Correct Option is (A)

Statement I: An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.

This statement is incorrect. Electrical resonance occurs in an AC circuit when the capacitive reactance and inductive reactance are equal, causing the impedance of the circuit to be minimum. This typically happens in a series RLC circuit or a parallel RLC circuit. If the circuit contains only a capacitor or an inductor, it cannot undergo electrical resonance as there is no counterpart reactance to balance the impedance.

Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

This statement is also incorrect. An AC circuit containing a pure capacitor or a pure inductor will have a power factor of 0, not a non-zero power factor. The power factor of a capacitor is -1, and the power factor of an inductor is +1, but when only considering the reactive components, the power factor is 0. In such a circuit, no real power is consumed, and the circuit only has reactive power. The energy is alternately stored and released by the capacitor and inductor, but no energy is dissipated as heat or used to perform work.

As both statements are incorrect, the correct answer would be an option that states both statements are false.

Correct Option is (C)

Statement I is true because in a series LCR circuit, the current first increases as the frequency increases, reaching a maximum value when the circuit is at resonance. At resonance, the inductive reactance (XL) and capacitive reactance (XC) cancel each other out, resulting in the lowest impedance (Z) and the highest current. As the frequency continues to increase beyond resonance, the current in the circuit decreases.

Statement II is also true because, at resonance in a series LCR circuit, the inductive reactance (XL) and capacitive reactance (XC) are equal and cancel each other out. This results in the impedance (Z) being purely resistive. The power factor at resonance is given by the cosine of the phase angle (θ), and since the phase angle is 0° at resonance, the power factor is 1.

Thus, both statements are true, and the correct answer is Both Statement I and Statement II are true.

Correct Option is (D)

In a series LCR circuit connected to an AC source, resonance occurs at a particular frequency at which the inductive reactance is equal to the capacitive reactance, resulting in the minimum impedance of the circuit. At this frequency, the circuit draws maximum current from the source, and thus, the maximum power is dissipated in the circuit. Therefore, Statement I is true.

In a circuit containing only a resistor, the power dissipated is given by P = VI = I${}^2$R, where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor. The voltage and current are in phase in a purely resistive circuit, which means that the power is maximized. Therefore, Statement II is also true.

Correct Answer is (4)

The Q factor (quality factor) in a series RLC circuit can be given by the formula:

$Q=\frac{X_L}{R}=\frac{\omega L}{R}$

where ($X_L$) is the inductive reactance, (R) is the resistance, (L) is the inductance, and ($\omega$) is the angular frequency.

In a series RLC circuit at resonance, the resonant frequency (f) is given by

$f=\frac{1}{2\pi \sqrt{LC}}$

or equivalently, the angular frequency ($\omega$) at resonance is

$\omega =\frac{1}{\sqrt{LC}}$

Substituting (L = 1H) and ($C=6.25\mu F=6.25\times {10}^{-6}F$) into the equation for (\omega) gives

$\omega =\frac{1}{\sqrt{1H\times 6.25\times {10}^{-6}F}}$ = 400 rad/s

Substituting ($\omega =400rad/s$), (L = 1H), and ($R=100\mathrm{\Omega }$) into the equation for the Q factor gives

$Q=\frac{\omega L}{R}=\frac{400rad/s\times 1H}{100\mathrm{\Omega }}=4$

So, the Q factor of the circuit is 4.

Correct Answer is (40)

To maximize the average rate at which energy supplied i.e. power will be maximum.

So in LCR circuit power will be maximum at the condition of resonance and in resonance condition

$\begin{array}{rl}& \therefore X_L=X_C\\ \\ & 79.6=\frac{1}{2\pi (50)\times C}\\ \\ & C=\frac{1}{79.6\times 2\pi (50)}\\ \\ & \approx 40\mu \mathrm{F}\end{array}$