Question

A coil has resistance 20 Ω and inductance 0.35 H. Compute its impedance to an alternating current of 25 cycles/s.

• A. 50.5 Ω
• B. 48.5 Ω
• C. 58.5 Ω
• D. 68.5 Ω

Hint:

The impedance of an inductive coil depends on both the resistance and the inductive reactance. The total impedance is the square root of the sum of the squares of these values.

Answer 1

The Correct Option is C

The impedance (Z) of a coil in an alternating current (AC) circuit is determined by its resistance (R) and its inductive reactance (X_L). This can be calculated using the formula:

Z = √(R² + X_L²) 

Given:

• Resistance, R = 20 Ω
• Inductance, L = 0.35 H
• Frequency, f = 25 cycles/s

The inductive reactance is given by:

X_L = 2πfL

Substituting the given values:

X_L = 2 × π × 25 × 0.35

X_L = 2 × 3.1416 × 25 × 0.35

X_L ≈ 55 Ω

Now, calculate the impedance:

Z = √(20² + 55²)

Z = √(400 + 3025)

Z = √3425

Z ≈ 58.5 Ω

Therefore, the impedance of the coil to an alternating current of 25 cycles/s is approximately 58.5 Ω, confirming the correct answer.

Answer 2

The impedance \( Z \) of a coil is given by: \[ Z = \sqrt{R^2 + (X_L)^2} \] Where:
- \( R = 20 \, \Omega \),
- \( X_L = 2\pi f L \) is the inductive reactance, with:
- \( f = 25 \, \text{Hz} \),
- \( L = 0.35 \, \text{H} \). First, calculate \( X_L \): \[ X_L = 2 \pi \times 25 \times 0.35 = 54.98 \, \Omega \] Then, the impedance is: \[ Z = \sqrt{20^2 + 54.98^2} = 58.5 \, \Omega \]