Alternating Current MCQs for NEET — Physics Questions with Answers

When 100V DC is applied, the resistance (R) of the coil can be found using Ohm's Law: $$ R = rac{V}{I} = rac{100V}{1A} = 100 ext{ ohms} $$ When 100V AC is applied, the current is 0.5A. The impedance (Z) can be found using the same formula: $$ Z = rac{V}{I} = rac{100V}{0.5A} = 200 ext{ ohms} $$ Therefore, the impedance of the coil is 200 ohms.

$ t = { T \over 4 } = { 2 \pi \sqrt{LC } \over 4 } $

The quality factor (Q) of an LCR circuit at resonance is given by the formula: $$ Q = \frac{\omega L}{R} $$ where \( \omega \) is the angular frequency, \( L \) is the inductance, and \( R \) is the resistance. This is because, at resonance, the impedance of the circuit is minimized, and the quality factor is defined as the ratio of the energy stored in the circuit to the energy dissipated per cycle.

The reactance of a capacitor is given by the formula: $$ X_C = \frac{1}{\omega C} $$ where \( X_C \) is the capacitive reactance, \( \omega \) is the angular frequency, and \( C \) is the capacitance. For high frequencies, the value of \( \omega \) increases, and as a result, the capacitive reactance \( X_C \) decreases. Therefore, a capacitor offers less reactance at high frequencies.

A choke coil, which is an inductor, opposes changes in current. This property is due to the inductive reactance, which is given by: $$ X_L = \omega L $$ where \( X_L \) is the inductive reactance, \( \omega \) is the angular frequency, and \( L \) is the inductance. As the inductive reactance increases, it resists the flow of current, effectively decreasing the current in the circuit.

The power factor (PF) of an AC circuit with resistance (R) and inductance (L) connected in series is given by the cosine of the phase angle (Ï•) between the voltage and the current. This phase angle is determined by the impedance of the circuit. The formula for power factor is:

$$ \text{Power Factor} = \cos(\varphi) = \frac{R}{\sqrt{R^2 + (\omega L)^2}} $$

Here, the impedance of the circuit is \( Z = \sqrt{R^2 + (\omega L)^2} \). Thus, the correct option is:

$$ \frac{R}{( R^2 + \omega^2 L^2 )^{1/2}} $$

The power dissipated in an AC circuit with a resistor (R) and inductor (L) in series is given by the formula:

$$ P = \frac{V^2 R}{Z^2} $$

where \( Z \) is the impedance of the circuit, which is given by \( Z = \sqrt{R^2 + (\omega L)^2} \). Substituting \( Z^2 \) into the formula, we get:

$$ P = \frac{V^2 R}{R^2 + (\omega L)^2} $$

Thus, the correct option is:

$$ \frac{V^2 R}{(R^2 + \omega^2 L^2)} $$

The resonant frequency (\( f_0 \)) of an LCR circuit is given by the formula:

$$ f_0 = \frac{1}{2\pi \sqrt{LC}} $$

To keep the resonant frequency unchanged when the capacitance is changed from \( C \) to \( 2C \), the inductance \( L \) must be adjusted such that the product \( LC \) remains constant. If \( C \) becomes \( 2C \), then \( L \) must become \( L/2 \) to maintain the same resonant frequency. Therefore, the correct option is:

$$ \frac{L}{2} $$

In an LCR series AC circuit, the voltages across the inductor (L) and capacitor (C) are 180 degrees out of phase. When the circuit is at resonance, the inductive reactance (XL) equals the capacitive reactance (XC), and their voltages cancel each other out. Therefore, the voltage across the LC combination is zero.

In an LCR circuit where the current leads the voltage by $45^ ext{°}$, the circuit is in resonance condition where the net reactance is zero. The capacitive reactance is equal to the inductive reactance ($X_C = X_L$). The condition $ an(45^ ext{°}) = 1$ gives $X_L - X_C = R$. Solving this using the given frequency, the value of capacitance can be found as $C = rac{1}{2 ext{π}f(2 ext{π}fL + R)}$.