1000 V.

Key Concept: RMS value, Phase difference

b) 32 degrees
[Solution Description]
First find the angular frequency $\omega = 100\pi$ rad/s. Then calculate inductive reactance $X_L = \omega L = 100\pi \times 0.1 = 10\pi$ $\Omega$. The impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{50^2 + (10\pi)^2}$ $\Omega$. The phase difference $\phi$ is given by $\tan^{-1}(X_L/R) = \tan^{-1}(10\pi/50)$. Calculating: $10\pi/50 \approx 0.628$, so $\phi = \tan^{-1}(0.628) \approx 32^\circ$.

Key Concept: Energy Losses

c) Eddy current losses
[Solution Description]
Eddy current losses occur due to circulating currents induced in the iron core. Laminating the core into thin layers insulated from each other reduces these losses by increasing resistance to the flow of eddy currents.

Key Concept: Phasor Diagram in AC Circuits

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a purely resistive AC circuit, the voltage across the resistor and the current through it are given by $V = V_0 \sin \omega t$ and $I = \frac{V_0}{R} \sin \omega t$, respectively. Here, both $V$ and $I$ have the same $\sin \omega t$ term, indicating they reach their maximum and zero values at the same time. Thus, they are in phase. This phase alignment occurs because Ohm’s law ($V = IR$) holds true at every instant for a resistor, making the Reason the correct explanation of the Assertion.

Key Concept: Disadvantages of AC

d) It is more dangerous than DC at the same voltage.
[Solution Description]
While AC has many advantages, one disadvantage is that it is more dangerous than DC at the same voltage due to its oscillatory nature causing severe muscular contractions. Additionally, AC tends to flow mostly on the surface of conductors (skin effect), requiring multiple thin wires instead of a single thick one for efficient conduction.

Key Concept: Transformer Construction

b) Secondary coil
[Solution Description]
In a step-up transformer, the secondary coil has more turns than the primary coil. This is because the voltage is increased by the ratio $\frac{N_s}{N_p}$ where $N_s$ is the number of turns in the secondary coil and $N_p$ is the number of turns in the primary coil.

Key Concept: Energy Storage in L-C Circuit

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In an ideal L-C circuit, the total energy remains conserved as there is no energy dissipation (since resistance is negligible). The energy continuously oscillates between the electric field of the capacitor ($E_e = \frac{q^2}{2C}$) and the magnetic field of the inductor ($E_m = \frac{1}{2} L I^2$). When the charge on the capacitor is zero ($q = 0$), all the energy is stored in the magnetic field of the inductor ($E_m$ is maximum at this point). Hence, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Electrical Oscillations in an L-C Circuit

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In an ideal L-C circuit, there is no resistance ($R = 0$), so energy is not lost as heat. The energy oscillates between the capacitor (electrical energy) and the inductor (magnetic energy) without any loss. Therefore, the total energy remains conserved. Both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion because energy conservation in an ideal L-C circuit is a direct consequence of the absence of resistive losses.

Key Concept: Effective Resistance Calculation

c) 50 $\Omega$
[Solution Description]
In a series connection, the effective resistance is the sum of individual resistances: $R_{\text{eff}} = R_1 + R_2 = 20 + 30 = 50 \Omega$.

Key Concept: Phasor Definition

c) A rotating vector representing sinusoidal AC quantities
[Solution Description]
A phasor is a rotating vector used to represent sinusoidal quantities like voltage and current in an AC circuit. It simplifies the analysis by converting time-domain signals into rotating vectors, where the angle between them represents the phase difference.

Key Concept:

b) It reduces current without significant energy loss
[Solution Description] A choke coil minimizes energy loss as heat since its power dissipation is nearly zero, whereas a resistor dissipates significant energy as heat ($I_{\text{rms}}^2 R$).

Key Concept: L-C Circuit Energy Conservation, Frequency

b) 0.125 J
[Solution Description]
The initial energy stored in the capacitor is given by:
$E = \frac{q_0^2}{2C} = \frac{(5 \times 10^{-3})^2}{2 \times 100 \times 10^{-6}} = 0.125 \text{ J}$
This energy is conserved in the L-C oscillations since there is no resistance.

Key Concept: Domestic AC Supply

b) 100 times
[Solution Description]
Since there are 50 cycles per second and the current becomes zero twice in each cycle, the total number of times the current becomes zero is:
$2 \times 50 = 100 \text{ times.}$

Key Concept: RMS Value of Alternating Current

d) 7.07 A
[Solution Description]
The rms value of an alternating current is given by $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$, where $I_0$ is the peak current. Here, $I_0 = 10$ A. Therefore, $I_{\text{rms}} = \frac{10}{\sqrt{2}} = 7.07$ A.

Key Concept: C-R Series Circuit

b) $112\,\Omega$
[Solution Description]
The impedance in an RC circuit is calculated using:
$Z = \sqrt{R^2 + X_C^2}$
Substituting the values:
$Z = \sqrt{100^2 + 50^2} = \sqrt{10000 + 2500} = \sqrt{12500} = 111.8\,\Omega$
So, the closest answer is approximately \$112\,\Omega\$.

Key Concept: L-C Oscillations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In an ideal L-C circuit, the resistance is negligible ($R = 0$), so no energy is lost as heat. The energy stored in the capacitor ($E_C = \frac{1}{2}CV^2$) is fully transferred to the inductor ($E_L = \frac{1}{2}LI^2$) and vice versa during oscillations. Thus, the total energy remains conserved, and both Assertion (A) and Reason (R) are true. Moreover, (R) correctly explains why the total energy remains constant in an ideal L-C circuit. Therefore, the correct answer is option (a).

Key Concept: Bandwidth

c) Double
[Solution Description]
The bandwidth ($\Delta \omega$) of a series LCR circuit is given by $\Delta \omega = \frac{R}{L}$. If $R$ is doubled, the new bandwidth becomes $\Delta \omega' = \frac{2R}{L} = 2\Delta \omega$. Thus, the bandwidth doubles when $R$ is doubled.

Key Concept: Transmission Efficiency, Voltage Control

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that AC is preferred for power transmission due to efficient voltage control with minimal energy loss, which is true. The reason explains that stepping up/down AC voltage reduces current, thus minimizing power loss ($P = I^2 R$). This is also true and directly supports the assertion by explaining why AC's voltage controllability leads to lower energy loss.

Key Concept: RMS Value, Power Dissipation

b) 490 W
[Solution Description]
First, calculate the RMS value of the current using the formula:
$$
I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{14}{\sqrt{2}} = 9.90 \text{A}
$$
Now, use the power dissipation formula for an AC circuit with a resistive load:
$$
P = I_{rms}^2 R = (9.90)^2 \times 5 = 490.05 \text{W}
$$
Rounding off, the power dissipated is approximately $490 \text{W}$.

Key Concept: Transformer theory and construction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a step-up transformer, the primary coil has fewer turns of thick wire compared to the secondary coil. This is because the primary coil carries higher current (as $\frac{I_s}{I_p} = \frac{N_p}{N_s}$), and thicker wire reduces resistance, minimizing copper losses ($P = I^2R$). Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Series L-C Circuit

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
At resonance in a series L-C circuit, the impedance is given by $Z = X_L - X_C$. Since at resonance $X_L = X_C$, the impedance becomes $Z = 0$. Therefore, the impedance is indeed minimum (zero) at resonance, making the Assertion true. The Reason states that resonance occurs when $X_L = X_C$, which is the correct definition of resonance and directly explains why the impedance is minimum (as it leads to $Z = 0$). Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Voltage Magnification

c) 400 V
[Solution Description]
At resonance, the current in the circuit is maximum and given by:
$I_{\text{max}} = \frac{V}{R} = \frac{100}{5} = 20 \, \text{A}$
The potential difference across the inductor is calculated using Ohm's law:
$V_L = I_{\text{max}} \cdot X_L = 20 \times 20 = 400 \, \text{V}$
Thus, the potential difference across the inductor is $400 \, \text{V}$.

Key Concept: Power in AC Circuits - Wattless Current

d) 0 W
[Solution Description]
In a purely inductive circuit, the phase difference between voltage and current is $90^\circ$. The average power is given by:
$P = V_{\text{rms}} \times I_{\text{rms}} \cos \phi$
Since $\cos 90^\circ = 0$, the power becomes:
$P = 220\,\text{V} \times 10\,\text{A} \times 0 = 0\,\text{W}$
Thus, the average power dissipated is zero, indicating wattless current.

Key Concept: Electromagnetic Induction, AC Generator, Transformer

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in a step-up transformer, the secondary voltage is higher due to more turns in the secondary coil, which is correct as per the transformation ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. The reason explains that the induced emf in the secondary is proportional to the turns ratio, which directly supports the assertion. Hence, both are true and the reason correctly explains the assertion.

Key Concept: Wattless Current

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a purely inductive AC circuit, the phase difference ($\phi$) between current and voltage is $90^\circ$. The average power consumed is given by:
$P_{\text{avg}} = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi$
Substituting $\phi = 90^\circ$:
$P_{\text{avg}} = V_{\text{rms}} \times I_{\text{rms}} \times 0 = 0$
Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Power Factor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the power factor of an L-R circuit is always less than 1. This is true because in an L-R circuit, the impedance $Z = \sqrt{R^2 + (\omega L)^2}$ is always greater than the resistance $R$ due to the presence of inductive reactance $\omega L$. Therefore, the power factor $\cos \phi = \frac{R}{\sqrt{R^2 + (\omega L)^2}}$ will always be less than 1 unless the inductive reactance $\omega L$ is zero (which would make it purely resistive).
The reason explains that the impedance $Z$ includes both resistive and inductive components, leading to a phase difference $\phi$ between voltage and current. This is also true and correctly explains why the power factor is less than 1, as the phase difference results in a reduction of the effective power delivered to the circuit (real power) compared to the apparent power.
Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Bandwidth and Q-Factor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Q-factor is given by $Q = \frac{\omega_0}{\Delta \omega} = \frac{1}{R} \sqrt{\frac{L}{C}}$. From this formula, it is clear that decreasing $R$ increases the Q-factor. The bandwidth $\Delta \omega = \frac{R}{L}$ shows that reducing $R$ decreases the bandwidth. Since Q-factor and bandwidth are inversely related ($Q = \frac{\omega_0}{\Delta \omega}$), the reason correctly explains why decreasing $R$ increases the Q-factor.

Key Concept: Impedance at resonance

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a series RLC circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance, $X_L = X_C$, so the term $(X_L - X_C)$ becomes zero. Therefore, the impedance reduces to $Z = R$, which is its minimum value. The current in the circuit is maximum at this point because $I = V/Z$, and $Z$ is minimum. Hence, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Resonant Frequency Calculation

c) 159.15 Hz
[Solution Description]
The resonant frequency of a series L-C circuit is given by:
$f_0 = \frac{1}{2 \pi \sqrt{L C}}$
Substituting the given values:
$f_0 = \frac{1}{2 \pi \sqrt{(10 \times 10^{-3}) (100 \times 10^{-6})}} = \frac{1}{2 \pi \sqrt{10^{-6}}} = \frac{1}{2 \pi \times 10^{-3}} = \frac{1000}{2 \pi} \approx 159.15 \, \text{Hz}$
Hence, the resonant frequency is approximately 159.15 Hz.

Key Concept: Power dissipation, RMS values

b) 360 W
[Solution Description]
The real power P in an AC circuit is given by $P = V_{rms} \times I_{rms} \times \text{power factor}$. Substituting the given values: $P = 120 \times 5 \times 0.6 = 360$ W.

Key Concept: Slip Rings, Brushes, Direction of Current

b) Both emf and current are zero.
[Solution Description]
When the plane of the coil becomes perpendicular to the magnetic field, the arms of the coil move parallel to the field lines. At this instant, there is no cutting of magnetic flux, so the rate of change of flux through the coil is zero. According to Faraday’s law, the induced emf is zero when the rate of change of flux is zero, and hence the current is also zero.

Key Concept: Phasor Representation in AC Circuits

d) Current lags voltage by $90^\circ$
[Solution Description]
In a purely inductive circuit, the current lags behind the voltage by $90^\circ$. This is because the inductor opposes changes in current, causing the current to reach its peak after the voltage.

Key Concept: Frequency, Time Period, Wave Equation

a) 25 Hz
[Solution Description]
Step 1: The standard form of a wave equation is $y = A \sin(\omega t + \phi)$, where $\omega$ is the angular frequency.
Step 2: From the given equation, $\omega = 50\pi$ rad/s.
Step 3: Use the formula $f = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} = 25$ Hz.

Key Concept: Transformer working principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that primary and secondary coils are wound on the same core to ensure maximum mutual induction, which is correct as this configuration allows efficient transfer of energy via changing magnetic flux. The reason explains that mutual induction occurs due to changing magnetic flux in the primary inducing emf in the secondary, which is also true and correctly explains why the coils are wound on the same core. Hence, both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Choke Coil in Fluorescent Lamps

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that choke coils are used in fluorescent lamps to minimize power dissipation, which is true because their high inductance and low resistance make the power factor $\cos \phi \approx 0$, resulting in negligible power loss ($P = V_{rms} I_{rms} \cos \phi \approx 0$). The reason correctly explains this by stating that $\cos \phi \approx 0$ due to the choke coil's properties. Thus, both the assertion and reason are true, and the reason explains the assertion.

Key Concept: Inductive Circuit

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) is true because in an inductor, the voltage always leads the current by $90^\circ$ due to the phase difference introduced by the inductive reactance $X_L = \omega L$. The reason (R) is also true as it correctly explains that the inductive reactance $X_L$ is responsible for this phase shift. Therefore, both are true and the reason correctly explains the assertion.

Key Concept: Periodic Time, Frequency, Angular Velocity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Given the angular frequency $\omega = 100\pi$ rad/s.
Step 2: Use the formula for periodic time $T = \frac{2\pi}{\omega}$.
Step 3: Substitute $\omega = 100\pi$ into the formula:
$T = \frac{2\pi}{100\pi} = \frac{2}{100} = 0.02 \text{ s} = 20 \text{ ms}$
Step 4: Thus, the assertion is true.
Step 5: The reason correctly states the relationship between $T$ and $\omega$, and it explains why the assertion is true.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Definitions and Properties of AC

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the RMS value of an alternating current is called the effective value, which is true because it represents the equivalent DC value that produces the same power dissipation. The Reason correctly explains why \$I_{rms}\$ is called the effective value by stating that it equals the DC current producing the same heating effect in a resistor. Therefore, both are true, and the Reason is the correct explanation for the Assertion.

Key Concept: Maximum Value or Peak Value

c) Peak value
[Solution Description]
The maximum value of an alternating current is referred to as the peak value or amplitude.

Key Concept: Power Factor in Purely Inductive Circuit

c) $0$
[Solution Description]
For a purely inductive circuit, the phase difference $\phi = 90^\circ$. The power factor is defined as $\cos \phi$, which equals $\cos 90^\circ = 0$.

Key Concept: Periodic Time Definition

b) The time taken for current to complete one full cycle
[Solution Description]
Periodic time is the time taken by the alternating current to complete one full cycle, which includes rising from zero to maximum, falling back to zero, rising to maximum in the opposite direction, and then falling to zero again.

Key Concept: Faraday's Law of Induction

c) 19.1 Hz
[Solution Description]
The general form of the induced emf is:
$V = V_0 \sin(\omega t)$
Comparing with the given equation $V = 15 \sin(120t)$ V, we get $\omega = 120$ rad/s. The frequency ($f$) is related to angular velocity by:
$f = \frac{\omega}{2\pi} = \frac{120}{2 \times 3.14} \approx 19.1 \, \text{Hz}$
Thus, the frequency is approximately 19.1 Hz.

Key Concept: Mean Value of AC over Full Cycle

c) 0
[Solution Description]
The mean value of an alternating current (AC) over a full cycle is zero because the positive and negative half-cycles cancel each other out.

Key Concept: L-C Oscillations

d) $$ f = \frac{1}{2\pi\sqrt{LC}}$$
[Solution Description]
The frequency of oscillations in an L-C circuit is determined by the formula:
$f = \frac{1}{2\pi\sqrt{LC}}$
where $L$ is the inductance and $C$ is the capacitance.

Key Concept: Wattless Current

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a purely capacitive AC circuit, the phase difference between current and voltage is indeed $90^\circ$ because current leads voltage by $90^\circ$ in such a circuit. The average power dissipated in an AC circuit is given by:
$P_{\text{avg}} = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi.$
Since $\phi = 90^\circ$, $\cos 90^\circ = 0$, which means $P_{\text{avg}} = 0$. Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Core material impact, Flux linkage

b) Reduces eddy currents; Flux linkage becomes negligible
[Solution Description]
A laminated soft-iron core minimizes eddy currents (energy loss) due to its high permeability and laminated structure, which disrupts current paths. Solid steel would introduce higher hysteresis loss and eddy currents.
If aluminum is used instead, flux linkage decreases drastically because aluminum has much lower permeability compared to soft iron. This reduces the transformer's efficiency as less magnetic flux couples the primary and secondary coils.

Key Concept: Phase Difference, Power Factor

b) 0.5
[Solution Description]
The power factor ($\cos \phi$) is defined as the cosine of the phase angle ($\phi$) between the current and voltage. Here, $\phi = 60^\circ$. Thus:
$$
\text{Power factor} = \cos(60^\circ) = 0.5
$$
Therefore, the power factor is $0.5$.

Key Concept: Resonance Frequency

d) $f_0 = \frac{1}{2\pi \sqrt{LC}}$
[Solution Description]
The resonant frequency $f_0$ of a series L-C-R circuit is given by the formula:
$f_0 = \frac{1}{2\pi \sqrt{LC}}$
where $L$ is the inductance and $C$ is the capacitance.

Key Concept: Disadvantages of AC

c) Increased safety hazards upon contact
[Solution Description]
One major disadvantage of AC is that it is more dangerous than DC when accidentally touched because it causes severe muscular contractions and nerve stimulation due to its oscillatory nature. Additionally, AC tends to flow on the surface of wires (skin effect), requiring special considerations for thick wire usage.

Key Concept: Resonant Frequency, Impedance Calculation

a) 376 \$\Omega\$
[Solution Description]
First, calculate the resonant frequency $\omega_0$ using:
$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 8 \times 10^{-6}}} = 500 \text{ rad/s}$
The given frequency is half of $\omega_0$, so $\omega = 250 \text{ rad/s}$. Next, compute $X_L$ and $X_C$ at this frequency:
$X_L = \omega L = 250 \times 0.5 = 125 \Omega$
$X_C = \frac{1}{\omega C} = \frac{1}{250 \times 8 \times 10^{-6}} = 500 \Omega$
Now, find the impedance $Z$:
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (125 - 500)^2} = \sqrt{900 + 140625} = 376 \Omega$

Key Concept: Energy Losses

c) Eddy current losses
[Solution Description]
Laminating the iron core reduces eddy currents, which are circulating currents induced in the core that cause energy loss as heat.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that an AC generator converts mechanical energy into electrical energy, which is correct as per the syllabus. The reason states that its working principle is based on electromagnetic induction, which is also true and directly explains why the assertion is correct. According to Faraday’s law of electromagnetic induction, a changing magnetic flux induces an emf in the coil, converting mechanical energy (rotation) into electrical energy (current).

Key Concept: Instantaneous Power, L-R Circuit

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the power factor $\cos \phi$ determines the average power dissipated in an L-R circuit, which is true because the average power is given by $P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi$. The reason provides the correct formula for the power factor in an L-R circuit: $\cos \phi = \frac{R}{\sqrt{R^2 + (\omega L)^2}}$, which is derived from the impedance triangle. Since the reason correctly explains how the power factor depends on the circuit parameters $R$ and $L$, it is the correct explanation of the assertion.

Key Concept: Need for Alternating Current

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that AC is preferred over DC for long-distance power transmission, which is true because AC can be transmitted at high voltages, reducing energy losses. The Reason given is that transformers can be used with AC to step-up or step-down voltage, which is also true and correctly explains why AC is preferred for long-distance transmission, as transformers are essential for efficient power distribution.
Step-by-Step Explanation:
1) AC can be easily transformed to high voltages using step-up transformers, which reduces the current in transmission lines.
2) Lower current results in reduced energy losses ($I^2R$) and less voltage drop ($IR$).
3) At the consumer end, the voltage can be stepped down to usable levels using step-down transformers.
4) This entire process is not possible with DC, making AC more efficient for long-distance transmission.
Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Time-Variation of Voltage, Faraday's Law

b) $\frac{\pi}{24}$ s
[Solution Description]
The induced voltage as a function of time is $V = V_0 \sin \omega t$.
Given $V = 20$ V and $V_0 = 40$ V, we have:
$20 = 40 \sin (4t)$.
Simplifying: $\sin (4t) = 0.5$.
Therefore, $4t = \frac{\pi}{6}$ or $4t = \frac{5\pi}{6}$.
Solving for $t$ gives $t = \frac{\pi}{24}$ s or $t = \frac{5\pi}{24}$ s.
The smallest positive time is $\frac{\pi}{24}$ s.

Key Concept: Series L-C Circuit, Resonance Condition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a series L-C circuit at resonance, the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$), so their magnitudes cancel each other out, making the total impedance purely resistive (theoretically zero in an ideal circuit). The voltage across the inductor ($V_L = I X_L$) and capacitor ($V_C = I X_C$) have equal magnitudes but are 180$^\circ$ out of phase because the inductor voltage leads the current by 90$^\circ$ while the capacitor voltage lags the current by 90$^\circ$. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Phasor Diagram

c) $53.13°$
[Solution Description]
The resultant voltage $V$ can be found using the phasor diagram. Since $V_R$ is in phase with current $I$ and $V_C$ lags behind $I$ by $90°$, the resultant voltage is the vector sum of $V_R$ and $V_C$.
Step 1: Calculate the magnitude of the resultant voltage.
$V = \sqrt{V_R^2 + V_C^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$ V.
Step 2: Find the phase angle $\phi$.
$\tan\phi = \frac{V_C}{V_R} = \frac{16}{12} = \frac{4}{3}$,
So, $\phi = \arctan\left(\frac{4}{3}\right)$.