JEE Physics Alternating Current MCQs Set 07

Answer: (a) 1

Explanation:
1. Quality factor \( Q=\frac{1}{R}\sqrt{\frac LC}=\frac{1}{10\pi}\sqrt{\frac{3}{27\times10^{-6}}}=\frac{333.3}{31.4}\approx10.6 \).
2. Bandwidth \( \Delta\omega=\frac RL=\frac{10\pi}{3}\approx10.5\ \text{rad/s} \).
3. \( \frac{Q}{\Delta\omega}\approx\frac{10.6}{10.5}\approx1\ \text{s} \).

Teacher's Note:
1. Q = ω0 L/R and bandwidth = R/L for a series resonant circuit.
2. Q is dimensionless, so Q divided by a frequency has units of seconds.

Answer: (a) \( \frac14 \)

Explanation:
1. \( \omega=\frac1{\sqrt{LC}}\Rightarrow\omega'=\frac1{\sqrt{2L\times8C}}=\frac{1}{4\sqrt{LC}}=\frac{\omega_0}{4} \).

Teacher's Note:
1. The resonant frequency is inversely proportional to √(LC).
2. Product LC becomes 16 times.

Answer: (a) 100 mH

Explanation:
1. Maximum current at resonance: \( \omega^2=\frac1{LC}\Rightarrow L=\frac1{4\pi^2f^2C} \).
2. \( L=\frac{1}{4\times10\times(2000)^2\times62.5\times10^{-9}}=\frac1{10}=0.1\ \text{H}=100\ \text{mH} \).

Teacher's Note:
1. The current amplitude is maximum when XL = XC.
2. Convert 62.5 nF to 62.5 x 10⁻⁹ F.

Answer: (a) 8

Explanation:
1. Impedance \( Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{80^2+60^2}=100\ \Omega \).
2. Power factor \( \cos\phi=\frac RZ=\frac{80}{100}=0.8=\frac{8}{10} \), so x = 8.

Teacher's Note:
1. The net reactance is the difference of XC and XL.
2. Power factor = R/Z.

Answer: (a) The rms current in circuit (b) can never be larger than that in (a)

Explanation:
1. Current in (a): \( I_a=\frac{V}{R} \). Current in (b): \( I_b=\frac{V}{\sqrt{R^2+(X_L-X_C)^2}}\le\frac VR \).
2. So (b) can at best equal (a) (at resonance), and is otherwise smaller.

Teacher's Note:
1. The impedance of a series LCR circuit is never smaller than R.
2. At resonance the reactances cancel and Z = R.

Answer: (a) 3872

Explanation:
1. Maximum current at resonance: \( C=\frac{1}{\omega^2L} \).
2. \( \omega=2\pi f=2\times\frac{22}7\times7000=44000\ \text{rad/s} \).
3. \( \omega^2L=(44000)^2\times2\times10^{-6}=3872 \), so \( C=\frac1{3872}\ \text{F} \) and x = 3872.

Teacher's Note:
1. Resonance requires ω²LC = 1.
2. Square 44000 carefully: 1.936 x 10⁹.

Answer: (b) 1 : \( \sqrt2 \)

Explanation:
1. LR circuit: \( P_1=\frac{R}{\sqrt{R^2+X_L^2}}=\frac1{\sqrt2} \).
2. With XC = XL the circuit is at resonance and Z = R: \( P_2=1 \).
3. \( P_1:P_2=\frac1{\sqrt2}:1=1:\sqrt2 \).

Teacher's Note:
1. At resonance the circuit is purely resistive.
2. Power factor = R/Z.

Answer: (a) 30 V

Explanation:
1. Across the resistor: \( V_A-V_x=IR=24\ \text{V} \).
2. Across the inductor (current direction from x to y): \( V_x-V_y=L\frac{dI}{dt}=6\times(-1)=-6\ \text{V} \).
3. Across the battery (current enters the positive terminal): \( V_y-V_B=+12\ \text{V} \).
4. \( V_{AB}=24-6+12=30\ \text{V} \).

Teacher's Note:
1. Add the potential differences along the path from A to B.
2. A decreasing current gives an inductor emf that tries to maintain the current (negative L dI/dt term).

Answer: (a) 2 A

Explanation:
1. \( Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{100^2+100^2}=100\sqrt2\ \Omega \).
2. \( I_{rms}=\frac{200\sqrt2}{100\sqrt2}=2\ \text{A} \).

Teacher's Note:
1. In series, the same current flows through every element.
2. Net reactance = 200 - 100 = 100 ohm.

Answer: (a) 1584 V

Explanation:
1. \( \omega=2\pi\times\frac{360}{60}=2\times\frac{22}{7}\times6=\frac{264}{7}\ \text{rad/s} \).
2. \( \varepsilon_{max}=NBA\omega=100\times3\times0.14\times\frac{264}7=1584\ \text{V} \).

Teacher's Note:
1. Convert rev/min into rev/s before finding ω.
2. Maximum emf = NBAω.

Answer: (b) \( \frac{R}{\sqrt{X_LX_C}} \)

Explanation:
1. R, XL and XC all have the dimensions of resistance (ohm).
2. \( \frac{R}{\sqrt{X_LX_C}} \) has dimensions \( \frac{\Omega}{\sqrt{\Omega^2}}=1 \), so it is dimensionless.

Teacher's Note:
1. Count the powers of ohm in each expression.
2. Options (a), (c) and (d) carry net powers of ohm.

Answer: (a) 242

Explanation:
1. Current in phase with emf means resonance: Z = R = 20 ohm.
2. The rms voltage is 220 V, so the peak voltage is \( 220\sqrt2 \) V.
3. \( I_0=\frac{220\sqrt2}{20}=11\sqrt2=\sqrt{242}\ \text{A} \), so x = 242.

Teacher's Note:
1. The 220 V of an ac source is the rms value.
2. Amplitude = √2 x rms.

Answer: (a) 3.14 ohm

Explanation:
1. Angular frequency \( \omega=628\ \text{rad/s} \).
2. \( X_L=\omega L=628\times5\times10^{-3}=3.14\ \Omega \).

Teacher's Note:
1. Read ω directly from the argument of the sine.
2. Convert mH to H.

Answer: (a) 25 A

Explanation:
1. \( Z=\sqrt{80^2+(100-40)^2}=100\ \Omega \).
2. \( I_0=\frac{V_0}{Z}=\frac{2500}{100}=25\ \text{A} \).

Teacher's Note:
1. 2500 is the amplitude (peak) of the voltage.
2. Z = √(R² + (XL - XC)²).

Answer: (c) A-II, B-IV, C-I, D-III

Explanation:
1. An AC generator works on electromagnetic induction (II).
2. A transformer works on mutual inductance (IV).
3. Resonance needs both an inductor and a capacitor (I).
4. The sharpness of resonance is measured by the quality factor (III).

Teacher's Note:
1. Match each device or phenomenon to its governing principle.
2. A high Q means a narrow, sharp resonance peak.

Answer: (a) 40 \( \mu \)F

Explanation:
1. Average power is maximum at resonance, when \( X_C=X_L=79.6\ \Omega \).
2. \( C=\frac{1}{\omega X_C}=\frac{1}{2\pi\times50\times79.6}=4.0\times10^{-5}\ \text{F}=40\ \mu\text{F} \).

Teacher's Note:
1. Power = Vrms Irms cosφ is largest when cosφ = 1.
2. Use ω = 2πf = 314 rad/s.