JEE (Advanced) 2023 Numerical:

1. Alternating Current Circuit with Inductor and Capacitor:

• Given:

  • Frequency (f) = 50 Hz
  • Inductance (L) = 100 mH = 0.1 H
  • Capacitance (C) = 10 μF = 10 × 10^-6 F
  • Maximum Current (I_max) = 1 A

• Maximum Voltage across the Capacitor (V_Cmax):

  Since the inductor and capacitor are in series, the maximum current flows through both components. Therefore, the voltage across the capacitor can be calculated using the formula:

  V_Cmax = I_max * X_C

  where X_C is the capacitive reactance, given by:

  X_C = 1 / (2πfC)

  Substituting the given values:

  X_C = 1 / (2π × 50 × 10 × 10^-6) ≈ 318.31 Ω

  V_Cmax = 1 A × 318.31 Ω ≈ 318.31 V

2. Series LCR Circuit:

• Given:
  • Voltage (V) = 220 V
  • Frequency (f) = 50 Hz
  • Inductance (L) = 0.1 H
  • Capacitance (C) = 100 μF = 100 × 10^-6 F
  • Resistance (R) = 10 Ω

a. Impedance (Z):

The impedance of the circuit is given by:

Z = √(R² + (X_L - X_C)²)

where X_L is the inductive reactance, given by:

X_L = 2πfL

Substituting the given values:

X_L = 2π × 50 × 0.1 ≈ 31.42 Ω

Z = √(10² + (31.42 - 318.31)²) ≈ 264.82 Ω

b. Phase Angle (φ):

The phase angle is given by:

φ = tan^-1[(X_L - X_C) / R]

Substituting the calculated values:

φ = tan^-1[(31.42 - 318.31) / 10] ≈ -1.51 radians

c. Power Factor (PF):

The power factor is given by:

PF = cos(φ)

Substituting the calculated phase angle:

PF = cos(-1.51) ≈ 0.196 (lagging)

d. Power Consumed (P):

The power consumed by the circuit is given by:

P = V² * PF / Z

Substituting the given values:

P = (220² × 0.196) / 264.82 ≈ 171.5 W

3. Power Factor Improvement of Light Bulb:

• Given:
  • Light Bulb Power Rating (P_bulb) = 100 W
  • Supply Voltage (V) = 220 V
  • Supply Frequency (f) = 50 Hz
  • Power Factor (PF) = 0.9 (desired)

a. Capacitance Required (C):

The power factor of a circuit can be improved by connecting a capacitor in series with the load. The capacitance required to achieve the desired power factor can be calculated using the formula:

C = (PF × I) / (2πfV)

First, we need to find the current drawn by the light bulb:

I = P_bulb / V = 100 W / 220 V ≈ 0.4545 A

Now, we can calculate the capacitance:

C = (0.9 × 0.4545 A) / (2π × 50 × 220 V) ≈ 3.96 μF

b. Average Power Consumed (P_avg):

The average power consumed by the circuit with the capacitor is given by:

P_avg = PF × V × I

Substituting the given values:

P_avg = 0.9 × 220 V × 0.4545 A ≈ 92.08 W

CBSE Class 12 Board Exam Numericals:

1. Alternating Current Circuit with Resistor, Inductor, and Capacitor:

• Given:
  • Emf Voltage (V) = 50 V
  • Frequency (f) = 50 Hz
  • Resistance (R) = 20 Ω
  • Inductance (L) = 50 mH = 50 × 10^-3 H
  • Capacitance (C) = 20 μF = 20 × 10^-6 F

a. Impedance (Z):

The impedance of the circuit is given by:

Z = √(R² + (X_L - X_C)²)

where X_L and X_C are the inductive and capacitive reactances, respectively.

X_L = 2πfL = 2π × 50 × 50 × 10^-3 ≈ 15.71 Ω

X_C = 1 / (2πfC) = 1 / (2π × 50 × 20 × 10^-6) ≈ 159.15 Ω

Z = √(20² + (15.71 - 159.15)²) ≈ 163.98 Ω

b. Current (I):

The current in the circuit is given by:

I = V / Z

Substituting the calculated impedance:

I = 50 V / 163.98 Ω ≈ 0.305 A

c. Voltages across Resistor, Inductor, and Capacitor:

V_R = I * R = 0.305 A × 20 Ω ≈ 6.11 V

V_L = I * X_L = 0.305 A × 15.71 Ω ≈ 4.79 V

V_C = I * X_C = 0.305 A × 159.15 Ω ≈ 49.03 V

d. Power Factor (PF):

The power factor is given by:

PF = cos(φ)

where φ is the phase angle between the voltage and the current.

φ = tan^-1[(X_L - X_C) / R] = tan^-1[(15.71 - 159.15) / 20] ≈ -1.49 radians

PF = cos(-1.49) ≈ 0.196 (lagging)

2. Alternating Current Circuit with Resistor and Capacitor:

• Given:
  • Voltage (V) = 100 V
  • Frequency (f) = 50 Hz
  • Resistance (R) = 50 Ω
  • Capacitance (C) = 100 μF = 100 × 10^-6 F

a. Impedance (Z):

The impedance of the circuit is given by:

Z = √(R² + X_C²)

where X_C is the capacitive reactance, given by:

X_C = 1 / (2πfC)

Substituting the given values:

X_C = 1 / (2π × 50 × 100 × 10^-6) ≈ 31.83 Ω

Z = √(50² + 31.83²) ≈ 58.82 Ω

b. Phase Angle (φ):

The phase angle is given by:

φ = tan^-1(-X_C / R)

Substituting the calculated values:

φ = tan^-1(-31.83 / 50) ≈ -0.58 radians

c. Power Factor (PF):

The power factor is given by:

PF = cos(φ)

Substituting the calculated phase angle:

PF = cos(-0.58) ≈ 0.81 (leading)